Integrand size = 20, antiderivative size = 78 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\frac {(2 a A-b B) x}{2 a^2}+\frac {B \cosh (x)}{2 a}+\frac {\left (2 a A b-a^2 B-b^2 B\right ) \log (a+b \cosh (x)-b \sinh (x))}{2 a^2 b}+\frac {B \sinh (x)}{2 a} \]
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Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, 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 = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\frac {\left (2 a A b+a^2 B-b^2 B\right ) x+2 a b B \cosh (x)-2 \left (-2 a A b+a^2 B+b^2 B\right ) \log \left ((a+b) \cosh \left (\frac {x}{2}\right )+(a-b) \sinh \left (\frac {x}{2}\right )\right )+2 a b B \sinh (x)}{4 a^2 b} \]
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Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {B \,{\mathrm e}^{x}}{2 a}+\frac {B x}{2 b}+\frac {\ln \left ({\mathrm e}^{x}+\frac {b}{a}\right ) A}{a}-\frac {\ln \left ({\mathrm e}^{x}+\frac {b}{a}\right ) B}{2 b}-\frac {b \ln \left ({\mathrm e}^{x}+\frac {b}{a}\right ) B}{2 a^{2}}\) | \(62\) |
default | \(\frac {B \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 b}-\frac {B}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-2 A a +B b \right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 a^{2}}+\frac {\left (2 A a b -B \,a^{2}-B \,b^{2}\right ) \ln \left (a \tanh \left (\frac {x}{2}\right )-b \tanh \left (\frac {x}{2}\right )+a +b \right )}{2 a^{2} b}\) | \(92\) |
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Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.72 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\frac {B a^{2} x + B a b \cosh \left (x\right ) + B a b \sinh \left (x\right ) - {\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}{2 \, a^{2} b} \]
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Leaf count of result is larger than twice the leaf count of optimal. 904 vs. \(2 (66) = 132\).
Time = 2.35 (sec) , antiderivative size = 904, normalized size of antiderivative = 11.59 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\text {Too large to display} \]
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Time = 0.21 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.79 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=A {\left (\frac {x}{a} + \frac {\log \left (b e^{\left (-x\right )} + a\right )}{a}\right )} - \frac {1}{2} \, B {\left (\frac {b x}{a^{2}} - \frac {e^{x}}{a} + \frac {{\left (a^{2} + b^{2}\right )} \log \left (b e^{\left (-x\right )} + a\right )}{a^{2} b}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.62 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\frac {B x}{2 \, b} + \frac {B e^{x}}{2 \, a} - \frac {{\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left ({\left | a e^{x} + b \right |}\right )}{2 \, a^{2} b} \]
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Time = 2.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.60 \[ \int \frac {A+B \cosh (x)}{a+b \cosh (x)-b \sinh (x)} \, dx=\frac {B\,{\mathrm {e}}^x}{2\,a}+\frac {B\,x}{2\,b}-\frac {\ln \left (b+a\,{\mathrm {e}}^x\right )\,\left (B\,a^2-2\,A\,a\,b+B\,b^2\right )}{2\,a^2\,b} \]
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