Integrand size = 23, antiderivative size = 144 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {x}{(a+b)^3}-\frac {\sqrt {a} \left (3 a^2+10 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{8 b^{5/2} (a+b)^3 d}+\frac {a \tanh ^3(c+d x)}{4 b (a+b) d \left (a+b \tanh ^2(c+d x)\right )^2}+\frac {a (3 a+7 b) \tanh (c+d x)}{8 b^2 (a+b)^2 d \left (a+b \tanh ^2(c+d x)\right )} \]
x/(a+b)^3-1/8*(3*a^2+10*a*b+15*b^2)*arctan(b^(1/2)*tanh(d*x+c)/a^(1/2))*a^ (1/2)/b^(5/2)/(a+b)^3/d+1/4*a*tanh(d*x+c)^3/b/(a+b)/d/(a+b*tanh(d*x+c)^2)^ 2+1/8*a*(3*a+7*b)*tanh(d*x+c)/b^2/(a+b)^2/d/(a+b*tanh(d*x+c)^2)
Time = 1.34 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\frac {8 (c+d x)-\frac {\sqrt {a} \left (3 a^2+10 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{b^{5/2}}-\frac {4 a^2 (a+b) \sinh (2 (c+d x))}{b (a-b+(a+b) \cosh (2 (c+d x)))^2}+\frac {3 a (a+b) (a+3 b) \sinh (2 (c+d x))}{b^2 (a-b+(a+b) \cosh (2 (c+d x)))}}{8 (a+b)^3 d} \]
(8*(c + d*x) - (Sqrt[a]*(3*a^2 + 10*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/b^(5/2) - (4*a^2*(a + b)*Sinh[2*(c + d*x)])/(b*(a - b + ( a + b)*Cosh[2*(c + d*x)])^2) + (3*a*(a + b)*(a + 3*b)*Sinh[2*(c + d*x)])/( b^2*(a - b + (a + b)*Cosh[2*(c + d*x)])))/(8*(a + b)^3*d)
Time = 0.41 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.20, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 25, 4153, 25, 372, 440, 397, 218, 219}
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.
| \(\displaystyle \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\tan (i c+i d x)^6}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\tan (i c+i d x)^6}{\left (a-b \tan (i c+i d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {\int -\frac {\tanh ^6(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\tanh ^6(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle -\frac {\frac {\int \frac {\tanh ^2(c+d x) \left (3 a-(3 a+4 b) \tanh ^2(c+d x)\right )}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )^2}d\tanh (c+d x)}{4 b (a+b)}-\frac {a \tanh ^3(c+d x)}{4 b (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle -\frac {\frac {\frac {\int \frac {a (3 a+7 b)-\left (3 a^2+7 b a+8 b^2\right ) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (b \tanh ^2(c+d x)+a\right )}d\tanh (c+d x)}{2 b (a+b)}-\frac {a (3 a+7 b) \tanh (c+d x)}{2 b (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{4 b (a+b)}-\frac {a \tanh ^3(c+d x)}{4 b (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle -\frac {\frac {\frac {\frac {a \left (3 a^2+10 a b+15 b^2\right ) \int \frac {1}{b \tanh ^2(c+d x)+a}d\tanh (c+d x)}{a+b}-\frac {8 b^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}}{2 b (a+b)}-\frac {a (3 a+7 b) \tanh (c+d x)}{2 b (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{4 b (a+b)}-\frac {a \tanh ^3(c+d x)}{4 b (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\sqrt {a} \left (3 a^2+10 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {b} (a+b)}-\frac {8 b^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a+b}}{2 b (a+b)}-\frac {a (3 a+7 b) \tanh (c+d x)}{2 b (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{4 b (a+b)}-\frac {a \tanh ^3(c+d x)}{4 b (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\frac {\frac {\frac {\sqrt {a} \left (3 a^2+10 a b+15 b^2\right ) \arctan \left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a}}\right )}{\sqrt {b} (a+b)}-\frac {8 b^2 \text {arctanh}(\tanh (c+d x))}{a+b}}{2 b (a+b)}-\frac {a (3 a+7 b) \tanh (c+d x)}{2 b (a+b) \left (a+b \tanh ^2(c+d x)\right )}}{4 b (a+b)}-\frac {a \tanh ^3(c+d x)}{4 b (a+b) \left (a+b \tanh ^2(c+d x)\right )^2}}{d}\) |
