Optimal. Leaf size=194 \[ \frac{b x}{8 \left (a^2-b^2\right )}-\frac{a^2 b x}{2 \left (a^2-b^2\right )^2}-\frac{a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac{a \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac{a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac{b \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}+\frac{b \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}+\frac{a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]
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Rubi [A] time = 0.34237, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3109, 2568, 2635, 8, 2564, 30, 3097, 3133} \[ \frac{b x}{8 \left (a^2-b^2\right )}-\frac{a^2 b x}{2 \left (a^2-b^2\right )^2}-\frac{a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac{a \sinh ^4(x)}{4 \left (a^2-b^2\right )}-\frac{a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}-\frac{b \sinh (x) \cosh ^3(x)}{4 \left (a^2-b^2\right )}+\frac{b \sinh (x) \cosh (x)}{8 \left (a^2-b^2\right )}+\frac{a^2 b \sinh (x) \cosh (x)}{2 \left (a^2-b^2\right )^2}+\frac{a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 3109
Rule 2568
Rule 2635
Rule 8
Rule 2564
Rule 30
Rule 3097
Rule 3133
Rubi steps
\begin{align*} \int \frac{\cosh ^2(x) \sinh ^3(x)}{a \cosh (x)+b \sinh (x)} \, dx &=\frac{a \int \cosh (x) \sinh ^3(x) \, dx}{a^2-b^2}-\frac{b \int \cosh ^2(x) \sinh ^2(x) \, dx}{a^2-b^2}+\frac{(a b) \int \frac{\cosh (x) \sinh ^2(x)}{a \cosh (x)+b \sinh (x)} \, dx}{a^2-b^2}\\ &=-\frac{b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac{\left (a^2 b\right ) \int \sinh ^2(x) \, dx}{\left (a^2-b^2\right )^2}-\frac{\left (a b^2\right ) \int \cosh (x) \sinh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac{\left (a^2 b^2\right ) \int \frac{\sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac{a \operatorname{Subst}\left (\int x^3 \, dx,x,i \sinh (x)\right )}{a^2-b^2}+\frac{b \int \cosh ^2(x) \, dx}{4 \left (a^2-b^2\right )}\\ &=-\frac{a^2 b^3 x}{\left (a^2-b^2\right )^3}+\frac{a^2 b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}+\frac{b \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}-\frac{b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}+\frac{a \sinh ^4(x)}{4 \left (a^2-b^2\right )}+\frac{\left (i a^3 b^2\right ) \int \frac{-i b \cosh (x)-i a \sinh (x)}{a \cosh (x)+b \sinh (x)} \, dx}{\left (a^2-b^2\right )^3}-\frac{\left (a^2 b\right ) \int 1 \, dx}{2 \left (a^2-b^2\right )^2}+\frac{\left (a b^2\right ) \operatorname{Subst}(\int x \, dx,x,i \sinh (x))}{\left (a^2-b^2\right )^2}+\frac{b \int 1 \, dx}{8 \left (a^2-b^2\right )}\\ &=-\frac{a^2 b^3 x}{\left (a^2-b^2\right )^3}-\frac{a^2 b x}{2 \left (a^2-b^2\right )^2}+\frac{b x}{8 \left (a^2-b^2\right )}+\frac{a^3 b^2 \log (a \cosh (x)+b \sinh (x))}{\left (a^2-b^2\right )^3}+\frac{a^2 b \cosh (x) \sinh (x)}{2 \left (a^2-b^2\right )^2}+\frac{b \cosh (x) \sinh (x)}{8 \left (a^2-b^2\right )}-\frac{b \cosh ^3(x) \sinh (x)}{4 \left (a^2-b^2\right )}-\frac{a b^2 \sinh ^2(x)}{2 \left (a^2-b^2\right )^2}+\frac{a \sinh ^4(x)}{4 \left (a^2-b^2\right )}\\ \end{align*}
Mathematica [A] time = 0.602485, size = 128, normalized size = 0.66 \[ \frac{a \left (a^2-b^2\right )^2 \cosh (4 x)-4 a \left (a^4-b^4\right ) \cosh (2 x)-b \left (-8 a^2 \left (a^2-b^2\right ) \sinh (2 x)+\left (a^2-b^2\right )^2 \sinh (4 x)+4 \left (6 a^2 b^2 x-8 a^3 b \log (a \cosh (x)+b \sinh (x))+3 a^4 x-b^4 x\right )\right )}{32 (a-b)^3 (a+b)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 321, normalized size = 1.7 \begin{align*} 4\,{\frac{1}{ \left ( 16\,a-16\,b \right ) \left ( \tanh \left ( x/2 \right ) +1 \right ) ^{4}}}-16\,{\frac{1}{ \left ( 32\,a-32\,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 ) ^{-1}}+{\frac{b}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{a}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,b}{8\, \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,ab}{8\, \left ( a-b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{b}^{2}}{8\, \left ( a-b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{{a}^{3}{b}^{2}}{ \left ( a-b \right ) ^{3} \left ( a+b \right ) ^{3}}\ln \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2} \right ) }+4\,{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \tanh \left ( x/2 \right ) -1 \right ) ^{4}}}+16\,{\frac{1}{ \left ( 32\,a+32\,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{3\,b}{8\, \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{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\,ab}{8\, \left ( a+b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+{\frac{{b}^{2}}{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 = 1.27258, size = 207, normalized size = 1.07 \begin{align*} \frac{a^{3} b^{2} \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 b + b^{2}\right )} x}{8 \,{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} - \frac{{\left (4 \, a 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 \, a 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 = 1.94103, size = 2612, normalized size = 13.46 \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 = 1.14632, size = 269, normalized size = 1.39 \begin{align*} \frac{a^{3} b^{2} \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 b - b^{2}\right )} x}{8 \,{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )}} + \frac{{\left (18 \, a b e^{\left (4 \, x\right )} - 6 \, b^{2} e^{\left (4 \, x\right )} - 4 \, a^{2} e^{\left (2 \, x\right )} + 4 \, a b 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 )} - 4 \, a 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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