Integrand size = 21, antiderivative size = 100 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {(a+4 b) \log (1-\tanh (c+d x))}{4 d}-\frac {(a-4 b) \log (1+\tanh (c+d x))}{4 d}+\frac {a \tanh (c+d x)}{2 d}+\frac {b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d} \]
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Time = 0.09 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3744, 1818, 1816, 647, 31} \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {(a+4 b) \log (1-\tanh (c+d x))}{4 d}-\frac {(a-4 b) \log (\tanh (c+d x)+1)}{4 d}+\frac {\sinh ^2(c+d x) (a \tanh (c+d x)+b)}{2 d}+\frac {a \tanh (c+d x)}{2 d}+\frac {b \tanh ^2(c+d x)}{2 d} \]
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Rule 31
Rule 647
Rule 1816
Rule 1818
Rule 3744
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^2 \left (a+b x^3\right )}{\left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d}+\frac {\text {Subst}\left (\int \frac {x \left (-2 b-a x-2 b x^2\right )}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d} \\ & = \frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d}+\frac {\text {Subst}\left (\int \left (a+2 b x-\frac {a+4 b x}{1-x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{2 d} \\ & = \frac {a \tanh (c+d x)}{2 d}+\frac {b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d}-\frac {\text {Subst}\left (\int \frac {a+4 b x}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 d} \\ & = \frac {a \tanh (c+d x)}{2 d}+\frac {b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d}+\frac {(a-4 b) \text {Subst}\left (\int \frac {1}{-1-x} \, dx,x,\tanh (c+d x)\right )}{4 d}-\frac {(a+4 b) \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\tanh (c+d x)\right )}{4 d} \\ & = \frac {(a+4 b) \log (1-\tanh (c+d x))}{4 d}-\frac {(a-4 b) \log (1+\tanh (c+d x))}{4 d}+\frac {a \tanh (c+d x)}{2 d}+\frac {b \tanh ^2(c+d x)}{2 d}+\frac {\sinh ^2(c+d x) (b+a \tanh (c+d x))}{2 d} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.69 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {a (-c-d x)}{2 d}-\frac {b \left (4 \log (\cosh (c+d x))+\text {sech}^2(c+d x)-\sinh ^2(c+d x)\right )}{2 d}+\frac {a \sinh (2 (c+d x))}{4 d} \]
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Time = 0.43 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.68
method | result | size |
derivativedivides | \(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )}{d}\) | \(68\) |
default | \(\frac {a \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+b \left (\frac {\sinh \left (d x +c \right )^{4}}{2 \cosh \left (d x +c \right )^{2}}-2 \ln \left (\cosh \left (d x +c \right )\right )+\tanh \left (d x +c \right )^{2}\right )}{d}\) | \(68\) |
risch | \(-\frac {a x}{2}+2 b x +\frac {{\mathrm e}^{2 d x +2 c} a}{8 d}+\frac {{\mathrm e}^{2 d x +2 c} b}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b}{8 d}+\frac {4 b c}{d}-\frac {2 b \,{\mathrm e}^{2 d x +2 c}}{d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{2}}-\frac {2 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(123\) |
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Leaf count of result is larger than twice the leaf count of optimal. 924 vs. \(2 (90) = 180\).
Time = 0.32 (sec) , antiderivative size = 924, normalized size of antiderivative = 9.24 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\text {Too large to display} \]
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\[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right ) \sinh ^{2}{\left (c + d x \right )}\, dx \]
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Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.41 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=-\frac {1}{8} \, a {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{8} \, b {\left (\frac {16 \, {\left (d x + c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} + \frac {16 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} - \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )} - 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}\right )} \]
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Time = 0.32 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.40 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=-\frac {4 \, {\left (d x + c\right )} {\left (a - 4 \, b\right )} - a e^{\left (2 \, d x + 2 \, c\right )} - b e^{\left (2 \, d x + 2 \, c\right )} - {\left (2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b e^{\left (2 \, d x + 2 \, c\right )} - a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 16 \, b \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac {8 \, {\left (3 \, b e^{\left (4 \, d x + 4 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{8 \, d} \]
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Time = 2.01 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.15 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^3(c+d x)\right ) \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+b\right )}{8\,d}-\frac {2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-x\,\left (\frac {a}{2}-2\,b\right )+\frac {2\,b}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (a-b\right )}{8\,d}-\frac {2\,b\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d} \]
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