Optimal. Leaf size=29 \[ \frac{\coth ^5(x)}{5 a}-\frac{2 \coth ^3(x)}{3 a}+\frac{\coth (x)}{a} \]
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Rubi [A] time = 0.0508984, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3175, 3767} \[ \frac{\coth ^5(x)}{5 a}-\frac{2 \coth ^3(x)}{3 a}+\frac{\coth (x)}{a} \]
Antiderivative was successfully verified.
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Rule 3175
Rule 3767
Rubi steps
\begin{align*} \int \frac{\text{csch}^4(x)}{a-a \cosh ^2(x)} \, dx &=-\frac{\int \text{csch}^6(x) \, dx}{a}\\ &=\frac{i \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-i \coth (x)\right )}{a}\\ &=\frac{\coth (x)}{a}-\frac{2 \coth ^3(x)}{3 a}+\frac{\coth ^5(x)}{5 a}\\ \end{align*}
Mathematica [A] time = 0.0037742, size = 32, normalized size = 1.1 \[ -\frac{-\frac{8 \coth (x)}{15}-\frac{1}{5} \coth (x) \text{csch}^4(x)+\frac{4}{15} \coth (x) \text{csch}^2(x)}{a} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.028, size = 53, normalized size = 1.8 \begin{align*}{\frac{1}{32\,a} \left ({\frac{1}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}-{\frac{5}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+10\,\tanh \left ( x/2 \right ) +10\, \left ( \tanh \left ( x/2 \right ) \right ) ^{-1}-{\frac{5}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}+{\frac{1}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-5}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.08527, size = 182, normalized size = 6.28 \begin{align*} \frac{16 \, e^{\left (-2 \, x\right )}}{3 \,{\left (5 \, a e^{\left (-2 \, x\right )} - 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} - 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} - a\right )}} - \frac{32 \, e^{\left (-4 \, x\right )}}{3 \,{\left (5 \, a e^{\left (-2 \, x\right )} - 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} - 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} - a\right )}} - \frac{16}{15 \,{\left (5 \, a e^{\left (-2 \, x\right )} - 10 \, a e^{\left (-4 \, x\right )} + 10 \, a e^{\left (-6 \, x\right )} - 5 \, a e^{\left (-8 \, x\right )} + a e^{\left (-10 \, x\right )} - a\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.80449, size = 694, normalized size = 23.93 \begin{align*} \frac{16 \,{\left (11 \, \cosh \left (x\right )^{2} + 18 \, \cosh \left (x\right ) \sinh \left (x\right ) + 11 \, \sinh \left (x\right )^{2} - 5\right )}}{15 \,{\left (a \cosh \left (x\right )^{8} + 8 \, a \cosh \left (x\right ) \sinh \left (x\right )^{7} + a \sinh \left (x\right )^{8} - 5 \, a \cosh \left (x\right )^{6} +{\left (28 \, a \cosh \left (x\right )^{2} - 5 \, a\right )} \sinh \left (x\right )^{6} + 2 \,{\left (28 \, a \cosh \left (x\right )^{3} - 15 \, a \cosh \left (x\right )\right )} \sinh \left (x\right )^{5} + 10 \, a \cosh \left (x\right )^{4} + 5 \,{\left (14 \, a \cosh \left (x\right )^{4} - 15 \, a \cosh \left (x\right )^{2} + 2 \, a\right )} \sinh \left (x\right )^{4} + 4 \,{\left (14 \, a \cosh \left (x\right )^{5} - 25 \, a \cosh \left (x\right )^{3} + 10 \, a \cosh \left (x\right )\right )} \sinh \left (x\right )^{3} - 11 \, a \cosh \left (x\right )^{2} +{\left (28 \, a \cosh \left (x\right )^{6} - 75 \, a \cosh \left (x\right )^{4} + 60 \, a \cosh \left (x\right )^{2} - 11 \, a\right )} \sinh \left (x\right )^{2} + 2 \,{\left (4 \, a \cosh \left (x\right )^{7} - 15 \, a \cosh \left (x\right )^{5} + 20 \, a \cosh \left (x\right )^{3} - 9 \, a \cosh \left (x\right )\right )} \sinh \left (x\right ) + 5 \, a\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.28108, size = 36, normalized size = 1.24 \begin{align*} \frac{16 \,{\left (10 \, e^{\left (4 \, x\right )} - 5 \, e^{\left (2 \, x\right )} + 1\right )}}{15 \, a{\left (e^{\left (2 \, x\right )} - 1\right )}^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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