Integrand size = 23, antiderivative size = 63 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=(a+b)^2 x-\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {a (a+2 b) \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d} \] Output:
(a+b)^2*x-(a+b)^2*coth(d*x+c)/d-1/3*a*(a+2*b)*coth(d*x+c)^3/d-1/5*a^2*coth (d*x+c)^5/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {a^2 \coth ^5(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\tanh ^2(c+d x)\right )}{5 d}-\frac {2 a b \coth ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(c+d x)\right )}{3 d}-\frac {b^2 \coth (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\tanh ^2(c+d x)\right )}{d} \] Input:
Integrate[Coth[c + d*x]^6*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-1/5*(a^2*Coth[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, Tanh[c + d*x]^2 ])/d - (2*a*b*Coth[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[c + d* x]^2])/(3*d) - (b^2*Coth[c + d*x]*Hypergeometric2F1[-1/2, 1, 1/2, Tanh[c + d*x]^2])/d
Time = 0.49 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 25, 4153, 25, 364, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (a-b \tan (i c+i d x)^2\right )^2}{\tan (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {\int -\frac {\coth ^6(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\coth ^6(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 364 |
| \(\displaystyle \frac {\int \left (a^2 \coth ^6(c+d x)+a (a+2 b) \coth ^4(c+d x)+(a+b)^2 \coth ^2(c+d x)-\frac {(a+b)^2}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {1}{5} a^2 \coth ^5(c+d x)-(a+b)^2 \text {arctanh}(\tanh (c+d x))+\frac {1}{3} a (a+2 b) \coth ^3(c+d x)+(a+b)^2 \coth (c+d x)}{d}\) |
Input:
Int[Coth[c + d*x]^6*(a + b*Tanh[c + d*x]^2)^2,x]
Output:
-((-((a + b)^2*ArcTanh[Tanh[c + d*x]]) + (a + b)^2*Coth[c + d*x] + (a*(a + 2*b)*Coth[c + d*x]^3)/3 + (a^2*Coth[c + d*x]^5)/5)/d)
Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x ] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In tegerQ[m] || IGtQ[2*(m + 1), 0] || !RationalQ[m])
Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[c*(ff/f) Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio nalQ[n]))
Time = 0.16 (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 | \(-\frac {-\frac {-a^{2}-2 a b -b^{2}}{\tanh \left (d x +c \right )}+\frac {a^{2}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a +2 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(110\) |
| default | \(-\frac {-\frac {-a^{2}-2 a b -b^{2}}{\tanh \left (d x +c \right )}+\frac {a^{2}}{5 \tanh \left (d x +c \right )^{5}}+\frac {a \left (a +2 b \right )}{3 \tanh \left (d x +c \right )^{3}}+\left (-\frac {1}{2} a^{2}-a b -\frac {1}{2} b^{2}\right ) \ln \left (1+\tanh \left (d x +c \right )\right )+\left (\frac {1}{2} a^{2}+a b +\frac {1}{2} b^{2}\right ) \ln \left (-1+\tanh \left (d x +c \right )\right )}{d}\) | \(110\) |
| risch | \(a^{2} x +2 a b x +b^{2} x -\frac {2 \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+60 a b \,{\mathrm e}^{8 d x +8 c}+15 b^{2} {\mathrm e}^{8 d x +8 c}-90 a^{2} {\mathrm e}^{6 d x +6 c}-180 a b \,{\mathrm e}^{6 d x +6 c}-60 b^{2} {\mathrm e}^{6 d x +6 c}+140 a^{2} {\mathrm e}^{4 d x +4 c}+220 a b \,{\mathrm e}^{4 d x +4 c}+90 b^{2} {\mathrm e}^{4 d x +4 c}-70 a^{2} {\mathrm e}^{2 d x +2 c}-140 a b \,{\mathrm e}^{2 d x +2 c}-60 b^{2} {\mathrm e}^{2 d x +2 c}+23 a^{2}+40 a b +15 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) | \(214\) |
Input:
int(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/15*(-3*coth(d*x+c)^5*a^2-5*a*coth(d*x+c)^3*(a+2*b)-15*(a+b)^2*coth(d*x+c )+15*d*x*(a+b)^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (59) = 118\).
