Integrand size = 15, antiderivative size = 68 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {(2 i A-3 B) \cosh (x)}{15 (i+\sinh (x))} \]
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Time = 0.04 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2829, 2729, 2727} \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {(-3 B+2 i A) \cosh (x)}{15 (\sinh (x)+i)}-\frac {(2 A+3 i B) \cosh (x)}{15 (\sinh (x)+i)^2}-\frac {(B+i A) \cosh (x)}{5 (\sinh (x)+i)^3} \]
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Rule 2727
Rule 2729
Rule 2829
Rubi steps \begin{align*} \text {integral}& = -\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}+\frac {1}{5} (-2 i A+3 B) \int \frac {1}{(i+\sinh (x))^2} \, dx \\ & = -\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {1}{15} (-2 A-3 i B) \int \frac {1}{i+\sinh (x)} \, dx \\ & = -\frac {(i A+B) \cosh (x)}{5 (i+\sinh (x))^3}-\frac {(2 A+3 i B) \cosh (x)}{15 (i+\sinh (x))^2}+\frac {(2 i A-3 B) \cosh (x)}{15 (i+\sinh (x))} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {\cosh (x) \left (-7 i A+3 B-3 (2 A+3 i B) \sinh (x)+(2 i A-3 B) \sinh ^2(x)\right )}{15 (i+\sinh (x))^3} \]
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Time = 1.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.75
method | result | size |
risch | \(-\frac {2 \left (15 B \,{\mathrm e}^{3 x}-15 B \,{\mathrm e}^{x}+15 i B \,{\mathrm e}^{2 x}-2 A +10 i A \,{\mathrm e}^{x}-3 i B +20 A \,{\mathrm e}^{2 x}\right )}{15 \left ({\mathrm e}^{x}+i\right )^{5}}\) | \(51\) |
default | \(-\frac {2 i B -4 A}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}-\frac {2 \left (8 i A +6 B \right )}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {-8 i B +8 A}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {2 \left (-4 i A -4 B \right )}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {2 i A}{\tanh \left (\frac {x}{2}\right )+i}\) | \(91\) |
parallelrisch | \(\frac {\operatorname {sech}\left (x \right ) \left (-10 i A \sinh \left (x \right )+2 i A \sinh \left (3 x \right )-28 i A \sinh \left (2 x \right )+15 i B \cosh \left (x \right )-12 i B \cosh \left (2 x \right )-3 i B \cosh \left (3 x \right )+35 A \cosh \left (x \right )-8 A \cosh \left (2 x \right )-7 A \cosh \left (3 x \right )-15 B \sinh \left (x \right )-3 B \sinh \left (3 x \right )+12 B \sinh \left (2 x \right )-20 A \right )}{120 i \sinh \left (x \right )+30 \cosh \left (2 x \right )-90}\) | \(105\) |
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Time = 0.31 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 5 \, {\left (4 \, A + 3 i \, B\right )} e^{\left (2 \, x\right )} + 5 \, {\left (2 i \, A - 3 \, B\right )} e^{x} - 2 \, A - 3 i \, B\right )}}{15 \, {\left (e^{\left (5 \, x\right )} + 5 i \, e^{\left (4 \, x\right )} - 10 \, e^{\left (3 \, x\right )} - 10 i \, e^{\left (2 \, x\right )} + 5 \, e^{x} + i\right )}} \]
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Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.21 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {4 A - 30 B e^{3 x} + 6 i B + \left (- 40 A - 30 i B\right ) e^{2 x} + \left (- 20 i A + 30 B\right ) e^{x}}{15 e^{5 x} + 75 i e^{4 x} - 150 e^{3 x} - 150 i e^{2 x} + 75 e^{x} + 15 i} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (50) = 100\).
Time = 0.20 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.93 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} + \frac {5 i \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {i}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i}\right )} - \frac {4}{15} \, A {\left (-\frac {5 i \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} + \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i} - \frac {1}{5 \, e^{\left (-x\right )} + 10 i \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-3 \, x\right )} - 5 i \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - i}\right )} \]
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Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.68 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=-\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} + 15 i \, B e^{\left (2 \, x\right )} + 10 i \, A e^{x} - 15 \, B e^{x} - 2 \, A - 3 i \, B\right )}}{15 \, {\left (e^{x} + i\right )}^{5}} \]
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Time = 1.57 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {A+B \sinh (x)}{(i+\sinh (x))^3} \, dx=\frac {\frac {A\,4{}\mathrm {i}}{15}-\frac {2\,B}{5}-\frac {A\,{\mathrm {e}}^{2\,x}\,8{}\mathrm {i}}{3}+{\mathrm {e}}^x\,\left (\frac {4\,A}{3}+B\,2{}\mathrm {i}\right )+2\,B\,{\mathrm {e}}^{2\,x}-B\,{\mathrm {e}}^{3\,x}\,2{}\mathrm {i}}{{\left (-1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )}^5} \]
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