\(\int \tanh ^3(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [157]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [B] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 23, antiderivative size = 107 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {(a+b)^3 \tanh ^2(c+d x)}{2 d}-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh ^4(c+d x)}{4 d}-\frac {b^2 (3 a+b) \tanh ^6(c+d x)}{6 d}-\frac {b^3 \tanh ^8(c+d x)}{8 d} \]

[Out]

(a+b)^3*ln(cosh(d*x+c))/d-1/2*(a+b)^3*tanh(d*x+c)^2/d-1/4*b*(3*a^2+3*a*b+b^2)*tanh(d*x+c)^4/d-1/6*b^2*(3*a+b)*
tanh(d*x+c)^6/d-1/8*b^3*tanh(d*x+c)^8/d

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 107, 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, 457, 78} \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh ^4(c+d x)}{4 d}-\frac {b^2 (3 a+b) \tanh ^6(c+d x)}{6 d}-\frac {(a+b)^3 \tanh ^2(c+d x)}{2 d}+\frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {b^3 \tanh ^8(c+d x)}{8 d} \]

[In]

Int[Tanh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

((a + b)^3*Log[Cosh[c + d*x]])/d - ((a + b)^3*Tanh[c + d*x]^2)/(2*d) - (b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x]^
4)/(4*d) - (b^2*(3*a + b)*Tanh[c + d*x]^6)/(6*d) - (b^3*Tanh[c + d*x]^8)/(8*d)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

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]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((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, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^3 \left (a+b x^2\right )^3}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \frac {x (a+b x)^3}{1-x} \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {\text {Subst}\left (\int \left (-(a+b)^3-\frac {(a+b)^3}{-1+x}-b \left (3 a^2+3 a b+b^2\right ) x-b^2 (3 a+b) x^2-b^3 x^3\right ) \, dx,x,\tanh ^2(c+d x)\right )}{2 d} \\ & = \frac {(a+b)^3 \log (\cosh (c+d x))}{d}-\frac {(a+b)^3 \tanh ^2(c+d x)}{2 d}-\frac {b \left (3 a^2+3 a b+b^2\right ) \tanh ^4(c+d x)}{4 d}-\frac {b^2 (3 a+b) \tanh ^6(c+d x)}{6 d}-\frac {b^3 \tanh ^8(c+d x)}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.92 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {2 (a+b)^3 \log (\cosh (c+d x))-(a+b)^3 \tanh ^2(c+d x)-\frac {1}{2} b \left (3 a^2+3 a b+b^2\right ) \tanh ^4(c+d x)-\frac {1}{3} b^2 (3 a+b) \tanh ^6(c+d x)-\frac {1}{4} b^3 \tanh ^8(c+d x)}{2 d} \]

[In]

Integrate[Tanh[c + d*x]^3*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(2*(a + b)^3*Log[Cosh[c + d*x]] - (a + b)^3*Tanh[c + d*x]^2 - (b*(3*a^2 + 3*a*b + b^2)*Tanh[c + d*x]^4)/2 - (b
^2*(3*a + b)*Tanh[c + d*x]^6)/3 - (b^3*Tanh[c + d*x]^8)/4)/(2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(204\) vs. \(2(99)=198\).

