Integrand size = 21, antiderivative size = 83 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {b (a+b)^2 \tanh ^2(c+d x)}{2 d}-\frac {(a+b) \left (a+b \tanh ^2(c+d x)\right )^2}{4 d}-\frac {\left (a+b \tanh ^2(c+d x)\right )^3}{6 d} \]
(a+b)^3*ln(cosh(d*x+c))/d-1/2*b*(a+b)^2*tanh(d*x+c)^2/d-1/4*(a+b)*(a+b*tan h(d*x+c)^2)^2/d-1/6*(a+b*tanh(d*x+c)^2)^3/d
Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {-2 (a+b)^3 \log (\cosh (c+d x))+b (a+b)^2 \tanh ^2(c+d x)+\frac {1}{2} (a+b) \left (a+b \tanh ^2(c+d x)\right )^2+\frac {1}{3} \left (a+b \tanh ^2(c+d x)\right )^3}{2 d} \]
-1/2*(-2*(a + b)^3*Log[Cosh[c + d*x]] + b*(a + b)^2*Tanh[c + d*x]^2 + ((a + b)*(a + b*Tanh[c + d*x]^2)^2)/2 + (a + b*Tanh[c + d*x]^2)^3/3)/d
Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 26, 4153, 26, 353, 49, 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.
| \(\displaystyle \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -i \tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle -i \int \tan (i c+i d x) \left (a-b \tan (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle -\frac {i \int \frac {i \tanh (c+d x) \left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle \frac {\int \frac {\tanh (c+d x) \left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 353 |
| \(\displaystyle \frac {\int \frac {\left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {\int \left (\frac {(a+b)^3}{1-\tanh ^2(c+d x)}-b (a+b)^2-b \left (b \tanh ^2(c+d x)+a\right ) (a+b)-b \left (b \tanh ^2(c+d x)+a\right )^2\right )d\tanh ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-b (a+b)^2 \tanh ^2(c+d x)-\frac {1}{2} (a+b) \left (a+b \tanh ^2(c+d x)\right )^2-\frac {1}{3} \left (a+b \tanh ^2(c+d x)\right )^3+(a+b)^3 \left (-\log \left (1-\tanh ^2(c+d x)\right )\right )}{2 d}\) |
(-((a + b)^3*Log[1 - Tanh[c + d*x]^2]) - b*(a + b)^2*Tanh[c + d*x]^2 - ((a + b)*(a + b*Tanh[c + d*x]^2)^2)/2 - (a + b*Tanh[c + d*x]^2)^3/3)/(2*d)
3.2.59.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[1/2 Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ {a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
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 = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82
| method | result | size |
| derivativedivides | \(\frac {-\frac {3 \tanh \left (d x +c \right )^{4} a \,b^{2}}{4}-\frac {3 \tanh \left (d x +c \right )^{2} a^{2} b}{2}-\frac {3 a \,b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{4}}{4}-\frac {b^{3} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{3}}{6}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{d}\) | \(151\) |
| default | \(\frac {-\frac {3 \tanh \left (d x +c \right )^{4} a \,b^{2}}{4}-\frac {3 \tanh \left (d x +c \right )^{2} a^{2} b}{2}-\frac {3 a \,b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{4}}{4}-\frac {b^{3} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{3}}{6}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{d}\) | \(151\) |
| parts | \(\frac {a^{3} \ln \left (\cosh \left (d x +c \right )\right )}{d}+\frac {b^{3} \left (-\frac {\tanh \left (d x +c \right )^{6}}{6}-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}\) | \(170\) |