-((-1/4*(a*Tanh[c + d*x]^3)/(b*(a + b)*(a + b*Tanh[c + d*x]^2)^2) + (((Sqr t[a]*(3*a^2 + 10*a*b + 15*b^2)*ArcTan[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a]])/(S qrt[b]*(a + b)) - (8*b^2*ArcTanh[Tanh[c + d*x]])/(a + b))/(2*b*(a + b)) - (a*(3*a + 7*b)*Tanh[c + d*x])/(2*b*(a + b)*(a + b*Tanh[c + d*x]^2)))/(4*b* (a + b)))/d)
3.2.90.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 )^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 )) Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 3] && IntBinomialQ[a , b, c, d, e, m, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
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 = 0.30 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {a \left (\frac {-\frac {\left (5 a^{2}+14 a b +9 b^{2}\right ) \tanh \left (d x +c \right )^{3}}{8 b}-\frac {a \left (3 a^{2}+10 a b +7 b^{2}\right ) \tanh \left (d x +c \right )}{8 b^{2}}}{\left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+10 a b +15 b^{2}\right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a +b \right )^{3}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(157\) |
| default | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2 \left (a +b \right )^{3}}-\frac {a \left (\frac {-\frac {\left (5 a^{2}+14 a b +9 b^{2}\right ) \tanh \left (d x +c \right )^{3}}{8 b}-\frac {a \left (3 a^{2}+10 a b +7 b^{2}\right ) \tanh \left (d x +c \right )}{8 b^{2}}}{\left (a +b \tanh \left (d x +c \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+10 a b +15 b^{2}\right ) \arctan \left (\frac {b \tanh \left (d x +c \right )}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a +b \right )^{3}}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2 \left (a +b \right )^{3}}}{d}\) | \(157\) |
| risch | \(\frac {x}{a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}}-\frac {a \left (3 a^{3} {\mathrm e}^{6 d x +6 c}+13 a^{2} b \,{\mathrm e}^{6 d x +6 c}+a \,b^{2} {\mathrm e}^{6 d x +6 c}-9 \,{\mathrm e}^{6 d x +6 c} b^{3}+9 a^{3} {\mathrm e}^{4 d x +4 c}+21 a^{2} b \,{\mathrm e}^{4 d x +4 c}-9 a \,b^{2} {\mathrm e}^{4 d x +4 c}+27 \,{\mathrm e}^{4 d x +4 c} b^{3}+9 a^{3} {\mathrm e}^{2 d x +2 c}+23 a^{2} b \,{\mathrm e}^{2 d x +2 c}-13 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}-27 \,{\mathrm e}^{2 d x +2 c} b^{3}+3 a^{3}+15 a^{2} b +21 a \,b^{2}+9 b^{3}\right )}{4 \left (a \,{\mathrm e}^{4 d x +4 c}+b \,{\mathrm e}^{4 d x +4 c}+2 \,{\mathrm e}^{2 d x +2 c} a -2 b \,{\mathrm e}^{2 d x +2 c}+a +b \right )^{2} \left (a +b \right ) \left (a^{2}+2 a b +b^{2}\right ) b^{2} d}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) a^{2}}{16 b^{3} \left (a +b \right )^{3} d}+\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right ) a}{8 b^{2} \left (a +b \right )^{3} d}+\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}-a +b}{a +b}\right )}{16 b \left (a +b \right )^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) a^{2}}{16 b^{3} \left (a +b \right )^{3} d}-\frac {5 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right ) a}{8 b^{2} \left (a +b \right )^{3} d}-\frac {15 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}+a -b}{a +b}\right )}{16 b \left (a +b \right )^{3} d}\) | \(604\) |
1/d*(-1/2/(a+b)^3*ln(tanh(d*x+c)-1)-a/(a+b)^3*((-1/8*(5*a^2+14*a*b+9*b^2)/ b*tanh(d*x+c)^3-1/8*a*(3*a^2+10*a*b+7*b^2)/b^2*tanh(d*x+c))/(a+b*tanh(d*x+ c)^2)^2+1/8*(3*a^2+10*a*b+15*b^2)/b^2/(a*b)^(1/2)*arctan(b*tanh(d*x+c)/(a* b)^(1/2)))+1/2/(a+b)^3*ln(tanh(d*x+c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 3603 vs. \(2 (130) = 260\).
Time = 0.40 (sec) , antiderivative size = 7528, normalized size of antiderivative = 52.28 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 3354 vs. \(2 (130) = 260\).
