Integrand size = 21, antiderivative size = 51 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {(a-2 b) \text {arctanh}(\cosh (c+d x))}{2 d}-\frac {a \coth (c+d x) \text {csch}(c+d x)}{2 d}+\frac {b \text {sech}(c+d x)}{d} \]
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Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3745, 466, 396, 213} \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {(a-2 b) \text {arctanh}(\cosh (c+d x))}{2 d}-\frac {a \coth (c+d x) \text {csch}(c+d x)}{2 d}+\frac {b \text {sech}(c+d x)}{d} \]
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Rule 213
Rule 396
Rule 466
Rule 3745
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^2 \left (a+b-b x^2\right )}{\left (-1+x^2\right )^2} \, dx,x,\text {sech}(c+d x)\right )}{d} \\ & = -\frac {a \coth (c+d x) \text {csch}(c+d x)}{2 d}+\frac {\text {Subst}\left (\int \frac {-a+2 b x^2}{-1+x^2} \, dx,x,\text {sech}(c+d x)\right )}{2 d} \\ & = -\frac {a \coth (c+d x) \text {csch}(c+d x)}{2 d}+\frac {b \text {sech}(c+d x)}{d}-\frac {(a-2 b) \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\text {sech}(c+d x)\right )}{2 d} \\ & = \frac {(a-2 b) \text {arctanh}(\cosh (c+d x))}{2 d}-\frac {a \coth (c+d x) \text {csch}(c+d x)}{2 d}+\frac {b \text {sech}(c+d x)}{d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(123\) vs. \(2(51)=102\).
Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 2.41 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=-\frac {a \text {csch}^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {a \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {b \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {a \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {b \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {a \text {sech}^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {b \text {sech}(c+d x)}{d} \]
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Time = 1.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.98
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\operatorname {csch}\left (d x +c \right ) \coth \left (d x +c \right )}{2}+\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )\right )+b \left (\frac {1}{\cosh \left (d x +c \right )}-2 \,\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )\right )}{d}\) | \(50\) |
default | \(\frac {a \left (-\frac {\operatorname {csch}\left (d x +c \right ) \coth \left (d x +c \right )}{2}+\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )\right )+b \left (\frac {1}{\cosh \left (d x +c \right )}-2 \,\operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )\right )}{d}\) | \(50\) |
risch | \(-\frac {{\mathrm e}^{d x +c} \left (a \,{\mathrm e}^{4 d x +4 c}-2 b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a +4 b \,{\mathrm e}^{2 d x +2 c}+a -2 b \right )}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{2} \left ({\mathrm e}^{2 d x +2 c}+1\right )}-\frac {a \ln \left ({\mathrm e}^{d x +c}-1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right ) b}{d}+\frac {a \ln \left ({\mathrm e}^{d x +c}+1\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right ) b}{d}\) | \(150\) |
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Leaf count of result is larger than twice the leaf count of optimal. 924 vs. \(2 (47) = 94\).
Time = 0.27 (sec) , antiderivative size = 924, normalized size of antiderivative = 18.12 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
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\[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right ) \operatorname {csch}^{3}{\left (c + d x \right )}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (47) = 94\).
Time = 0.21 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.98 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {1}{2} \, a {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} + \frac {2 \, {\left (e^{\left (-d x - c\right )} + e^{\left (-3 \, d x - 3 \, c\right )}\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} - e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}}\right )} - b {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {2 \, e^{\left (-d x - c\right )}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (47) = 94\).
Time = 0.29 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.78 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {{\left (a - 2 \, b\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} + 2\right ) - {\left (a - 2 \, b\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} - 2\right ) - \frac {4 \, {\left (a {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} - 2 \, b {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 8 \, b\right )}}{{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3} - 4 \, e^{\left (d x + c\right )} - 4 \, e^{\left (-d x - c\right )}}}{4 \, d} \]
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Time = 1.86 (sec) , antiderivative size = 156, normalized size of antiderivative = 3.06 \[ \int \text {csch}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right ) \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (a\,\sqrt {-d^2}-2\,b\,\sqrt {-d^2}\right )}{d\,\sqrt {a^2-4\,a\,b+4\,b^2}}\right )\,\sqrt {a^2-4\,a\,b+4\,b^2}}{\sqrt {-d^2}}-\frac {a\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}+\frac {2\,b\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]
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