Time = 0.14 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.92

method result size
derivativedivides \(\frac {-\frac {\tanh \left (d x +c \right )^{6} a \,b^{2}}{2}-\frac {3 \tanh \left (d x +c \right )^{4} a^{2} b}{4}-\frac {3 \tanh \left (d x +c \right )^{4} a \,b^{2}}{4}-\frac {3 \tanh \left (d x +c \right )^{2} a^{2} b}{2}-\frac {3 a \,b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{3}}{6}-\frac {b^{3} \tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2} a^{3}}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{8} b^{3}}{8}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{d}\) \(205\)
default \(\frac {-\frac {\tanh \left (d x +c \right )^{6} a \,b^{2}}{2}-\frac {3 \tanh \left (d x +c \right )^{4} a^{2} b}{4}-\frac {3 \tanh \left (d x +c \right )^{4} a \,b^{2}}{4}-\frac {3 \tanh \left (d x +c \right )^{2} a^{2} b}{2}-\frac {3 a \,b^{2} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{6} b^{3}}{6}-\frac {b^{3} \tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2} a^{3}}{2}-\frac {b^{3} \tanh \left (d x +c \right )^{2}}{2}-\frac {\tanh \left (d x +c \right )^{8} b^{3}}{8}+\frac {\left (-a^{3}-3 a^{2} b -3 a \,b^{2}-b^{3}\right ) \ln \left (\tanh \left (d x +c \right )+1\right )}{2}-\frac {\left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) \ln \left (\tanh \left (d x +c \right )-1\right )}{2}}{d}\) \(205\)
parts \(\frac {a^{3} \left (-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {b^{3} \left (-\frac {\tanh \left (d x +c \right )^{8}}{8}-\frac {\tanh \left (d x +c \right )^{6}}{6}-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a \,b^{2} \left (-\frac {\tanh \left (d x +c \right )^{6}}{6}-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (-\frac {\tanh \left (d x +c \right )^{4}}{4}-\frac {\tanh \left (d x +c \right )^{2}}{2}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{2}-\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{2}\right )}{d}\) \(226\)
parallelrisch \(-\frac {3 \tanh \left (d x +c \right )^{8} b^{3}+12 \tanh \left (d x +c \right )^{6} a \,b^{2}+4 \tanh \left (d x +c \right )^{6} b^{3}+18 \tanh \left (d x +c \right )^{4} a^{2} b +18 \tanh \left (d x +c \right )^{4} a \,b^{2}+6 b^{3} \tanh \left (d x +c \right )^{4}+24 a^{3} d x +72 a^{2} b d x +72 a \,b^{2} d x +24 b^{3} d x +12 \tanh \left (d x +c \right )^{2} a^{3}+36 \tanh \left (d x +c \right )^{2} a^{2} b +36 a \,b^{2} \tanh \left (d x +c \right )^{2}+12 b^{3} \tanh \left (d x +c \right )^{2}+24 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{3}+72 \ln \left (1-\tanh \left (d x +c \right )\right ) a^{2} b +72 \ln \left (1-\tanh \left (d x +c \right )\right ) a \,b^{2}+24 \ln \left (1-\tanh \left (d x +c \right )\right ) b^{3}}{24 d}\) \(238\)
risch \(-a^{3} x -3 b \,a^{2} x -3 a \,b^{2} x -b^{3} x -\frac {2 a^{3} c}{d}-\frac {6 b c \,a^{2}}{d}-\frac {6 a \,b^{2} c}{d}-\frac {2 b^{3} c}{d}+\frac {2 \,{\mathrm e}^{2 d x +2 c} \left (27 a \,b^{2}+3 a^{3}+18 a^{2} b \,{\mathrm e}^{12 d x +12 c}+108 a \,b^{2} {\mathrm e}^{10 d x +10 c}+90 a^{2} b \,{\mathrm e}^{10 d x +10 c}+27 a \,b^{2} {\mathrm e}^{12 d x +12 c}+100 \,{\mathrm e}^{4 d x +4 c} b^{3}+18 a^{2} b +108 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}+90 a^{2} b \,{\mathrm e}^{2 d x +2 c}+237 a \,b^{2} {\mathrm e}^{4 d x +4 c}+198 a^{2} b \,{\mathrm e}^{4 d x +4 c}+252 a^{2} b \,{\mathrm e}^{6 d x +6 c}+312 a \,b^{2} {\mathrm e}^{6 d x +6 c}+12 b^{3}+198 a^{2} b \,{\mathrm e}^{8 d x +8 c}+237 a \,b^{2} {\mathrm e}^{8 d x +8 c}+12 b^{3} {\mathrm e}^{12 d x +12 c}+60 a^{3} {\mathrm e}^{6 d x +6 c}+104 \,{\mathrm e}^{6 d x +6 c} b^{3}+45 a^{3} {\mathrm e}^{4 d x +4 c}+18 a^{3} {\mathrm e}^{2 d x +2 c}+36 \,{\mathrm e}^{2 d x +2 c} b^{3}+36 b^{3} {\mathrm e}^{10 d x +10 c}+100 b^{3} {\mathrm e}^{8 d x +8 c}+18 a^{3} {\mathrm e}^{10 d x +10 c}+3 a^{3} {\mathrm e}^{12 d x +12 c}+45 a^{3} {\mathrm e}^{8 d x +8 c}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{8}}+\frac {\ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{3}}{d}+\frac {3 b \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a^{2}}{d}+\frac {3 \ln \left ({\mathrm e}^{2 d x +2 c}+1\right ) a \,b^{2}}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\) \(544\)