Time = 1.24 (sec) , antiderivative size = 3354, normalized size of antiderivative = 23.29 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
-1/512*(3*a^5 + 25*a^4*b + 150*a^3*b^2 - 150*a^2*b^3 - 25*a*b^4 - 3*b^5)*a rctan(1/2*((a + b)*e^(2*d*x + 2*c) + a - b)/sqrt(a*b))/((a^5*b^2 + 3*a^4*b ^3 + 3*a^3*b^4 + a^2*b^5)*sqrt(a*b)*d) + 1/512*(3*a^5 + 25*a^4*b + 150*a^3 *b^2 - 150*a^2*b^3 - 25*a*b^4 - 3*b^5)*arctan(1/2*((a + b)*e^(-2*d*x - 2*c ) + a - b)/sqrt(a*b))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*sqrt(a* b)*d) - 1/256*(3*a^6 + 30*a^5*b - 99*a^4*b^2 - 252*a^3*b^3 - 99*a^2*b^4 + 30*a*b^5 + 3*b^6 + (3*a^6 + 28*a^5*b - 465*a^4*b^2 + 465*a^2*b^4 - 28*a*b^ 5 - 3*b^6)*e^(6*d*x + 6*c) + (9*a^6 + 66*a^5*b - 905*a^4*b^2 + 1148*a^3*b^ 3 - 905*a^2*b^4 + 66*a*b^5 + 9*b^6)*e^(4*d*x + 4*c) + (9*a^6 + 68*a^5*b - 659*a^4*b^2 + 659*a^2*b^4 - 68*a*b^5 - 9*b^6)*e^(2*d*x + 2*c))/((a^7*b^2 + 5*a^6*b^3 + 10*a^5*b^4 + 10*a^4*b^5 + 5*a^3*b^6 + a^2*b^7 + (a^7*b^2 + 5* a^6*b^3 + 10*a^5*b^4 + 10*a^4*b^5 + 5*a^3*b^6 + a^2*b^7)*e^(8*d*x + 8*c) + 4*(a^7*b^2 + 3*a^6*b^3 + 2*a^5*b^4 - 2*a^4*b^5 - 3*a^3*b^6 - a^2*b^7)*e^( 6*d*x + 6*c) + 2*(3*a^7*b^2 + 7*a^6*b^3 + 6*a^5*b^4 + 6*a^4*b^5 + 7*a^3*b^ 6 + 3*a^2*b^7)*e^(4*d*x + 4*c) + 4*(a^7*b^2 + 3*a^6*b^3 + 2*a^5*b^4 - 2*a^ 4*b^5 - 3*a^3*b^6 - a^2*b^7)*e^(2*d*x + 2*c))*d) + 1/256*(3*a^6 + 30*a^5*b - 99*a^4*b^2 - 252*a^3*b^3 - 99*a^2*b^4 + 30*a*b^5 + 3*b^6 + (9*a^6 + 68* a^5*b - 659*a^4*b^2 + 659*a^2*b^4 - 68*a*b^5 - 9*b^6)*e^(-2*d*x - 2*c) + ( 9*a^6 + 66*a^5*b - 905*a^4*b^2 + 1148*a^3*b^3 - 905*a^2*b^4 + 66*a*b^5 + 9 *b^6)*e^(-4*d*x - 4*c) + (3*a^6 + 28*a^5*b - 465*a^4*b^2 + 465*a^2*b^4 ...
Leaf count of result is larger than twice the leaf count of optimal. 408 vs. \(2 (130) = 260\).
