MathAction #440 cannot take first of empty list in integral

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fricas

(1) -> integrate(log(1-z^3)*(%i*z)^(1/2), z)

$\label{eq1}\frac{{{\sqrt{{16}\ i}}\ {\log \left({\frac{{4 \ i \ z \ {\sqrt{i \ z}}}+{i \ {\sqrt{{16}\ i}}}}{4}}\right)}}-{{\sqrt{{16}\ i}}\ {\log \left({\frac{{4 \ i \ z \ {\sqrt{i \ z}}}-{i \ {\sqrt{{1 6}\ i}}}}{4}}\right)}}+{{\left({4 \ z \ {\log \left({-{{z}^{3}}+ 1}\right)}}-{8 \ z}\right)}\ {\sqrt{i \ z}}}}{6}$

(1)

Type: Union(Expression(Complex(Integer)),...)

another instance --kratt6, Tue, 13 Jan 2009 02:09:57 -0800 reply

fricas

integrate(sin(z)*csc(z)*(1-1/(%i*z)^(1/2))^(1/2), z)

$\label{eq2}\frac{{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}+ 1}\right)}}-{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}- 1}\right)}}+{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}}}{4 \ {\sqrt{i \ z}}}$

(2)

Type: Union(Expression(Complex(Integer)),...)

simpler instance --kratt6, Tue, 13 Jan 2009 02:12:25 -0800 reply

fricas

integrate((1-1/(%i*z)^(1/2))^(1/2), z)

$\label{eq3}\frac{{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}+ 1}\right)}}-{i \ {\sqrt{i \ z}}\ {\log \left({{\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}- 1}\right)}}+{{\left({4 \ z \ {\sqrt{i \ z}}}-{2 \ z}\right)}\ {\sqrt{\frac{{\sqrt{i \ z}}- 1}{\sqrt{i \ z}}}}}}{4 \ {\sqrt{i \ z}}}$

(3)

Type: Union(Expression(Complex(Integer)),...)

fixed in FriCAS revision 486 --kratt6, Fri, 16 Jan 2009 05:07:16 -0800 reply

Status: open => fixed somewhere

Subject: Be Bold !!

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