Optimal. Leaf size=133 \[ -\frac{2 b^3 \left (a+b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac{x}{2}\right )+a+2 b \tanh \left (\frac{x}{2}\right )\right )}-\frac{3 a b^2 \tan ^{-1}\left (\frac{a \sinh (x)+b \cosh (x)}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (\tanh \left (\frac{x}{2}\right )+1\right )} \]
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Rubi [A] time = 0.781512, antiderivative size = 193, normalized size of antiderivative = 1.45, number of steps used = 8, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6742, 638, 618, 204} \[ -\frac{2 b^3 \left (a+b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a \tanh ^2\left (\frac{x}{2}\right )+a+2 b \tanh \left (\frac{x}{2}\right )\right )}-\frac{2 b^4 \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac{2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (\tanh \left (\frac{x}{2}\right )+1\right )} \]
Antiderivative was successfully verified.
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Rule 6742
Rule 638
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\cosh ^3(x)}{(a \cosh (x)+b \sinh (x))^2} \, dx &=2 \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^3}{\left (1-x^2\right )^2 \left (a+2 b x+a x^2\right )^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{1}{2 (a+b)^2 (-1+x)^2}+\frac{1}{2 (a-b)^2 (1+x)^2}-\frac{2 b^3 x}{a \left (-a^2+b^2\right ) \left (a+2 b x+a x^2\right )^2}+\frac{-3 a^2 b^2+b^4}{a \left (a^2-b^2\right )^2 \left (a+2 b x+a x^2\right )}\right ) \, dx,x,\tanh \left (\frac{x}{2}\right )\right )\\ &=\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (1+\tanh \left (\frac{x}{2}\right )\right )}+\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a+2 b x+a x^2\right )^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )}-\frac{\left (2 b^2 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (1+\tanh \left (\frac{x}{2}\right )\right )}-\frac{2 b^3 \left (a+b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac{x}{2}\right )+a \tanh ^2\left (\frac{x}{2}\right )\right )}-\frac{\left (2 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}+\frac{\left (4 b^2 \left (3 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=-\frac{2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (1+\tanh \left (\frac{x}{2}\right )\right )}-\frac{2 b^3 \left (a+b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac{x}{2}\right )+a \tanh ^2\left (\frac{x}{2}\right )\right )}+\frac{\left (4 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2}\\ &=-\frac{2 b^4 \tan ^{-1}\left (\frac{b+a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}-\frac{2 b^2 \left (3 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{5/2}}+\frac{1}{(a+b)^2 \left (1-\tanh \left (\frac{x}{2}\right )\right )}-\frac{1}{(a-b)^2 \left (1+\tanh \left (\frac{x}{2}\right )\right )}-\frac{2 b^3 \left (a+b \tanh \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )^2 \left (a+2 b \tanh \left (\frac{x}{2}\right )+a \tanh ^2\left (\frac{x}{2}\right )\right )}\\ \end{align*}
Mathematica [A] time = 0.349257, size = 204, normalized size = 1.53 \[ \frac{b \sqrt{a-b} \left (a^2 b+a^3+a b^2+b^3\right ) \sinh ^2(x)-2 a^2 b \sqrt{a-b} (a+b) \cosh ^2(x)-6 a b^3 \sqrt{a+b} \sinh (x) \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )+a \cosh (x) \left ((a-b)^{3/2} (a+b)^2 \sinh (x)-6 a b^2 \sqrt{a+b} \tan ^{-1}\left (\frac{a \tanh \left (\frac{x}{2}\right )+b}{\sqrt{a-b} \sqrt{a+b}}\right )\right )+b^3 \left (-\sqrt{a-b}\right ) (a+b)}{(a-b)^{5/2} (a+b)^3 (a \cosh (x)+b \sinh (x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.072, size = 167, normalized size = 1.3 \begin{align*} -{\frac{1}{ \left ( a-b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{{b}^{4}\tanh \left ( x/2 \right ) }{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}a \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{{b}^{3}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2} \left ( a+2\,\tanh \left ( x/2 \right ) b+a \left ( \tanh \left ( x/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{a{b}^{2}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{ \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.0218, size = 3667, normalized size = 27.57 \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 = 1.14985, size = 235, normalized size = 1.77 \begin{align*} -\frac{6 \, a b^{2} \arctan \left (\frac{a e^{x} + b e^{x}}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt{a^{2} - b^{2}}} + \frac{e^{x}}{2 \,{\left (a^{2} + 2 \, a b + b^{2}\right )}} - \frac{a^{3} e^{\left (2 \, x\right )} + 3 \, a^{2} b e^{\left (2 \, x\right )} + 3 \, a b^{2} e^{\left (2 \, x\right )} + 5 \, b^{3} e^{\left (2 \, x\right )} + a^{3} + a^{2} b - a b^{2} - b^{3}}{2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (a e^{\left (3 \, x\right )} + b e^{\left (3 \, x\right )} + a e^{x} - b e^{x}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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