Integrand size = 13, antiderivative size = 37 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=-\frac {i \text {sech}^3(x)}{5 (i+\sinh (x))}-\frac {4}{5} i \tanh (x)+\frac {4}{15} i \tanh ^3(x) \]
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Time = 0.03 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2751, 3852} \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=\frac {4}{15} i \tanh ^3(x)-\frac {4}{5} i \tanh (x)-\frac {i \text {sech}^3(x)}{5 (\sinh (x)+i)} \]
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Rule 2751
Rule 3852
Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {sech}^3(x)}{5 (i+\sinh (x))}-\frac {4}{5} i \int \text {sech}^4(x) \, dx \\ & = -\frac {i \text {sech}^3(x)}{5 (i+\sinh (x))}+\frac {4}{5} \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-i \tanh (x)\right ) \\ & = -\frac {i \text {sech}^3(x)}{5 (i+\sinh (x))}-\frac {4}{5} i \tanh (x)+\frac {4}{15} i \tanh ^3(x) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=-\frac {1}{15} i \left (\frac {3 \text {sech}^3(x)}{i+\sinh (x)}+12 \text {sech}^2(x) \tanh (x)+8 \tanh ^3(x)\right ) \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (27 ) = 54\).
Time = 0.43 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.51
\[\frac {i}{6 \left (-i+\tanh \left (\frac {x}{2}\right )\right )^{3}}-\frac {5 i}{8 \left (-i+\tanh \left (\frac {x}{2}\right )\right )}+\frac {1}{4 \left (-i+\tanh \left (\frac {x}{2}\right )\right )^{2}}-\frac {2 i}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}+\frac {5 i}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {11 i}{8 \left (\tanh \left (\frac {x}{2}\right )+i\right )}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {3}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}\]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.68 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=-\frac {16 \, {\left (6 \, e^{\left (3 \, x\right )} + 2 i \, e^{\left (2 \, x\right )} + 2 \, e^{x} + i\right )}}{15 \, {\left (e^{\left (8 \, x\right )} + 2 i \, e^{\left (7 \, x\right )} + 2 \, e^{\left (6 \, x\right )} + 6 i \, e^{\left (5 \, x\right )} + 6 i \, e^{\left (3 \, x\right )} - 2 \, e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )}} \]
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\[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=\int \frac {\operatorname {sech}^{4}{\left (x \right )}}{\sinh {\left (x \right )} + i}\, dx \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (23) = 46\).
Time = 0.20 (sec) , antiderivative size = 205, normalized size of antiderivative = 5.54 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=-\frac {32 \, e^{\left (-x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac {32 i \, e^{\left (-2 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} - \frac {96 \, e^{\left (-3 \, x\right )}}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} + \frac {16 i}{-30 i \, e^{\left (-x\right )} - 30 \, e^{\left (-2 \, x\right )} - 90 i \, e^{\left (-3 \, x\right )} - 90 i \, e^{\left (-5 \, x\right )} + 30 \, e^{\left (-6 \, x\right )} - 30 i \, e^{\left (-7 \, x\right )} + 15 \, e^{\left (-8 \, x\right )} - 15} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=\frac {9 \, e^{\left (2 \, x\right )} - 24 i \, e^{x} - 11}{24 \, {\left (e^{x} - i\right )}^{3}} - \frac {45 \, e^{\left (4 \, x\right )} + 240 i \, e^{\left (3 \, x\right )} - 490 \, e^{\left (2 \, x\right )} - 320 i \, e^{x} + 73}{120 \, {\left (e^{x} + i\right )}^{5}} \]
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Time = 1.78 (sec) , antiderivative size = 231, normalized size of antiderivative = 6.24 \[ \int \frac {\text {sech}^4(x)}{i+\sinh (x)} \, dx=-\frac {1}{6\,\left ({\mathrm {e}}^{2\,x}\,3{}\mathrm {i}-{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x-\mathrm {i}\right )}-\frac {\frac {3\,{\mathrm {e}}^x}{40}+\frac {1}{8}{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1+{\mathrm {e}}^x\,2{}\mathrm {i}}-\frac {\frac {3\,{\mathrm {e}}^{2\,x}}{40}-\frac {5}{24}+\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{4}}{{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}+{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x-\mathrm {i}}+\frac {1{}\mathrm {i}}{4\,\left (1-{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,2{}\mathrm {i}\right )}+\frac {3}{8\,\left ({\mathrm {e}}^x-\mathrm {i}\right )}-\frac {3}{40\,\left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}-\frac {\frac {{\mathrm {e}}^{2\,x}\,3{}\mathrm {i}}{8}+\frac {3\,{\mathrm {e}}^{3\,x}}{40}-\frac {5\,{\mathrm {e}}^x}{8}-\frac {1}{8}{}\mathrm {i}}{{\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1+{\mathrm {e}}^{3\,x}\,4{}\mathrm {i}-{\mathrm {e}}^x\,4{}\mathrm {i}}-\frac {\frac {3\,{\mathrm {e}}^{4\,x}}{40}-\frac {5\,{\mathrm {e}}^{2\,x}}{4}+\frac {3}{40}+\frac {{\mathrm {e}}^{3\,x}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^x\,1{}\mathrm {i}}{2}}{{\mathrm {e}}^{5\,x}-10\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}\,5{}\mathrm {i}-{\mathrm {e}}^{2\,x}\,10{}\mathrm {i}+5\,{\mathrm {e}}^x+1{}\mathrm {i}} \]
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