[In]

int(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/2*tanh(d*x+c)^6*a*b^2-3/4*tanh(d*x+c)^4*a^2*b-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)^2-1/6*tanh(d*x+c)^6*b^3-1/4*b^3*tanh(d*x+c)^4-1/2*tanh(d*x+c)^2*a^3-1/2*b^3*tanh(d*x+c)^2-1/8*ta
nh(d*x+c)^8*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))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7502 vs. \(2 (99) = 198\).

Time = 0.34 (sec) , antiderivative size = 7502, normalized size of antiderivative = 70.11 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \]

[In]

integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

Too large to include

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (94) = 188\).

Time = 0.27 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.61 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\begin {cases} a^{3} x - \frac {a^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a^{3} \tanh ^{2}{\left (c + d x \right )}}{2 d} + 3 a^{2} b x - \frac {3 a^{2} b \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {3 a^{2} b \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {3 a^{2} b \tanh ^{2}{\left (c + d x \right )}}{2 d} + 3 a b^{2} x - \frac {3 a b^{2} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a b^{2} \tanh ^{6}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {3 a b^{2} \tanh ^{2}{\left (c + d x \right )}}{2 d} + b^{3} x - \frac {b^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {b^{3} \tanh ^{8}{\left (c + d x \right )}}{8 d} - \frac {b^{3} \tanh ^{6}{\left (c + d x \right )}}{6 d} - \frac {b^{3} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \tanh ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tanh ^{2}{\left (c \right )}\right )^{3} \tanh ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \]

[In]

integrate(tanh(d*x+c)**3*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Piecewise((a**3*x - a**3*log(tanh(c + d*x) + 1)/d - a**3*tanh(c + d*x)**2/(2*d) + 3*a**2*b*x - 3*a**2*b*log(ta
nh(c + d*x) + 1)/d - 3*a**2*b*tanh(c + d*x)**4/(4*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 - a*b**2*tanh(c + d*x)**6/(2*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)**8/(8*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
)**3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (99) = 198\).

Time = 0.29 (sec) , antiderivative size = 540, normalized size of antiderivative = 5.05 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=a b^{2} {\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 )} + \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 {8 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + 25 \, e^{\left (-6 \, d x - 6 \, c\right )} + 26 \, e^{\left (-8 \, d x - 8 \, c\right )} + 25 \, e^{\left (-10 \, d x - 10 \, c\right )} + 9 \, e^{\left (-12 \, d x - 12 \, c\right )} + 3 \, e^{\left (-14 \, d x - 14 \, c\right )}\right )}}{d {\left (8 \, e^{\left (-2 \, d x - 2 \, c\right )} + 28 \, e^{\left (-4 \, d x - 4 \, c\right )} + 56 \, e^{\left (-6 \, d x - 6 \, c\right )} + 70 \, e^{\left (-8 \, d x - 8 \, c\right )} + 56 \, e^{\left (-10 \, d x - 10 \, c\right )} + 28 \, e^{\left (-12 \, d x - 12 \, c\right )} + 8 \, e^{\left (-14 \, d x - 14 \, c\right )} + e^{\left (-16 \, d x - 16 \, 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 {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 )} + a^{3} {\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 )} \]

