Integrand size = 19, antiderivative size = 71 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {(2 a A+b C) x}{2 a^2}+\frac {C \cosh (x)}{2 a}-\frac {1}{2} \left (\frac {2 A}{a}-\frac {C}{b}+\frac {b C}{a^2}\right ) \log (a+b \cosh (x)+b \sinh (x))-\frac {C \sinh (x)}{2 a} \]
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Time = 0.03 (sec) , antiderivative size = 71, 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 = {3210} \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {x (2 a A+b C)}{2 a^2}-\frac {1}{2} \left (\frac {b C}{a^2}+\frac {2 A}{a}-\frac {C}{b}\right ) \log (a+b \sinh (x)+b \cosh (x))-\frac {C \sinh (x)}{2 a}+\frac {C \cosh (x)}{2 a} \]
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Rule 3210
Rubi steps \begin{align*} \text {integral}& = \frac {(2 a A+b C) x}{2 a^2}+\frac {C \cosh (x)}{2 a}-\frac {1}{2} \left (\frac {2 A}{a}-\frac {C}{b}+\frac {b C}{a^2}\right ) \log (a+b \cosh (x)+b \sinh (x))-\frac {C \sinh (x)}{2 a} \\ \end{align*}
Time = 0.35 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.21 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {\left (2 a A b+a^2 C+b^2 C\right ) x+2 a b C \cosh (x)+2 \left (-2 a A b+a^2 C-b^2 C\right ) \log \left ((a+b) \cosh \left (\frac {x}{2}\right )+(-a+b) \sinh \left (\frac {x}{2}\right )\right )-2 a b C \sinh (x)}{4 a^2 b} \]
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Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {C \,{\mathrm e}^{-x}}{2 a}+\frac {x A}{a}+\frac {b x C}{2 a^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) A}{a}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) C}{2 b}-\frac {b \ln \left ({\mathrm e}^{x}+\frac {a}{b}\right ) C}{2 a^{2}}\) | \(72\) |
default | \(-\frac {\left (2 A a b -C \,a^{2}+C \,b^{2}\right ) \ln \left (a \tanh \left (\frac {x}{2}\right )-b \tanh \left (\frac {x}{2}\right )-a -b \right )}{2 a^{2} b}+\frac {C}{a \left (1+\tanh \left (\frac {x}{2}\right )\right )}+\frac {\left (2 A a +b C \right ) \ln \left (1+\tanh \left (\frac {x}{2}\right )\right )}{2 a^{2}}-\frac {C \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b}\) | \(94\) |
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Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.51 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {C a b + {\left (2 \, A a b + C b^{2}\right )} x \cosh \left (x\right ) + {\left (2 \, A a b + C b^{2}\right )} x \sinh \left (x\right ) + {\left ({\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \cosh \left (x\right ) + {\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \sinh \left (x\right )\right )} \log \left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}{2 \, {\left (a^{2} b \cosh \left (x\right ) + a^{2} b \sinh \left (x\right )\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 753 vs. \(2 (66) = 132\).
Time = 2.21 (sec) , antiderivative size = 753, normalized size of antiderivative = 10.61 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\text {Too large to display} \]
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Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {1}{2} \, C {\left (\frac {x}{b} + \frac {e^{\left (-x\right )}}{a} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (a e^{\left (-x\right )} + b\right )}{a^{2} b}\right )} - \frac {A \log \left (a e^{\left (-x\right )} + b\right )}{a} \]
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Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.82 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {C e^{\left (-x\right )}}{2 \, a} + \frac {{\left (2 \, A a + C b\right )} x}{2 \, a^{2}} + \frac {{\left (C a^{2} - 2 \, A a b - C b^{2}\right )} \log \left ({\left | b e^{x} + a \right |}\right )}{2 \, a^{2} b} \]
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Time = 2.33 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.80 \[ \int \frac {A+C \sinh (x)}{a+b \cosh (x)+b \sinh (x)} \, dx=\frac {C\,{\mathrm {e}}^{-x}}{2\,a}+\frac {x\,\left (2\,A\,a+C\,b\right )}{2\,a^2}-\frac {\ln \left (a+b\,{\mathrm {e}}^x\right )\,\left (-C\,a^2+2\,A\,a\,b+C\,b^2\right )}{2\,a^2\,b} \]
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