Integrand size = 20, antiderivative size = 110 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {(3 i A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}} \]
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Time = 0.07 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2829, 2729, 2728, 212} \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {(5 B+3 i A) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(5 B+3 i A) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(-B+i A) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}} \]
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Rule 212
Rule 2728
Rule 2729
Rule 2829
Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 A-5 i B) \int \frac {1}{(a+i a \sinh (x))^{3/2}} \, dx}{8 a} \\ & = \frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(3 A-5 i B) \int \frac {1}{\sqrt {a+i a \sinh (x)}} \, dx}{32 a^2} \\ & = \frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}}+\frac {(3 i A+5 B) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cosh (x)}{\sqrt {a+i a \sinh (x)}}\right )}{16 a^2} \\ & = \frac {(3 i A+5 B) \text {arctanh}\left (\frac {\sqrt {a} \cosh (x)}{\sqrt {2} \sqrt {a+i a \sinh (x)}}\right )}{16 \sqrt {2} a^{5/2}}+\frac {(i A-B) \cosh (x)}{4 (a+i a \sinh (x))^{5/2}}+\frac {(3 i A+5 B) \cosh (x)}{16 a (a+i a \sinh (x))^{3/2}} \\ \end{align*}
Time = 0.32 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right ) \left (4 i (A+i B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )+(3 i A+5 B) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^3+(1-i) \sqrt [4]{-1} (3 A-5 i B) \arctan \left (\frac {i+\tanh \left (\frac {x}{4}\right )}{\sqrt {2}}\right ) \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )^4+8 (A+i B) \sinh \left (\frac {x}{2}\right )+2 (3 i A+5 B) \sinh \left (\frac {x}{2}\right ) (-i+\sinh (x))\right )}{16 (a+i a \sinh (x))^{5/2}} \]
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\[\int \frac {A +B \sinh \left (x \right )}{\left (a +i a \sinh \left (x \right )\right )^{\frac {5}{2}}}d x\]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 347 vs. \(2 (77) = 154\).
Time = 0.30 (sec) , antiderivative size = 347, normalized size of antiderivative = 3.15 \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) - \sqrt {\frac {1}{2}} {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} \log \left (-\frac {\sqrt {\frac {1}{2}} a^{3} \sqrt {-\frac {9 \, A^{2} - 30 i \, A B - 25 \, B^{2}}{a^{5}}} - \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (3 i \, A + 5 \, B\right )}}{3 i \, A + 5 \, B}\right ) + 2 \, {\left ({\left (-3 i \, A - 5 \, B\right )} e^{\left (4 \, x\right )} - {\left (11 \, A + 3 i \, B\right )} e^{\left (3 \, x\right )} + {\left (-11 i \, A + 3 \, B\right )} e^{\left (2 \, x\right )} - {\left (3 \, A - 5 i \, B\right )} e^{x}\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}}}{16 \, {\left (a^{3} e^{\left (4 \, x\right )} - 4 i \, a^{3} e^{\left (3 \, x\right )} - 6 \, a^{3} e^{\left (2 \, x\right )} + 4 i \, a^{3} e^{x} + a^{3}\right )}} \]
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Timed out. \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\int { \frac {B \sinh \left (x\right ) + A}{{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\int { \frac {B \sinh \left (x\right ) + A}{{\left (i \, a \sinh \left (x\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {A+B \sinh (x)}{(a+i a \sinh (x))^{5/2}} \, dx=\int \frac {A+B\,\mathrm {sinh}\left (x\right )}{{\left (a+a\,\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]
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