Integrand size = 11, antiderivative size = 38 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))} \]
[Out]
Time = 0.05 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3597, 862, 90, 213} \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}+\frac {1}{4 (\tanh (x)+1)}-\frac {1}{8 (\tanh (x)+1)^2} \]
[In]
[Out]
Rule 90
Rule 213
Rule 862
Rule 3597
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {x^2}{(1+x) \left (-1+x^2\right )^2} \, dx,x,\tanh (x)\right ) \\ & = \text {Subst}\left (\int \frac {x^2}{(-1+x)^2 (1+x)^3} \, dx,x,\tanh (x)\right ) \\ & = \text {Subst}\left (\int \left (\frac {1}{8 (-1+x)^2}+\frac {1}{4 (1+x)^3}-\frac {1}{4 (1+x)^2}+\frac {1}{8 \left (-1+x^2\right )}\right ) \, dx,x,\tanh (x)\right ) \\ & = \frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (x)\right ) \\ & = -\frac {x}{8}+\frac {1}{8 (1-\tanh (x))}-\frac {1}{8 (1+\tanh (x))^2}+\frac {1}{4 (1+\tanh (x))} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.63 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{32} (-4 x+4 \cosh (2 x)-\cosh (4 x)+\sinh (4 x)) \]
[In]
[Out]
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.61
method | result | size |
risch | \(-\frac {x}{8}+\frac {{\mathrm e}^{2 x}}{16}+\frac {{\mathrm e}^{-2 x}}{16}-\frac {{\mathrm e}^{-4 x}}{32}\) | \(23\) |
parallelrisch | \(-\frac {x}{8}+\frac {\sinh \left (4 x \right )}{32}-\frac {\cosh \left (4 x \right )}{32}+\frac {\cosh \left (2 x \right )}{8}-\frac {3}{32}\) | \(24\) |
default | \(-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{4}}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {1}{2 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{8}+\frac {1}{4 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {1}{4 \tanh \left (\frac {x}{2}\right )-4}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{8}\) | \(68\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.34 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 3 \, \sinh \left (x\right )^{3} - 2 \, {\left (2 \, x - 1\right )} \cosh \left (x\right ) + {\left (9 \, \cosh \left (x\right )^{2} - 4 \, x - 2\right )} \sinh \left (x\right )}{32 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \]
[In]
[Out]
\[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\int \frac {\sinh ^{2}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \]
[In]
[Out]
none
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.58 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=-\frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} + \frac {1}{16} \, e^{\left (-2 \, x\right )} - \frac {1}{32} \, e^{\left (-4 \, x\right )} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.79 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{32} \, {\left (3 \, e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-4 \, x\right )} - \frac {1}{8} \, x + \frac {1}{16} \, e^{\left (2 \, x\right )} \]
[In]
[Out]
Time = 1.70 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.58 \[ \int \frac {\sinh ^2(x)}{1+\tanh (x)} \, dx=\frac {{\mathrm {e}}^{-2\,x}}{16}-\frac {x}{8}+\frac {{\mathrm {e}}^{2\,x}}{16}-\frac {{\mathrm {e}}^{-4\,x}}{32} \]
[In]
[Out]