Time = 0.09 (sec) , antiderivative size = 473, normalized size of antiderivative = 7.51 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {{\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{5} - 5 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 2 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 3 \, {\left (5 \, a^{2} + 16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (5 \, a^{2} + 4 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x - 3 \, {\left (15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d x + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 46 \, a^{2} + 80 \, a b + 30 \, b^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \] Input:
integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
Output:
-1/15*((23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^5 + 5*(23*a^2 + 40*a*b + 1 5*b^2)*cosh(d*x + c)*sinh(d*x + c)^4 - (15*(a^2 + 2*a*b + b^2)*d*x + 23*a^ 2 + 40*a*b + 15*b^2)*sinh(d*x + c)^5 - 5*(5*a^2 + 16*a*b + 9*b^2)*cosh(d*x + c)^3 + 5*(15*(a^2 + 2*a*b + b^2)*d*x - 2*(15*(a^2 + 2*a*b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^2 + 23*a^2 + 40*a*b + 15*b^2)*sinh (d*x + c)^3 + 5*(2*(23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^3 - 3*(5*a^2 + 16*a*b + 9*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^2 + 4*a*b + 3*b^ 2)*cosh(d*x + c) - 5*((15*(a^2 + 2*a*b + b^2)*d*x + 23*a^2 + 40*a*b + 15*b ^2)*cosh(d*x + c)^4 + 30*(a^2 + 2*a*b + b^2)*d*x - 3*(15*(a^2 + 2*a*b + b^ 2)*d*x + 23*a^2 + 40*a*b + 15*b^2)*cosh(d*x + c)^2 + 46*a^2 + 80*a*b + 30* b^2)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh( d*x + c)^3 + 5*(d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))
\[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \coth ^{6}{\left (c + d x \right )}\, dx \] Input:
integrate(coth(d*x+c)**6*(a+b*tanh(d*x+c)**2)**2,x)
Output:
Integral((a + b*tanh(c + d*x)**2)**2*coth(c + d*x)**6, x)
Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (59) = 118\).
Time = 0.05 (sec) , antiderivative size = 231, normalized size of antiderivative = 3.67 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {1}{15} \, a^{2} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} - 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} - 45 \, e^{\left (-8 \, d x - 8 \, c\right )} - 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {2}{3} \, a b {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} + b^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} \] Input:
integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
Output:
1/15*a^2*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) - 140*e^(-4*d*x - 4*c) + 90*e^(-6*d*x - 6*c) - 45*e^(-8*d*x - 8*c) - 23)/(d*(5*e^(-2*d*x - 2*c) - 1 0*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))) + 2/3*a*b*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d *x - 4*c) - 2)/(d*(3*e^(-2*d*x - 2*c) - 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6 *c) - 1))) + b^2*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1)))
Leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (59) = 118\).
Time = 0.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 3.46 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (45 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 90 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} - 180 \, a b e^{\left (6 \, d x + 6 \, c\right )} - 60 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 140 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 220 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 70 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 140 \, a b e^{\left (2 \, d x + 2 \, c\right )} - 60 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{2} + 40 \, a b + 15 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{15 \, d} \] Input:
integrate(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/15*(15*(a^2 + 2*a*b + b^2)*(d*x + c) - 2*(45*a^2*e^(8*d*x + 8*c) + 60*a* b*e^(8*d*x + 8*c) + 15*b^2*e^(8*d*x + 8*c) - 90*a^2*e^(6*d*x + 6*c) - 180* a*b*e^(6*d*x + 6*c) - 60*b^2*e^(6*d*x + 6*c) + 140*a^2*e^(4*d*x + 4*c) + 2 20*a*b*e^(4*d*x + 4*c) + 90*b^2*e^(4*d*x + 4*c) - 70*a^2*e^(2*d*x + 2*c) - 140*a*b*e^(2*d*x + 2*c) - 60*b^2*e^(2*d*x + 2*c) + 23*a^2 + 40*a*b + 15*b ^2)/(e^(2*d*x + 2*c) - 1)^5)/d
Time = 0.19 (sec) , antiderivative size = 529, normalized size of antiderivative = 8.40 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1}+\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}+x\,{\left (a+b\right )}^2-\frac {\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}-\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {\frac {2\,\left (5\,a^2+4\,a\,b+3\,b^2\right )}{15\,d}-\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {2\,\left (3\,a^2+4\,a\,b+b^2\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \] Input:
int(coth(c + d*x)^6*(a + b*tanh(c + d*x)^2)^2,x)
Output:
((2*(2*a*b + b^2))/(5*d) - (2*exp(2*c + 2*d*x)*(4*a*b + 3*a^2 + b^2))/(5*d ))/(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1) + ((2*(2*a*b + b^2))/(5*d) + (6*exp(4*c + 4*d*x)*(2*a*b + b^2))/(5*d) - (2*exp(6*c + 6*d*x)*(4*a*b + 3*a^2 + b^2))/(5*d) - (2*exp(2*c + 2*d*x)*(4*a*b + 5*a^2 + 3*b^2))/(5*d))/ (6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) + x*(a + b)^2 - ((2*(4*a*b + 3*a^2 + b^2))/(5*d) - (8*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) - (8*exp(6*c + 6*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(8*c + 8*d*x)*(4*a*b + 3*a^2 + b^2))/(5*d) + (4*exp(4*c + 4*d*x)*(4 *a*b + 5*a^2 + 3*b^2))/(5*d))/(5*exp(2*c + 2*d*x) - 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6*d*x) - 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) - 1) - ((2*( 4*a*b + 5*a^2 + 3*b^2))/(15*d) - (4*exp(2*c + 2*d*x)*(2*a*b + b^2))/(5*d) + (2*exp(4*c + 4*d*x)*(4*a*b + 3*a^2 + b^2))/(5*d))/(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (2*(4*a*b + 3*a^2 + b^2))/(5* d*(exp(2*c + 2*d*x) - 1))
Time = 0.57 (sec) , antiderivative size = 513, normalized size of antiderivative = 8.14 \[ \int \coth ^6(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {-18 e^{10 d x +10 c} a^{2}-6 e^{10 d x +10 c} b^{2}+75 e^{2 d x +2 c} a^{2} d x +15 e^{10 d x +10 c} a^{2} d x +15 e^{10 d x +10 c} b^{2} d x -75 e^{8 d x +8 c} a^{2} d x -75 e^{8 d x +8 c} b^{2} d x +150 e^{6 d x +6 c} a^{2} d x +150 e^{6 d x +6 c} b^{2} d x -150 e^{4 d x +4 c} a^{2} d x -150 e^{4 d x +4 c} b^{2} d x -28 a^{2}+150 e^{2 d x +2 c} a b d x +30 e^{10 d x +10 c} a b d x -150 e^{8 d x +8 c} a b d x +300 e^{6 d x +6 c} a b d x -300 e^{4 d x +4 c} a b d x +90 e^{2 d x +2 c} b^{2}-30 a b d x -15 a^{2} d x -15 b^{2} d x +75 e^{2 d x +2 c} b^{2} d x -24 b^{2}-24 e^{10 d x +10 c} a b +120 e^{6 d x +6 c} a b -200 e^{4 d x +4 c} a b +160 e^{2 d x +2 c} a b -56 a b +60 e^{6 d x +6 c} b^{2}-100 e^{4 d x +4 c} a^{2}-120 e^{4 d x +4 c} b^{2}+50 e^{2 d x +2 c} a^{2}}{15 d \left (e^{10 d x +10 c}-5 e^{8 d x +8 c}+10 e^{6 d x +6 c}-10 e^{4 d x +4 c}+5 e^{2 d x +2 c}-1\right )} \] Input:
int(coth(d*x+c)^6*(a+b*tanh(d*x+c)^2)^2,x)
Output:
(15*e**(10*c + 10*d*x)*a**2*d*x - 18*e**(10*c + 10*d*x)*a**2 + 30*e**(10*c + 10*d*x)*a*b*d*x - 24*e**(10*c + 10*d*x)*a*b + 15*e**(10*c + 10*d*x)*b** 2*d*x - 6*e**(10*c + 10*d*x)*b**2 - 75*e**(8*c + 8*d*x)*a**2*d*x - 150*e** (8*c + 8*d*x)*a*b*d*x - 75*e**(8*c + 8*d*x)*b**2*d*x + 150*e**(6*c + 6*d*x )*a**2*d*x + 300*e**(6*c + 6*d*x)*a*b*d*x + 120*e**(6*c + 6*d*x)*a*b + 150 *e**(6*c + 6*d*x)*b**2*d*x + 60*e**(6*c + 6*d*x)*b**2 - 150*e**(4*c + 4*d* x)*a**2*d*x - 100*e**(4*c + 4*d*x)*a**2 - 300*e**(4*c + 4*d*x)*a*b*d*x - 2 00*e**(4*c + 4*d*x)*a*b - 150*e**(4*c + 4*d*x)*b**2*d*x - 120*e**(4*c + 4* d*x)*b**2 + 75*e**(2*c + 2*d*x)*a**2*d*x + 50*e**(2*c + 2*d*x)*a**2 + 150* e**(2*c + 2*d*x)*a*b*d*x + 160*e**(2*c + 2*d*x)*a*b + 75*e**(2*c + 2*d*x)* b**2*d*x + 90*e**(2*c + 2*d*x)*b**2 - 15*a**2*d*x - 28*a**2 - 30*a*b*d*x - 56*a*b - 15*b**2*d*x - 24*b**2)/(15*d*(e**(10*c + 10*d*x) - 5*e**(8*c + 8 *d*x) + 10*e**(6*c + 6*d*x) - 10*e**(4*c + 4*d*x) + 5*e**(2*c + 2*d*x) - 1 ))