[In]

integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

a*b^2*(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))) + 1/3*b^3*(3*x + 3*c/d +
3*log(e^(-2*d*x - 2*c) + 1)/d + 8*(3*e^(-2*d*x - 2*c) + 9*e^(-4*d*x - 4*c) + 25*e^(-6*d*x - 6*c) + 26*e^(-8*d*
x - 8*c) + 25*e^(-10*d*x - 10*c) + 9*e^(-12*d*x - 12*c) + 3*e^(-14*d*x - 14*c))/(d*(8*e^(-2*d*x - 2*c) + 28*e^
(-4*d*x - 4*c) + 56*e^(-6*d*x - 6*c) + 70*e^(-8*d*x - 8*c) + 56*e^(-10*d*x - 10*c) + 28*e^(-12*d*x - 12*c) + 8
*e^(-14*d*x - 14*c) + e^(-16*d*x - 16*c) + 1))) + 3*a^2*b*(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))) + a^3*(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)))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 309 vs. \(2 (99) = 198\).

Time = 0.46 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.89 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - 3 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - \frac {2 \, {\left (3 \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} e^{\left (14 \, d x + 14 \, c\right )} + 18 \, {\left (a^{3} + 5 \, a^{2} b + 6 \, a b^{2} + 2 \, b^{3}\right )} e^{\left (12 \, d x + 12 \, c\right )} + {\left (45 \, a^{3} + 198 \, a^{2} b + 237 \, a b^{2} + 100 \, b^{3}\right )} e^{\left (10 \, d x + 10 \, c\right )} + 4 \, {\left (15 \, a^{3} + 63 \, a^{2} b + 78 \, a b^{2} + 26 \, b^{3}\right )} e^{\left (8 \, d x + 8 \, c\right )} + {\left (45 \, a^{3} + 198 \, a^{2} b + 237 \, a b^{2} + 100 \, b^{3}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 18 \, {\left (a^{3} + 5 \, a^{2} b + 6 \, a b^{2} + 2 \, b^{3}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, {\left (a^{3} + 6 \, a^{2} b + 9 \, a b^{2} + 4 \, b^{3}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{8}}}{3 \, d} \]

[In]

integrate(tanh(d*x+c)^3*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-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*(3*(a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*e^(14*d*x + 14*c) + 18*(a^3 + 5*a^2*b + 6*a*b^2 + 2*b^3)*e^(12*d*x +
 12*c) + (45*a^3 + 198*a^2*b + 237*a*b^2 + 100*b^3)*e^(10*d*x + 10*c) + 4*(15*a^3 + 63*a^2*b + 78*a*b^2 + 26*b
^3)*e^(8*d*x + 8*c) + (45*a^3 + 198*a^2*b + 237*a*b^2 + 100*b^3)*e^(6*d*x + 6*c) + 18*(a^3 + 5*a^2*b + 6*a*b^2
 + 2*b^3)*e^(4*d*x + 4*c) + 3*(a^3 + 6*a^2*b + 9*a*b^2 + 4*b^3)*e^(2*d*x + 2*c))/(e^(2*d*x + 2*c) + 1)^8)/d

Mupad [B] (verification not implemented)

Time = 1.88 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.45 \[ \int \tanh ^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=x\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^4\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{4\,d}-\frac {\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^6\,\left (b^3+3\,a\,b^2\right )}{6\,d}-\frac {b^3\,{\mathrm {tanh}\left (c+d\,x\right )}^8}{8\,d}-\frac {{\mathrm {tanh}\left (c+d\,x\right )}^2\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{2\,d} \]

[In]

int(tanh(c + d*x)^3*(a + b*tanh(c + d*x)^2)^3,x)

[Out]

x*(3*a*b^2 + 3*a^2*b + a^3 + b^3) - (tanh(c + d*x)^4*(3*a*b^2 + 3*a^2*b + b^3))/(4*d) - (log(tanh(c + d*x) + 1
)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/d - (tanh(c + d*x)^6*(3*a*b^2 + b^3))/(6*d) - (b^3*tanh(c + d*x)^8)/(8*d) -
 (tanh(c + d*x)^2*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(2*d)