Optimal. Leaf size=147 \[ -\frac{a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac{\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac{a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac{a (3 a-b) \log (\tanh (x)+1)}{16 (a-b)^3} \]
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Rubi [A] time = 0.345582, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3516, 1647, 801} \[ -\frac{a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac{\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac{a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac{a (3 a-b) \log (\tanh (x)+1)}{16 (a-b)^3} \]
Antiderivative was successfully verified.
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Rule 3516
Rule 1647
Rule 801
Rubi steps
\begin{align*} \int \frac{\sinh ^4(x)}{a+b \tanh (x)} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{x^4}{(a+x) \left (-b^2+x^2\right )^3} \, dx,x,b \tanh (x)\right )\right )\\ &=-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^4}{a^2-b^2}-\frac{3 a b^4 x}{a^2-b^2}+4 b^2 x^2}{(a+x) \left (-b^2+x^2\right )^2} \, dx,x,b \tanh (x)\right )}{4 b}\\ &=-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac{\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac{\operatorname{Subst}\left (\int \frac{\frac{a^2 b^4 \left (3 a^2+b^2\right )}{\left (a^2-b^2\right )^2}-\frac{a b^4 \left (5 a^2-b^2\right ) x}{\left (a^2-b^2\right )^2}}{(a+x) \left (-b^2+x^2\right )} \, dx,x,b \tanh (x)\right )}{8 b^3}\\ &=-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac{\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}-\frac{\operatorname{Subst}\left (\int \left (-\frac{a b^3 (3 a+b)}{2 (a+b)^3 (b-x)}+\frac{8 a^4 b^4}{(a-b)^3 (a+b)^3 (a+x)}-\frac{a (3 a-b) b^3}{2 (a-b)^3 (b+x)}\right ) \, dx,x,b \tanh (x)\right )}{8 b^3}\\ &=-\frac{a (3 a+b) \log (1-\tanh (x))}{16 (a+b)^3}+\frac{a (3 a-b) \log (1+\tanh (x))}{16 (a-b)^3}-\frac{a^4 b \log (a+b \tanh (x))}{\left (a^2-b^2\right )^3}-\frac{\cosh ^4(x) (b-a \tanh (x))}{4 \left (a^2-b^2\right )}+\frac{\cosh ^2(x) \left (4 b \left (2 a^2-b^2\right )-a \left (5 a^2-b^2\right ) \tanh (x)\right )}{8 \left (a^2-b^2\right )^2}\\ \end{align*}
Mathematica [A] time = 0.539949, size = 144, normalized size = 0.98 \[ \frac{24 a^3 b^2 x-8 a^3 \left (a^2-b^2\right ) \sinh (2 x)-2 a^3 b^2 \sinh (4 x)+4 b \left (-4 a^2 b^2+3 a^4+b^4\right ) \cosh (2 x)-b \left (a^2-b^2\right )^2 \cosh (4 x)-32 a^4 b \log (a \cosh (x)+b \sinh (x))+12 a^5 x+a^5 \sinh (4 x)-4 a b^4 x+a b^4 \sinh (4 x)}{32 (a-b)^3 (a+b)^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.046, size = 320, normalized size = 2.2 \begin{align*} -8\,{\frac{1}{ \left ( 32\,a-32\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{4}}}+32\,{\frac{1}{ \left ( 64\,a-64\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{3}}}+{\frac{a}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}+{\frac{b}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,a}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{b}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,{a}^{2}}{8\, \left ( a-b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{ab}{8\, \left ( a-b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{{a}^{4}b}{ \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+2\,\tanh \left ( x/2 \right ) b+a \right ) }+8\,{\frac{1}{ \left ( 32\,a+32\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{4}}}+32\,{\frac{1}{ \left ( 64\,a+64\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{3}}}-{\frac{a}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}+{\frac{b}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{3\,a}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{b}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{3\,{a}^{2}}{8\, \left ( a+b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{ab}{8\, \left ( a+b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.16127, size = 220, normalized size = 1.5 \begin{align*} -\frac{a^{4} b \log \left (-{\left (a - b\right )} e^{\left (-2 \, x\right )} - a - b\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{{\left (3 \, a^{2} + a b\right )} x}{8 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{{\left (4 \,{\left (2 \, a + b\right )} e^{\left (-2 \, x\right )} - a - b\right )} e^{\left (4 \, x\right )}}{64 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac{4 \,{\left (2 \, a - b\right )} e^{\left (-2 \, x\right )} -{\left (a - b\right )} e^{\left (-4 \, x\right )}}{64 \,{\left (a^{2} - 2 \, a b + b^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.54086, size = 2747, normalized size = 18.69 \begin{align*} \text{result too large to display} \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 = 1.18694, size = 289, normalized size = 1.97 \begin{align*} -\frac{a^{4} b \log \left ({\left | a e^{\left (2 \, x\right )} + b e^{\left (2 \, x\right )} + a - b \right |}\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac{{\left (3 \, a^{2} - a b\right )} x}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} - \frac{{\left (18 \, a^{2} e^{\left (4 \, x\right )} - 6 \, a b e^{\left (4 \, x\right )} - 8 \, a^{2} e^{\left (2 \, x\right )} + 12 \, a b e^{\left (2 \, x\right )} - 4 \, b^{2} e^{\left (2 \, x\right )} + a^{2} - 2 \, a b + b^{2}\right )} e^{\left (-4 \, x\right )}}{64 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac{a e^{\left (4 \, x\right )} + b e^{\left (4 \, x\right )} - 8 \, a e^{\left (2 \, x\right )} - 4 \, b e^{\left (2 \, x\right )}}{64 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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