| parallelrisch | \(-\frac {2 \tanh \left (d x +c \right )^{6} b^{3}+9 \tanh \left (d x +c \right )^{4} a \,b^{2}+3 b^{3} \tanh \left (d x +c \right )^{4}+12 a^{3} d x +36 a^{2} b d x +36 a \,b^{2} d x +12 b^{3} d x +18 \tanh \left (d x +c \right )^{2} a^{2} b +18 a \,b^{2} \tanh \left (d x +c \right )^{2}+6 b^{3} \tanh \left (d x +c \right )^{2}+12 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{3}+36 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2} b +36 \ln \left (1-\tanh \left (d x +c \right )\right ) a \,b^{2}+12 \ln \left (1-\tanh \left (d x +c \right )\right ) b^{3}}{12 d}\) | \(184\) |
| risch | \(-a^{3} x -3 b \,a^{2} x -3 a \,b^{2} x -b^{3} x -\frac {2 a^{3} c}{d}-\frac {6 b c \,a^{2}}{d}-\frac {6 a \,b^{2} c}{d}-\frac {2 b^{3} c}{d}+\frac {2 b \,{\mathrm e}^{2 d x +2 c} \left (9 a^{2} {\mathrm e}^{8 d x +8 c}+18 a b \,{\mathrm e}^{8 d x +8 c}+9 b^{2} {\mathrm e}^{8 d x +8 c}+36 a^{2} {\mathrm e}^{6 d x +6 c}+54 a b \,{\mathrm e}^{6 d x +6 c}+18 b^{2} {\mathrm e}^{6 d x +6 c}+54 a^{2} {\mathrm e}^{4 d x +4 c}+72 a b \,{\mathrm e}^{4 d x +4 c}+34 \,{\mathrm e}^{4 d x +4 c} b^{2}+36 a^{2} {\mathrm e}^{2 d x +2 c}+54 a b \,{\mathrm e}^{2 d x +2 c}+18 \,{\mathrm e}^{2 d x +2 c} b^{2}+9 a^{2}+18 a b +9 b^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{6}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{3}}{d}+\frac {3 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a \,b^{2}}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) | \(353\) |
1/d*(-3/4*tanh(d*x+c)^4*a*b^2-3/2*tanh(d*x+c)^2*a^2*b-3/2*a*b^2*tanh(d*x+c )^2-1/4*b^3*tanh(d*x+c)^4-1/2*b^3*tanh(d*x+c)^2-1/6*tanh(d*x+c)^6*b^3+1/2* (-a^3-3*a^2*b-3*a*b^2-b^3)*ln(tanh(d*x+c)+1)-1/2*(a^3+3*a^2*b+3*a*b^2+b^3) *ln(tanh(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 4298 vs. \(2 (77) = 154\).
Time = 0.30 (sec) , antiderivative size = 4298, normalized size of antiderivative = 51.78 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]
-1/3*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^12 + 36*(a^3 + 3 *a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)*sinh(d*x + c)^11 + 3*(a^3 + 3*a^ 2*b + 3*a*b^2 + b^3)*d*x*sinh(d*x + c)^12 - 18*(a^2*b + 2*a*b^2 + b^3 - (a ^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^10 + 18*(11*(a^3 + 3*a^2* b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^2 - a^2*b - 2*a*b^2 - b^3 + (a^3 + 3* a^2*b + 3*a*b^2 + b^3)*d*x)*sinh(d*x + c)^10 + 60*(11*(a^3 + 3*a^2*b + 3*a *b^2 + b^3)*d*x*cosh(d*x + c)^3 - 3*(a^2*b + 2*a*b^2 + b^3 - (a^3 + 3*a^2* b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^9 - 9*(8*a^2*b + 12*a *b^2 + 4*b^3 - 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^8 + 9* (165*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^4 - 8*a^2*b - 12*a* b^2 - 4*b^3 + 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 90*(a^2*b + 2*a*b^2 + b^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c )^8 + 72*(33*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x*cosh(d*x + c)^5 - 30*(a^2 *b + 2*a*b^2 + b^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^3 - (8*a^2*b + 12*a*b^2 + 4*b^3 - 5*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cos h(d*x + c))*sinh(d*x + c)^7 - 4*(27*a^2*b + 36*a*b^2 + 17*b^3 - 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^6 + 4*(693*(a^3 + 3*a^2*b + 3* a*b^2 + b^3)*d*x*cosh(d*x + c)^6 - 945*(a^2*b + 2*a*b^2 + b^3 - (a^3 + 3*a ^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^4 - 27*a^2*b - 36*a*b^2 - 17*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x - 63*(8*a^2*b + 12*a*b^2 + 4*b...
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (71) = 142\).