Time = 0.52 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.83 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (3 \, a^{3} + 10 \, a^{2} b + 15 \, a b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + b e^{\left (2 \, d x + 2 \, c\right )} + a - b}{2 \, \sqrt {a b}}\right )}{{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \sqrt {a b}} - \frac {8 \, {\left (d x + c\right )}}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (3 \, a^{4} e^{\left (6 \, d x + 6 \, c\right )} + 13 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} + a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 9 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 9 \, a^{4} e^{\left (4 \, d x + 4 \, c\right )} + 21 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} - 9 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 27 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, a^{4} e^{\left (2 \, d x + 2 \, c\right )} + 23 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} - 13 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 27 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a^{4} + 15 \, a^{3} b + 21 \, a^{2} b^{2} + 9 \, a b^{3}\right )}}{{\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + b e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b e^{\left (2 \, d x + 2 \, c\right )} + a + b\right )}^{2}}}{8 \, d} \]
-1/8*((3*a^3 + 10*a^2*b + 15*a*b^2)*arctan(1/2*(a*e^(2*d*x + 2*c) + b*e^(2 *d*x + 2*c) + a - b)/sqrt(a*b))/((a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*sqr t(a*b)) - 8*(d*x + c)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3) + 2*(3*a^4*e^(6*d*x + 6*c) + 13*a^3*b*e^(6*d*x + 6*c) + a^2*b^2*e^(6*d*x + 6*c) - 9*a*b^3*e^(6 *d*x + 6*c) + 9*a^4*e^(4*d*x + 4*c) + 21*a^3*b*e^(4*d*x + 4*c) - 9*a^2*b^2 *e^(4*d*x + 4*c) + 27*a*b^3*e^(4*d*x + 4*c) + 9*a^4*e^(2*d*x + 2*c) + 23*a ^3*b*e^(2*d*x + 2*c) - 13*a^2*b^2*e^(2*d*x + 2*c) - 27*a*b^3*e^(2*d*x + 2* c) + 3*a^4 + 15*a^3*b + 21*a^2*b^2 + 9*a*b^3)/((a^3*b^2 + 3*a^2*b^3 + 3*a* b^4 + b^5)*(a*e^(4*d*x + 4*c) + b*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) - 2*b*e^(2*d*x + 2*c) + a + b)^2))/d
Time = 2.55 (sec) , antiderivative size = 2669, normalized size of antiderivative = 18.53 \[ \int \frac {\tanh ^6(c+d x)}{\left (a+b \tanh ^2(c+d x)\right )^3} \, dx=\text {Too large to display} \]
log(tanh(c + d*x) + 1)/(2*a^3*d + 2*b^3*d + 6*a*b^2*d + 6*a^2*b*d) + ((tan h(c + d*x)^3*(9*a*b + 5*a^2))/(8*b*(2*a*b + a^2 + b^2)) + (a*tanh(c + d*x) *(7*a*b + 3*a^2))/(8*b^2*(2*a*b + a^2 + b^2)))/(a^2*d + b^2*d*tanh(c + d*x )^4 + 2*a*b*d*tanh(c + d*x)^2) - log(tanh(c + d*x) - 1)/(2*d*(a + b)^3) - (atan((((-a*b^5)^(1/2)*((tanh(c + d*x)*(60*a^5*b + 9*a^6 + 64*b^6 + 225*a^ 2*b^4 + 300*a^3*b^3 + 190*a^4*b^2))/(32*(b^7*d^2 + 4*a*b^6*d^2 + 6*a^2*b^5 *d^2 + 4*a^3*b^4*d^2 + a^4*b^3*d^2)) + (((224*a*b^10*d^2 + 1440*a^2*b^9*d^ 2 + 3936*a^3*b^8*d^2 + 5920*a^4*b^7*d^2 + 5280*a^5*b^6*d^2 + 2784*a^6*b^5* d^2 + 800*a^7*b^4*d^2 + 96*a^8*b^3*d^2)/(64*(b^9*d^3 + 6*a*b^8*d^3 + 15*a^ 2*b^7*d^3 + 20*a^3*b^6*d^3 + 15*a^4*b^5*d^3 + 6*a^5*b^4*d^3 + a^6*b^3*d^3) ) - (tanh(c + d*x)*(-a*b^5)^(1/2)*(10*a*b + 3*a^2 + 15*b^2)*(256*b^12*d^2 + 1280*a*b^11*d^2 + 2304*a^2*b^10*d^2 + 1280*a^3*b^9*d^2 - 1280*a^4*b^8*d^ 2 - 2304*a^5*b^7*d^2 - 1280*a^6*b^6*d^2 - 256*a^7*b^5*d^2))/(512*(b^8*d + 3*a^2*b^6*d + a^3*b^5*d + 3*a*b^7*d)*(b^7*d^2 + 4*a*b^6*d^2 + 6*a^2*b^5*d^ 2 + 4*a^3*b^4*d^2 + a^4*b^3*d^2)))*(-a*b^5)^(1/2)*(10*a*b + 3*a^2 + 15*b^2 ))/(16*(b^8*d + 3*a^2*b^6*d + a^3*b^5*d + 3*a*b^7*d)))*(10*a*b + 3*a^2 + 1 5*b^2)*1i)/(16*(b^8*d + 3*a^2*b^6*d + a^3*b^5*d + 3*a*b^7*d)) + ((-a*b^5)^ (1/2)*((tanh(c + d*x)*(60*a^5*b + 9*a^6 + 64*b^6 + 225*a^2*b^4 + 300*a^3*b ^3 + 190*a^4*b^2))/(32*(b^7*d^2 + 4*a*b^6*d^2 + 6*a^2*b^5*d^2 + 4*a^3*b^4* d^2 + a^4*b^3*d^2)) - (((224*a*b^10*d^2 + 1440*a^2*b^9*d^2 + 3936*a^3*b...