Integrand size = 14, antiderivative size = 74 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=(a+b)^3 x-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)}{d}-\frac {b^2 (3 a+b) \tanh ^3(c+d x)}{3 d}-\frac {b^3 \tanh ^5(c+d x)}{5 d} \]
(a+b)^3*x-b*(3*a^2+3*a*b+b^2)*tanh(d*x+c)/d-1/3*b^2*(3*a+b)*tanh(d*x+c)^3/ d-1/5*b^3*tanh(d*x+c)^5/d
Time = 0.72 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.28 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\tanh (c+d x) \left (\frac {15 (a+b)^3 \text {arctanh}\left (\sqrt {\tanh ^2(c+d x)}\right )}{\sqrt {\tanh ^2(c+d x)}}-b \left (45 a^2+15 a b \left (3+\tanh ^2(c+d x)\right )+b^2 \left (15+5 \tanh ^2(c+d x)+3 \tanh ^4(c+d x)\right )\right )\right )}{15 d} \]
(Tanh[c + d*x]*((15*(a + b)^3*ArcTanh[Sqrt[Tanh[c + d*x]^2]])/Sqrt[Tanh[c + d*x]^2] - b*(45*a^2 + 15*a*b*(3 + Tanh[c + d*x]^2) + b^2*(15 + 5*Tanh[c + d*x]^2 + 3*Tanh[c + d*x]^4))))/(15*d)
Time = 0.25 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4144, 300, 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.
| \(\displaystyle \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a-b \tan (i c+i d x)^2\right )^3dx\) |
| \(\Big \downarrow \) 4144 |
| \(\displaystyle \frac {\int \frac {\left (b \tanh ^2(c+d x)+a\right )^3}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 300 |
| \(\displaystyle \frac {\int \left (-b^3 \tanh ^4(c+d x)-b^2 (3 a+b) \tanh ^2(c+d x)-b \left (3 a^2+3 b a+b^2\right )+\frac {(a+b)^3}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-b \left (3 a^2+3 a b+b^2\right ) \tanh (c+d x)+(a+b)^3 \text {arctanh}(\tanh (c+d x))-\frac {1}{3} b^2 (3 a+b) \tanh ^3(c+d x)-\frac {1}{5} b^3 \tanh ^5(c+d x)}{d}\) |
((a + b)^3*ArcTanh[Tanh[c + d*x]] - b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x] - (b^2*(3*a + b)*Tanh[c + d*x]^3)/3 - (b^3*Tanh[c + d*x]^5)/5)/d
3.2.60.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
Int[((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[(a + b* (ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])
Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.51
| method | result | size |
| parallelrisch | \(-\frac {3 b^{3} \tanh \left (d x +c \right )^{5}+15 a \,b^{2} \tanh \left (d x +c \right )^{3}+5 b^{3} \tanh \left (d x +c \right )^{3}-15 a^{3} d x -45 a^{2} b d x -45 a \,b^{2} d x -15 b^{3} d x +45 a^{2} b \tanh \left (d x +c \right )+45 a \,b^{2} \tanh \left (d x +c \right )+15 b^{3} \tanh \left (d x +c \right )}{15 d}\) | \(112\) |
| derivativedivides | \(\frac {-3 a^{2} b \tanh \left (d x +c \right )-3 a \,b^{2} \tanh \left (d x +c \right )-a \,b^{2} \tanh \left (d x +c \right )^{3}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{3}}{3}-b^{3} \tanh \left (d x +c \right )-\frac {b^{3} \tanh \left (d x +c \right )^{5}}{5}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}}{d}\) | \(141\) |
| default | \(\frac {-3 a^{2} b \tanh \left (d x +c \right )-3 a \,b^{2} \tanh \left (d x +c \right )-a \,b^{2} \tanh \left (d x +c \right )^{3}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{3}}{3}-b^{3} \tanh \left (d x +c \right )-\frac {b^{3} \tanh \left (d x +c \right )^{5}}{5}+\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}}{d}\) | \(141\) |
| parts | \(a^{3} x +\frac {b^{3} \left (-\frac {\tanh \left (d x +c \right )^{5}}{5}-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\tanh \left (d x +c \right )^{3}}{3}-\tanh \left (d x +c \right )-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}\) | \(155\) |
| risch | \(a^{3} x +3 b \,a^{2} x +3 a \,b^{2} x +b^{3} x +\frac {2 b \left (45 a^{2} {\mathrm e}^{8 d x +8 c}+90 a b \,{\mathrm e}^{8 d x +8 c}+45 b^{2} {\mathrm e}^{8 d x +8 c}+180 a^{2} {\mathrm e}^{6 d x +6 c}+270 a b \,{\mathrm e}^{6 d x +6 c}+90 b^{2} {\mathrm e}^{6 d x +6 c}+270 a^{2} {\mathrm e}^{4 d x +4 c}+330 a b \,{\mathrm e}^{4 d x +4 c}+140 \,{\mathrm e}^{4 d x +4 c} b^{2}+180 a^{2} {\mathrm e}^{2 d x +2 c}+210 a b \,{\mathrm e}^{2 d x +2 c}+70 \,{\mathrm e}^{2 d x +2 c} b^{2}+45 a^{2}+60 a b +23 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(224\) |
-1/15*(3*b^3*tanh(d*x+c)^5+15*a*b^2*tanh(d*x+c)^3+5*b^3*tanh(d*x+c)^3-15*a ^3*d*x-45*a^2*b*d*x-45*a*b^2*d*x-15*b^3*d*x+45*a^2*b*tanh(d*x+c)+45*a*b^2* tanh(d*x+c)+15*b^3*tanh(d*x+c))/d
Leaf count of result is larger than twice the leaf count of optimal. 567 vs. \(2 (70) = 140\).
