Integrand size = 23, antiderivative size = 63 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d} \]
[Out]
Time = 0.07 (sec) , antiderivative size = 63, 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} \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth ^5(c+d x)}{5 d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {(a+b)^2 \coth (c+d x)}{d}+x (a+b)^2 \]
[In]
[Out]
Rule 213
Rule 472
Rule 3751
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+b x^2\right )^2}{x^6 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2}{x^6}+\frac {a (a+2 b)}{x^4}+\frac {(a+b)^2}{x^2}-\frac {(a+b)^2}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{d} \\ & = -\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}-\frac {(a+b)^2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = (a+b)^2 x-\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\tanh ^2(c+d x)\right )}{5 d}-\frac {2 a b \coth ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(c+d x)\right )}{3 d}-\frac {b^2 \coth (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\tanh ^2(c+d x)\right )}{d} \]
[In]
[Out]
Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92
method | result | size |
parallelrisch | \(\frac {-3 \coth \left (d x +c \right )^{5} a^{2}-5 a \coth \left (d x +c \right )^{3} \left (a +2 b \right )-15 \left (a +b \right )^{2} \coth \left (d x +c \right )+15 d x \left (a +b \right )^{2}}{15 d}\) | \(58\) |
derivativedivides | \(-\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )-\frac {-a^{2}-2 a b -b^{2}}{\tanh \left (d x +c \right )}+\frac {a^{2}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a +2 b \right )}{3 \tanh \left (d x +c \right )^{3}}}{d}\) | \(110\) |
default | \(-\frac {\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )-\frac {-a^{2}-2 a b -b^{2}}{\tanh \left (d x +c \right )}+\frac {a^{2}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a +2 b \right )}{3 \tanh \left (d x +c \right )^{3}}}{d}\) | \(110\) |
risch | \(a^{2} x +2 a b x +b^{2} x -\frac {2 \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+60 a b \,{\mathrm e}^{8 d x +8 c}+15 b^{2} {\mathrm e}^{8 d x +8 c}-90 a^{2} {\mathrm e}^{6 d x +6 c}-180 a b \,{\mathrm e}^{6 d x +6 c}-60 b^{2} {\mathrm e}^{6 d x +6 c}+140 a^{2} {\mathrm e}^{4 d x +4 c}+220 a b \,{\mathrm e}^{4 d x +4 c}+90 \,{\mathrm e}^{4 d x +4 c} b^{2}-70 a^{2} {\mathrm e}^{2 d x +2 c}-140 a b \,{\mathrm e}^{2 d x +2 c}-60 \,{\mathrm e}^{2 d x +2 c} b^{2}+23 a^{2}+40 a b +15 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) | \(214\) |
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (59) = 118\).
Time = 0.26 (sec) , antiderivative size = 473, normalized size of antiderivative = 7.51 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {{\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 2 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 3 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 46 \, a^{2} + 80 \, a b + 30 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
[In]
[Out]
\[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{6}{\left (c + d x \right )}\, dx \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (59) = 118\).
Time = 0.19 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.67 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{15} \, a^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {2}{3} \, a b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (59) = 118\).
Time = 0.42 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.46 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 60 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 60 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \]
[In]
[Out]
Time = 0.18 (sec) , antiderivative size = 529, normalized size of antiderivative = 8.40 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+x\,{\left (a+b\right )}^2-\frac {\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{15\,d}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
[In]
[Out]