Integrand size = 23, antiderivative size = 74 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=(a+b)^3 x-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}-\frac {a^3 \coth ^5(c+d x)}{5 d} \]
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Time = 0.07 (sec) , antiderivative size = 74, 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, 472, 213} \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a^3 \coth ^5(c+d x)}{5 d}-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}+x (a+b)^3 \]
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Rule 213
Rule 472
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^3}{x^6 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^3}{x^6}+\frac {a^2 (a+3 b)}{x^4}+\frac {a \left (a^2+3 a b+3 b^2\right )}{x^2}-\frac {(a+b)^3}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}-\frac {a^3 \coth ^5(c+d x)}{5 d}-\frac {(a+b)^3 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = (a+b)^3 x-\frac {a \left (a^2+3 a b+3 b^2\right ) \coth (c+d x)}{d}-\frac {a^2 (a+3 b) \coth ^3(c+d x)}{3 d}-\frac {a^3 \coth ^5(c+d x)}{5 d} \\ \end{align*}
Time = 1.62 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a \coth (c+d x) \left (15 \left (a^2+3 a b+3 b^2\right )+5 a (a+3 b) \coth ^2(c+d x)+3 a^2 \coth ^4(c+d x)\right )}{15 d}+\frac {(a+b)^3 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right ) \tanh (c+d x)}{d \sqrt {\tanh ^2(c+d x)}} \]
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Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.97
method | result | size |
parallelrisch | \(\frac {-3 \coth \left (d x +c \right )^{5} a^{3}-5 a^{2} \coth \left (d x +c \right )^{3} \left (a +3 b \right )+\left (-15 a^{3}-45 a^{2} b -45 a \,b^{2}\right ) \coth \left (d x +c \right )+15 d x \left (a +b \right )^{3}}{15 d}\) | \(72\) |
derivativedivides | \(-\frac {\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\frac {a^{3}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{\tanh \left (d x +c \right )}+\frac {a^{2} \left (a +3 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{3}-\frac {3}{2} a^{2} b -\frac {3}{2} a \,b^{2}-\frac {1}{2} b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{d}\) | \(127\) |
default | \(-\frac {\left (\frac {1}{2} a^{3}+\frac {3}{2} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {1}{2} b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\frac {a^{3}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a^{2}+3 a b +3 b^{2}\right )}{\tanh \left (d x +c \right )}+\frac {a^{2} \left (a +3 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{3}-\frac {3}{2} a^{2} b -\frac {3}{2} a \,b^{2}-\frac {1}{2} b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{d}\) | \(127\) |
risch | \(a^{3} x +3 b \,a^{2} x +3 a \,b^{2} x +b^{3} x -\frac {2 a \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+90 a b \,{\mathrm e}^{8 d x +8 c}+45 b^{2} {\mathrm e}^{8 d x +8 c}-90 a^{2} {\mathrm e}^{6 d x +6 c}-270 a b \,{\mathrm e}^{6 d x +6 c}-180 b^{2} {\mathrm e}^{6 d x +6 c}+140 a^{2} {\mathrm e}^{4 d x +4 c}+330 a b \,{\mathrm e}^{4 d x +4 c}+270 \,{\mathrm e}^{4 d x +4 c} b^{2}-70 a^{2} {\mathrm e}^{2 d x +2 c}-210 a b \,{\mathrm e}^{2 d x +2 c}-180 \,{\mathrm e}^{2 d x +2 c} b^{2}+23 a^{2}+60 a b +45 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) | \(224\) |
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Leaf count of result is larger than twice the leaf count of optimal. 557 vs. \(2 (70) = 140\).
Time = 0.26 (sec) , antiderivative size = 557, normalized size of antiderivative = 7.53 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {{\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{3} + 24 \, a^{2} b + 27 \, a b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{4} + 46 \, a^{3} + 120 \, a^{2} b + 90 \, a b^{2} + 30 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x - 3 \, {\left (23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
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Timed out. \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (70) = 140\).
Time = 0.21 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.23 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{15} \, a^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + a^{2} b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + b^{3} x \]
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Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (70) = 140\).
Time = 0.57 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.26 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} - 270 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 270 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 210 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 180 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{3} + 60 \, a^{2} b + 45 \, a b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]
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Time = 1.89 (sec) , antiderivative size = 568, normalized size of antiderivative = 7.68 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,{\left (a+b\right )}^3-\frac {\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{15\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}-\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^3+6\,a^2\,b+9\,a\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+\frac {\frac {6\,\left (a^2\,b+a\,b^2\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}-\frac {6\,\left (a^3+2\,a^2\,b+a\,b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
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