Integrand size = 23, antiderivative size = 109 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \]
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Time = 0.13 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3751, 457, 90} \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {a^2}{4 b^2 d (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 d (a+b)^2 \left (a+b \tanh ^2(c+d x)\right )}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 d (a+b)^3}+\frac {\log (\cosh (c+d x))}{d (a+b)^3} \]
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Rule 90
Rule 457
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^5}{\left (1-x^2\right ) \left (a+b x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x^2}{(1-x) (a+b x)^3} \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {1}{(a+b)^3 (-1+x)}+\frac {a^2}{b (a+b) (a+b x)^3}-\frac {a (a+2 b)}{b (a+b)^2 (a+b x)^2}+\frac {b}{(a+b)^3 (a+b x)}\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {\log (\cosh (c+d x))}{(a+b)^3 d}+\frac {\log \left (a+b \tanh ^2(c+d x)\right )}{2 (a+b)^3 d}-\frac {a^2}{4 b^2 (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (a+2 b)}{2 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \\ \end{align*}
Time = 1.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.83 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {-4 \log (\cosh (c+d x))-2 \log \left (a+b \tanh ^2(c+d x)\right )+\frac {a^2 (a+b)^2}{b^2 \left (a+b \tanh ^2(c+d x)\right )^2}-\frac {2 a (a+b) (a+2 b)}{b^2 \left (a+b \tanh ^2(c+d x)\right )}}{4 (a+b)^3 d} \]
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Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.17
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(127\) |
default | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {-\frac {a \left (a^{2}+3 a b +2 b^{2}\right )}{b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )}+\frac {a^{2} \left (a^{2}+2 a b +b^{2}\right )}{2 b^{2} \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}-\ln \left (a +b \tanh \left (d x +c \right )^{2}\right )}{2 \left (a +b \right )^{3}}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(127\) |
risch | \(-\frac {x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {2 c}{d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {4 \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right ) a \,{\mathrm e}^{2 d x +2 c}}{\left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} d \left (a +b \right )^{3}}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a -b \right ) {\mathrm e}^{2 d x +2 c}}{a +b}+1\right )}{2 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}\) | \(231\) |
parallelrisch | \(-\frac {4 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2} b^{2}+4 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{4} b^{4}-2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{4} b^{4}-4 a \,b^{3} \tanh \left (d x +c \right )^{2}-2 \tanh \left (d x +c \right )^{2} a^{3} b -2 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) a^{2} b^{2}+8 \ln \left (1-\tanh \left (d x +c \right )\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}+4 b^{4} \tanh \left (d x +c \right )^{4} x d -a^{4}+4 a^{2} b^{2} d x -3 a^{2} b^{2}-4 a^{3} b -6 a^{2} b^{2} \tanh \left (d x +c \right )^{2}+8 a \,b^{3} \tanh \left (d x +c \right )^{2} x d -4 \ln \left (a +b \tanh \left (d x +c \right )^{2}\right ) \tanh \left (d x +c \right )^{2} a \,b^{3}}{4 \left (a +b \tanh \left (d x +c \right )^{2}\right )^{2} \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) b^{2} d}\) | \(291\) |
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Leaf count of result is larger than twice the leaf count of optimal. 2584 vs. \(2 (103) = 206\).
Time = 0.30 (sec) , antiderivative size = 2584, normalized size of antiderivative = 23.71 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
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Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 376 vs. \(2 (103) = 206\).
Time = 0.24 (sec) , antiderivative size = 376, normalized size of antiderivative = 3.45 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {d x + c}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac {4 \, {\left ({\left (a^{2} + a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} - 2 \, a b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{2} + a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} + 7 \, a^{4} b + 6 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 7 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} + 3 \, a^{4} b + 2 \, a^{3} b^{2} - 2 \, a^{2} b^{3} - 3 \, a b^{4} - b^{5}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {\log \left (2 \, {\left (a - b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} + a + b\right )}{2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} \]
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Leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (103) = 206\).
Time = 0.47 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.25 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {\frac {2 \, \log \left ({\left | a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} - \frac {3 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} + 3 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )}^{2} - 4 \, a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 12 \, b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} - 4 \, a + 12 \, b}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (a {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + b {\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a - 2 \, b\right )}^{2}}}{4 \, d} \]
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Time = 0.74 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.82 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {a^4+a^3\,b\,\left (2\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\right )-a\,b^3\,\left (-4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+{\mathrm {tanh}\left (c+d\,x\right )}^2\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,8{}\mathrm {i}\right )+a^2\,b^2\,\left (6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+3-\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}\right )-b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4\,\mathrm {atan}\left (\frac {a\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{2\,a-a\,{\mathrm {tanh}\left (c+d\,x\right )}^2+b\,{\mathrm {tanh}\left (c+d\,x\right )}^2}\right )\,4{}\mathrm {i}}{4\,d\,a^5\,b^2+8\,d\,a^4\,b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^4\,b^3+4\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^3\,b^4\,{\mathrm {tanh}\left (c+d\,x\right )}^2+12\,d\,a^3\,b^4+12\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^4+24\,d\,a^2\,b^5\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,a^2\,b^5+12\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^4+8\,d\,a\,b^6\,{\mathrm {tanh}\left (c+d\,x\right )}^2+4\,d\,b^7\,{\mathrm {tanh}\left (c+d\,x\right )}^4} \]
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