Time = 0.27 (sec) , antiderivative size = 567, normalized size of antiderivative = 7.66 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {{\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} - {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 5 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3} + 2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3} + 15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) - 5 \, {\left ({\left (45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 18 \, a^{2} b + 12 \, a b^{2} + 10 \, b^{3} + 3 \, {\left (27 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \]
1/15*((45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d *x)*cosh(d*x + c)^5 + 5*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^4 - (45*a^2*b + 60*a*b^2 + 23*b^3)*sinh(d*x + c)^5 + 5*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3 *a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c)^3 - 5*(27*a^2*b + 24*a*b^2 + 5* b^3 + 2*(45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^3 + 5*(2*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d* x)*cosh(d*x + c)^3 + 3*(45*a^2*b + 60*a*b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(45*a^2*b + 60*a* b^2 + 23*b^3 + 15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d*x)*cosh(d*x + c) - 5*( (45*a^2*b + 60*a*b^2 + 23*b^3)*cosh(d*x + c)^4 + 18*a^2*b + 12*a*b^2 + 10* b^3 + 3*(27*a^2*b + 24*a*b^2 + 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*c osh(d*x + c)^5 + 5*d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d *x + c))
Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x + 3 a^{2} b x - \frac {3 a^{2} b \tanh {\left (c + d x \right )}}{d} + 3 a b^{2} x - \frac {a b^{2} \tanh ^{3}{\left (c + d x \right )}}{d} - \frac {3 a b^{2} \tanh {\left (c + d x \right )}}{d} + b^{3} x - \frac {b^{3} \tanh ^{5}{\left (c + d x \right )}}{5 d} - \frac {b^{3} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{3} & \text {otherwise} \end {cases} \]
Piecewise((a**3*x + 3*a**2*b*x - 3*a**2*b*tanh(c + d*x)/d + 3*a*b**2*x - a *b**2*tanh(c + d*x)**3/d - 3*a*b**2*tanh(c + d*x)/d + b**3*x - b**3*tanh(c + d*x)**5/(5*d) - b**3*tanh(c + d*x)**3/(3*d) - b**3*tanh(c + d*x)/d, Ne( d, 0)), (x*(a + b*tanh(c)**2)**3, True))
Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (70) = 140\).
Time = 0.20 (sec) , antiderivative size = 239, normalized size of antiderivative = 3.23 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {1}{15} \, b^{3} {\left (15 \, x + \frac {15 \, c}{d} - \frac {2 \, {\left (70 \, e^{\left (-2 \, d x - 2 \, c\right )} + 140 \, e^{\left (-4 \, d x - 4 \, c\right )} + 90 \, e^{\left (-6 \, d x - 6 \, c\right )} + 45 \, e^{\left (-8 \, d x - 8 \, c\right )} + 23\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + 3 \, a^{2} b {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{3} x \]
1/15*b^3*(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))) + a*b^2*(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))) + 3*a^2*b*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + a^3*x
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (70) = 140\).
Time = 0.31 (sec) , antiderivative size = 241, normalized size of antiderivative = 3.26 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 330 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 210 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b + 60 \, a b^{2} + 23 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]
1/15*(15*(a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(d*x + c) + 2*(45*a^2*b*e^(8*d*x + 8*c) + 90*a*b^2*e^(8*d*x + 8*c) + 45*b^3*e^(8*d*x + 8*c) + 180*a^2*b*e^( 6*d*x + 6*c) + 270*a*b^2*e^(6*d*x + 6*c) + 90*b^3*e^(6*d*x + 6*c) + 270*a^ 2*b*e^(4*d*x + 4*c) + 330*a*b^2*e^(4*d*x + 4*c) + 140*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e^(2*d*x + 2*c) + 210*a*b^2*e^(2*d*x + 2*c) + 70*b^3*e^(2*d*x + 2*c) + 45*a^2*b + 60*a*b^2 + 23*b^3)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^3\,\left (b^3+3\,a\,b^2\right )}{3\,d}-\frac {b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^5}{5\,d}-\frac {b\,\mathrm {tanh}\left (c+d\,x\right )\,\left (3\,a^2+3\,a\,b+b^2\right )}{d} \]