Time = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.54 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x - \frac {a^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} + 3 a^{2} b x - \frac {3 a^{2} b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {3 a^{2} b \tanh ^{2}{\left (c + d x \right )}}{2 d} + 3 a b^{2} x - \frac {3 a b^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {3 a b^{2} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {3 a b^{2} \tanh ^{2}{\left (c + d x \right )}}{2 d} + b^{3} x - \frac {b^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {b^{3} \tanh ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{3} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \tanh ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{3} \tanh {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((a**3*x - a**3*log(tanh(c + d*x) + 1)/d + 3*a**2*b*x - 3*a**2*b* log(tanh(c + d*x) + 1)/d - 3*a**2*b*tanh(c + d*x)**2/(2*d) + 3*a*b**2*x - 3*a*b**2*log(tanh(c + d*x) + 1)/d - 3*a*b**2*tanh(c + d*x)**4/(4*d) - 3*a* b**2*tanh(c + d*x)**2/(2*d) + b**3*x - b**3*log(tanh(c + d*x) + 1)/d - b** 3*tanh(c + d*x)**6/(6*d) - b**3*tanh(c + d*x)**4/(4*d) - b**3*tanh(c + d*x )**2/(2*d), Ne(d, 0)), (x*(a + b*tanh(c)**2)**3*tanh(c), True))
Leaf count of result is larger than twice the leaf count of optimal. 351 vs. \(2 (77) = 154\).
Time = 0.28 (sec) , antiderivative size = 351, normalized size of antiderivative = 4.23 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{3} \, b^{3} {\left (3 \, x + \frac {3 \, c}{d} + \frac {3 \, \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, {\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} + 18 \, e^{\left (-4 \, d x - 4 \, c\right )} + 34 \, e^{\left (-6 \, d x - 6 \, c\right )} + 18 \, e^{\left (-8 \, d x - 8 \, c\right )} + 9 \, e^{\left (-10 \, d x - 10 \, c\right )}\right )}}{d {\left (6 \, e^{\left (-2 \, d x - 2 \, c\right )} + 15 \, e^{\left (-4 \, d x - 4 \, c\right )} + 20 \, e^{\left (-6 \, d x - 6 \, c\right )} + 15 \, e^{\left (-8 \, d x - 8 \, c\right )} + 6 \, e^{\left (-10 \, d x - 10 \, c\right )} + e^{\left (-12 \, d x - 12 \, c\right )} + 1\right )}}\right )} + 3 \, a b^{2} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {a^{3} \log \left (\cosh \left (d x + c\right )\right )}{d} \]
1/3*b^3*(3*x + 3*c/d + 3*log(e^(-2*d*x - 2*c) + 1)/d + 2*(9*e^(-2*d*x - 2* c) + 18*e^(-4*d*x - 4*c) + 34*e^(-6*d*x - 6*c) + 18*e^(-8*d*x - 8*c) + 9*e ^(-10*d*x - 10*c))/(d*(6*e^(-2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e^(-6 *d*x - 6*c) + 15*e^(-8*d*x - 8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12 *c) + 1))) + 3*a*b^2*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/(d*(4*e^(-2*d*x - 2*c) + 6* e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) + 3*a^2*b* (x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d* x - 2*c) + e^(-4*d*x - 4*c) + 1))) + a^3*log(cosh(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 216 vs. \(2 (77) = 154\).
Time = 0.40 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.60 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac {2 \, {\left (9 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} e^{\left (10 \, d x + 10 \, c\right )} + 18 \, {\left (2 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )} + 2 \, {\left (27 \, a^{2} b + 36 \, a b^{2} + 17 \, b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 18 \, {\left (2 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{6}}}{3 \, d} \]
-1/3*(3*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) - 3*(a^3 + 3*a^2*b + 3*a *b^2 + b^3)*log(e^(2*d*x + 2*c) + 1) - 2*(9*(a^2*b + 2*a*b^2 + b^3)*e^(10* d*x + 10*c) + 18*(2*a^2*b + 3*a*b^2 + b^3)*e^(8*d*x + 8*c) + 2*(27*a^2*b + 36*a*b^2 + 17*b^3)*e^(6*d*x + 6*c) + 18*(2*a^2*b + 3*a*b^2 + b^3)*e^(4*d* x + 4*c) + 9*(a^2*b + 2*a*b^2 + b^3)*e^(2*d*x + 2*c))/(e^(2*d*x + 2*c) + 1 )^6)/d
Time = 1.85 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.48 \[ \int \tanh (c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,d}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^4\,\left (b^3+3\,a\,b^2\right )}{4\,d}-\frac {b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^6}{6\,d} \]