Integrand size = 19, antiderivative size = 77 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {(2 a A-b B) x}{2 a^2}-\frac {B \cosh (x)}{2 a}-\frac {\left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cosh (x)+b \sinh (x))}{2 a^2 b}+\frac {B \sinh (x)}{2 a} \]
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Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {3211} \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=-\frac {\left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a+b \sinh (x)+b \cosh (x))}{2 a^2 b}+\frac {x (2 a A-b B)}{2 a^2}+\frac {B \sinh (x)}{2 a}-\frac {B \cosh (x)}{2 a} \]
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Rule 3211
Rubi steps \begin{align*} \text {integral}& = \frac {(2 a A-b B) x}{2 a^2}-\frac {B \cosh (x)}{2 a}-\frac {\left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cosh (x)+b \sinh (x))}{2 a^2 b}+\frac {B \sinh (x)}{2 a} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {\left (2 a A+\frac {a^2 B}{b}-b B\right ) x-2 a B \cosh (x)+\frac {2 \left (-2 a A b+a^2 B+b^2 B\right ) \log \left ((a+b) \cosh \left (\frac {x}{2}\right )+(-a+b) \sinh \left (\frac {x}{2}\right )\right )}{b}+2 a B \sinh (x)}{4 a^2} \]
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Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.94
method | result | size |
risch | \(-\frac {B \,{\mathrm e}^{-x}}{2 a}+\frac {x A}{a}-\frac {b x B}{2 a^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) A}{a}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) B}{2 b}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) B}{2 a^{2}}\) | \(72\) |
default | \(-\frac {B}{a \left (1+\tanh \left (\frac {x}{2}\right )\right )}+\frac {\left (2 A a -B b \right ) \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}}-\frac {\left (2 A a b -B \,a^{2}-B \,b^{2}\right ) \ln \left (a \tanh \left (\frac {x}{2}\right )-b \tanh \left (\frac {x}{2}\right )-a -b \right )}{2 a^{2} b}-\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b}\) | \(97\) |
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Time = 0.26 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.43 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=-\frac {B a b - {\left (2 \, A a b - B b^{2}\right )} x \cosh \left (x\right ) - {\left (2 \, A a b - B b^{2}\right )} x \sinh \left (x\right ) - {\left ({\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \cosh \left (x\right ) + {\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \sinh \left (x\right )\right )} \log \left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}{2 \, {\left (a^{2} b \cosh \left (x\right ) + a^{2} b \sinh \left (x\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 806 vs. \(2 (66) = 132\).
Time = 2.26 (sec) , antiderivative size = 806, normalized size of antiderivative = 10.47 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\text {Too large to display} \]
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Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {1}{2} \, B {\left (\frac {x}{b} - \frac {e^{\left (-x\right )}}{a} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (a e^{\left (-x\right )} + b\right )}{a^{2} b}\right )} - \frac {A \log \left (a e^{\left (-x\right )} + b\right )}{a} \]
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.75 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=-\frac {B e^{\left (-x\right )}}{2 \, a} + \frac {{\left (2 \, A a - B b\right )} x}{2 \, a^{2}} + \frac {{\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left ({\left | b e^{x} + a \right |}\right )}{2 \, a^{2} b} \]
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Time = 2.42 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.74 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {x\,\left (2\,A\,a-B\,b\right )}{2\,a^2}-\frac {B\,{\mathrm {e}}^{-x}}{2\,a}+\frac {\ln \left (a+b\,{\mathrm {e}}^x\right )\,\left (B\,a^2-2\,A\,a\,b+B\,b^2\right )}{2\,a^2\,b} \]
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