Integrand size = 16, antiderivative size = 25 \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=-\frac {\log (a+b \cosh (x))}{b}+\frac {x \sinh (x)}{a+b \cosh (x)} \]
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Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {5756, 2747, 31} \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\log (a+b \cosh (x))}{b} \]
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Rule 31
Rule 2747
Rule 5756
Rubi steps \begin{align*} \text {integral}& = \frac {x \sinh (x)}{a+b \cosh (x)}-\int \frac {\sinh (x)}{a+b \cosh (x)} \, dx \\ & = \frac {x \sinh (x)}{a+b \cosh (x)}-\frac {\text {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \cosh (x)\right )}{b} \\ & = -\frac {\log (a+b \cosh (x))}{b}+\frac {x \sinh (x)}{a+b \cosh (x)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=-\frac {\log (a+b \cosh (x))}{b}+\frac {x \sinh (x)}{a+b \cosh (x)} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(54\) vs. \(2(25)=50\).
Time = 0.56 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.20
method | result | size |
risch | \(\frac {2 x}{b}-\frac {2 x \left (a \,{\mathrm e}^{x}+b \right )}{b \left (b \,{\mathrm e}^{2 x}+2 a \,{\mathrm e}^{x}+b \right )}-\frac {\ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{b}\) | \(55\) |
parallelrisch | \(\frac {\left (-b \cosh \left (x \right )-a \right ) \ln \left (\frac {-2 b \cosh \left (x \right )-2 a}{\cosh \left (x \right )+1}\right )+\left (2 b \cosh \left (x \right )+2 a \right ) \ln \left (1-\coth \left (x \right )+\operatorname {csch}\left (x \right )\right )+x \left (a +b \cosh \left (x \right )+b \sinh \left (x \right )\right )}{b \left (a +b \cosh \left (x \right )\right )}\) | \(72\) |
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Leaf count of result is larger than twice the leaf count of optimal. 129 vs. \(2 (25) = 50\).
Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 5.16 \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\frac {2 \, b x \cosh \left (x\right )^{2} + 2 \, b x \sinh \left (x\right )^{2} + 2 \, a x \cosh \left (x\right ) - {\left (b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b\right )} \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, {\left (2 \, b x \cosh \left (x\right ) + a x\right )} \sinh \left (x\right )}{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right )} \]
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Timed out. \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\text {Exception raised: ValueError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (25) = 50\).
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 4.00 \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\frac {2 \, b x e^{\left (2 \, x\right )} - b e^{\left (2 \, x\right )} \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right ) - 2 \, a e^{x} \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right ) - 2 \, b x - b \log \left (-b e^{\left (2 \, x\right )} - 2 \, a e^{x} - b\right )}{b^{2} e^{\left (2 \, x\right )} + 2 \, a b e^{x} + b^{2}} \]
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Time = 2.43 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.20 \[ \int \frac {x (b+a \cosh (x))}{(a+b \cosh (x))^2} \, dx=\frac {2\,x}{b}+\frac {\frac {2\,\left (b^3\,x-a^2\,b\,x\right )}{a^2\,b-b^3}+\frac {2\,{\mathrm {e}}^x\,\left (a\,b^3\,x-a^3\,b\,x\right )}{b\,\left (a^2\,b-b^3\right )}}{b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}}-\frac {\ln \left (b+2\,a\,{\mathrm {e}}^x+b\,{\mathrm {e}}^{2\,x}\right )}{b} \]
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