Integrand size = 13, antiderivative size = 37 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {2}{3} i \text {sech}^3(x)-\frac {2}{5} i \text {sech}^5(x)-\frac {\tanh ^3(x)}{3}+\frac {2 \tanh ^5(x)}{5} \]
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Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2790, 2687, 14, 2686, 30} \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {2 \tanh ^5(x)}{5}-\frac {\tanh ^3(x)}{3}-\frac {2}{5} i \text {sech}^5(x)+\frac {2}{3} i \text {sech}^3(x) \]
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Rule 14
Rule 30
Rule 2686
Rule 2687
Rule 2790
Rubi steps \begin{align*} \text {integral}& = -\int \left (\text {sech}^4(x) \tanh ^2(x)+2 i \text {sech}^3(x) \tanh ^3(x)-\text {sech}^2(x) \tanh ^4(x)\right ) \, dx \\ & = -\left (2 i \int \text {sech}^3(x) \tanh ^3(x) \, dx\right )-\int \text {sech}^4(x) \tanh ^2(x) \, dx+\int \text {sech}^2(x) \tanh ^4(x) \, dx \\ & = -\left (i \text {Subst}\left (\int x^4 \, dx,x,i \tanh (x)\right )\right )-i \text {Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,i \tanh (x)\right )-2 i \text {Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\text {sech}(x)\right ) \\ & = \frac {\tanh ^5(x)}{5}-i \text {Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,i \tanh (x)\right )-2 i \text {Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\text {sech}(x)\right ) \\ & = \frac {2}{3} i \text {sech}^3(x)-\frac {2}{5} i \text {sech}^5(x)-\frac {\tanh ^3(x)}{3}+\frac {2 \tanh ^5(x)}{5} \\ \end{align*}
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(84\) vs. \(2(37)=74\).
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.27 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {80 i-55 i \cosh (x)-16 i \cosh (2 x)+11 i \cosh (3 x)+140 \sinh (x)-44 \sinh (2 x)-4 \sinh (3 x)}{240 \left (\cosh \left (\frac {x}{2}\right )-i \sinh \left (\frac {x}{2}\right )\right )^5 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )} \]
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Time = 13.90 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.16
method | result | size |
risch | \(-\frac {2 \left (-20 \,{\mathrm e}^{2 x}-4 i {\mathrm e}^{x}+1+20 i {\mathrm e}^{3 x}+15 \,{\mathrm e}^{4 x}\right )}{15 \left ({\mathrm e}^{x}-i\right ) \left ({\mathrm e}^{x}+i\right )^{5}}\) | \(43\) |
default | \(\frac {1}{-4 i+4 \tanh \left (\frac {x}{2}\right )}+\frac {2 i}{\left (\tanh \left (\frac {x}{2}\right )+i\right )^{4}}-\frac {i}{2 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{2}}+\frac {4}{5 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{5}}-\frac {5}{3 \left (\tanh \left (\frac {x}{2}\right )+i\right )^{3}}-\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )+i\right )}\) | \(70\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=-\frac {2 \, {\left (15 \, e^{\left (4 \, x\right )} + 20 i \, e^{\left (3 \, x\right )} - 20 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1\right )}}{15 \, {\left (e^{\left (6 \, x\right )} + 4 i \, e^{\left (5 \, x\right )} - 5 \, e^{\left (4 \, x\right )} - 5 \, e^{\left (2 \, x\right )} - 4 i \, e^{x} + 1\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.78 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {- 30 e^{4 x} - 40 i e^{3 x} + 40 e^{2 x} + 8 i e^{x} - 2}{15 e^{6 x} + 60 i e^{5 x} - 75 e^{4 x} - 75 e^{2 x} - 60 i e^{x} + 15} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (25) = 50\).
Time = 0.22 (sec) , antiderivative size = 197, normalized size of antiderivative = 5.32 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {8 i \, e^{\left (-x\right )}}{60 i \, e^{\left (-x\right )} - 75 \, e^{\left (-2 \, x\right )} - 75 \, e^{\left (-4 \, x\right )} - 60 i \, e^{\left (-5 \, x\right )} + 15 \, e^{\left (-6 \, x\right )} + 15} - \frac {40 \, e^{\left (-2 \, x\right )}}{60 i \, e^{\left (-x\right )} - 75 \, e^{\left (-2 \, x\right )} - 75 \, e^{\left (-4 \, x\right )} - 60 i \, e^{\left (-5 \, x\right )} + 15 \, e^{\left (-6 \, x\right )} + 15} - \frac {40 i \, e^{\left (-3 \, x\right )}}{60 i \, e^{\left (-x\right )} - 75 \, e^{\left (-2 \, x\right )} - 75 \, e^{\left (-4 \, x\right )} - 60 i \, e^{\left (-5 \, x\right )} + 15 \, e^{\left (-6 \, x\right )} + 15} + \frac {30 \, e^{\left (-4 \, x\right )}}{60 i \, e^{\left (-x\right )} - 75 \, e^{\left (-2 \, x\right )} - 75 \, e^{\left (-4 \, x\right )} - 60 i \, e^{\left (-5 \, x\right )} + 15 \, e^{\left (-6 \, x\right )} + 15} + \frac {2}{60 i \, e^{\left (-x\right )} - 75 \, e^{\left (-2 \, x\right )} - 75 \, e^{\left (-4 \, x\right )} - 60 i \, e^{\left (-5 \, x\right )} + 15 \, e^{\left (-6 \, x\right )} + 15} \]
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Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.11 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {i}{4 \, {\left (e^{x} - i\right )}} - \frac {15 i \, e^{\left (4 \, x\right )} + 30 \, e^{\left (3 \, x\right )} + 40 i \, e^{\left (2 \, x\right )} - 50 \, e^{x} - 7 i}{60 \, {\left (e^{x} + i\right )}^{5}} \]
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Time = 0.31 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.76 \[ \int \frac {\tanh ^2(x)}{(i+\sinh (x))^2} \, dx=\frac {\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,\left (2\,{\mathrm {e}}^{4\,x}-\frac {8\,{\mathrm {e}}^{2\,x}}{3}+\frac {2}{15}\right )\,1{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}-\frac {\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )\,\left (2\,{\mathrm {e}}^{4\,x}-\frac {8\,{\mathrm {e}}^{2\,x}}{3}+\frac {2}{15}\right )}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}-\frac {\left (4\,{\mathrm {e}}^{3\,x}-4\,{\mathrm {e}}^x\right )\,\left (\frac {8\,{\mathrm {e}}^{3\,x}}{3}-\frac {8\,{\mathrm {e}}^x}{15}\right )}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5}-\frac {\left (\frac {8\,{\mathrm {e}}^{3\,x}}{3}-\frac {8\,{\mathrm {e}}^x}{15}\right )\,\left ({\mathrm {e}}^{4\,x}-6\,{\mathrm {e}}^{2\,x}+1\right )\,1{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}+1\right )}^5} \]
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