16.35 Problem number 169

\[ \int \frac {\left (d^2-e^2 x^2\right )^{5/2}}{x^7 (d+e x)^2} \, dx \]

Optimal antiderivative \[ -\frac {\left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{6 x^{6}}+\frac {2 e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{5 d \,x^{5}}-\frac {3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{8 d^{2} x^{4}}+\frac {4 e^{3} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{15 d^{3} x^{3}}+\frac {3 e^{6} \arctanh \left (\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{d}\right )}{16 d^{3}}-\frac {3 e^{4} \sqrt {-e^{2} x^{2}+d^{2}}}{16 d^{2} x^{2}} \]

command

integrate((-e^2*x^2+d^2)^(5/2)/x^7/(e*x+d)^2,x, algorithm="giac")

Giac 1.9.0-11 via sagemath 9.6 output

\[ \frac {1}{7680} \, {\left (\frac {1440 \, e^{5} \log \left (\sqrt {\frac {2 \, d}{x e + d} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{3}} - \frac {1440 \, e^{5} \log \left ({\left | \sqrt {\frac {2 \, d}{x e + d} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{3}} + \frac {16 \, {\left (45 \, e^{5} \log \left (2\right ) - 90 \, e^{5} \log \left (i + 1\right ) + 128 i \, e^{5}\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{3}} - \frac {45 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {11}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + 1025 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {9}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 174 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {7}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + 594 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {5}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - 255 \, {\left (\frac {2 \, d}{x e + d} - 1\right )}^{\frac {3}{2}} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) + 45 \, \sqrt {\frac {2 \, d}{x e + d} - 1} e^{5} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{d^{3} {\left (\frac {d}{x e + d} - 1\right )}^{6}}\right )} e \]

Giac 1.7.0 via sagemath 9.3 output

\[ \text {Exception raised: TypeError} \]________________________________________________________________________________________