Optimal. Leaf size=36 \[ \frac {1}{5} i \coth ^5(x)-\frac {3}{8} \tanh ^{-1}(\cosh (x))-\frac {1}{4} \coth ^3(x) \text {csch}(x)-\frac {3}{8} \coth (x) \text {csch}(x) \]
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Rubi [A] time = 0.10, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac {1}{5} i \coth ^5(x)-\frac {3}{8} \tanh ^{-1}(\cosh (x))-\frac {1}{4} \coth ^3(x) \text {csch}(x)-\frac {3}{8} \coth (x) \text {csch}(x) \]
Antiderivative was successfully verified.
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Rule 30
Rule 2607
Rule 2611
Rule 2706
Rule 3770
Rubi steps
\begin {align*} \int \frac {\coth ^6(x)}{i+\sinh (x)} \, dx &=-\left (i \int \coth ^4(x) \text {csch}^2(x) \, dx\right )+\int \coth ^4(x) \text {csch}(x) \, dx\\ &=-\frac {1}{4} \coth ^3(x) \text {csch}(x)+\frac {3}{4} \int \coth ^2(x) \text {csch}(x) \, dx+\operatorname {Subst}\left (\int x^4 \, dx,x,i \coth (x)\right )\\ &=\frac {1}{5} i \coth ^5(x)-\frac {3}{8} \coth (x) \text {csch}(x)-\frac {1}{4} \coth ^3(x) \text {csch}(x)+\frac {3}{8} \int \text {csch}(x) \, dx\\ &=-\frac {3}{8} \tanh ^{-1}(\cosh (x))+\frac {1}{5} i \coth ^5(x)-\frac {3}{8} \coth (x) \text {csch}(x)-\frac {1}{4} \coth ^3(x) \text {csch}(x)\\ \end {align*}
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Mathematica [B] time = 0.04, size = 164, normalized size = 4.56 \[ \frac {1}{10} i \tanh \left (\frac {x}{2}\right )+\frac {1}{10} i \coth \left (\frac {x}{2}\right )-\frac {1}{64} \text {csch}^4\left (\frac {x}{2}\right )-\frac {5}{32} \text {csch}^2\left (\frac {x}{2}\right )+\frac {1}{64} \text {sech}^4\left (\frac {x}{2}\right )-\frac {5}{32} \text {sech}^2\left (\frac {x}{2}\right )+\frac {3}{8} \log \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {1}{160} i \coth \left (\frac {x}{2}\right ) \text {csch}^4\left (\frac {x}{2}\right )+\frac {7}{160} i \coth \left (\frac {x}{2}\right ) \text {csch}^2\left (\frac {x}{2}\right )+\frac {1}{160} i \tanh \left (\frac {x}{2}\right ) \text {sech}^4\left (\frac {x}{2}\right )-\frac {7}{160} i \tanh \left (\frac {x}{2}\right ) \text {sech}^2\left (\frac {x}{2}\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 1.87, size = 144, normalized size = 4.00 \[ -\frac {15 \, {\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) - 15 \, {\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) + 50 \, e^{\left (9 \, x\right )} - 80 i \, e^{\left (8 \, x\right )} - 20 \, e^{\left (7 \, x\right )} - 160 i \, e^{\left (4 \, x\right )} + 20 \, e^{\left (3 \, x\right )} - 50 \, e^{x} - 16 i}{40 \, {\left (e^{\left (10 \, x\right )} - 5 \, e^{\left (8 \, x\right )} + 10 \, e^{\left (6 \, x\right )} - 10 \, e^{\left (4 \, x\right )} + 5 \, e^{\left (2 \, x\right )} - 1\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.20, size = 62, normalized size = 1.72 \[ -\frac {25 \, e^{\left (9 \, x\right )} - 40 i \, e^{\left (8 \, x\right )} - 10 \, e^{\left (7 \, x\right )} - 80 i \, e^{\left (4 \, x\right )} + 10 \, e^{\left (3 \, x\right )} - 25 \, e^{x} - 8 i}{20 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} - \frac {3}{8} \, \log \left (e^{x} + 1\right ) + \frac {3}{8} \, \log \left ({\left | e^{x} - 1 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.10, size = 93, normalized size = 2.58 \[ \frac {i \tanh \left (\frac {x}{2}\right )}{16}+\frac {i \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{160}+\frac {\left (\tanh ^{4}\left (\frac {x}{2}\right )\right )}{64}+\frac {i \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{32}+\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {i}{160 \tanh \left (\frac {x}{2}\right )^{5}}+\frac {i}{32 \tanh \left (\frac {x}{2}\right )^{3}}-\frac {1}{64 \tanh \left (\frac {x}{2}\right )^{4}}+\frac {i}{16 \tanh \left (\frac {x}{2}\right )}-\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{8} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.31, size = 91, normalized size = 2.53 \[ \frac {25 \, e^{\left (-x\right )} - 10 \, e^{\left (-3 \, x\right )} - 80 i \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-7 \, x\right )} - 40 i \, e^{\left (-8 \, x\right )} - 25 \, e^{\left (-9 \, x\right )} - 8 i}{20 \, {\left (5 \, e^{\left (-2 \, x\right )} - 10 \, e^{\left (-4 \, x\right )} + 10 \, e^{\left (-6 \, x\right )} - 5 \, e^{\left (-8 \, x\right )} + e^{\left (-10 \, x\right )} - 1\right )}} - \frac {3}{8} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac {3}{8} \, \log \left (e^{\left (-x\right )} - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.97, size = 124, normalized size = 3.44 \[ \frac {3\,\ln \left (\frac {3}{4}-\frac {3\,{\mathrm {e}}^x}{4}\right )}{8}-\frac {3\,\ln \left (\frac {3\,{\mathrm {e}}^x}{4}+\frac {3}{4}\right )}{8}-\frac {5\,{\mathrm {e}}^x}{4\,\left ({\mathrm {e}}^{2\,x}-1\right )}-\frac {9\,{\mathrm {e}}^x}{2\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^2}-\frac {6\,{\mathrm {e}}^x}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^3}-\frac {4\,{\mathrm {e}}^x}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^4}+\frac {2{}\mathrm {i}}{{\mathrm {e}}^{2\,x}-1}+\frac {8{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^2}+\frac {16{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^3}+\frac {16{}\mathrm {i}}{{\left ({\mathrm {e}}^{2\,x}-1\right )}^4}+\frac {32{}\mathrm {i}}{5\,{\left ({\mathrm {e}}^{2\,x}-1\right )}^5} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.31, size = 100, normalized size = 2.78 \[ \frac {3 \log {\left (e^{x} - 1 \right )}}{8} - \frac {3 \log {\left (e^{x} + 1 \right )}}{8} + \frac {25 e^{9 x} - 40 i e^{8 x} - 10 e^{7 x} - 80 i e^{4 x} + 10 e^{3 x} - 25 e^{x} - 8 i}{- 20 e^{10 x} + 100 e^{8 x} - 200 e^{6 x} + 200 e^{4 x} - 100 e^{2 x} + 20} \]
Verification of antiderivative is not currently implemented for this CAS.
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