\(\int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx\) [58]

Optimal result

Integrand size = 20, antiderivative size = 277 \[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=-\frac {b^2 (c+d x)^3}{f}+\frac {a^2 (c+d x)^4}{4 d}-\frac {a b (c+d x)^4}{2 d}+\frac {b^2 (c+d x)^4}{4 d}+\frac {3 b^2 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 b^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac {3 a b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}-\frac {3 b^2 d^3 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^4}-\frac {3 a b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac {3 a b d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{2 f^4}-\frac {b^2 (c+d x)^3 \tanh (e+f x)}{f} \] Output:

-b^2*(d*x+c)^3/f+1/4*a^2*(d*x+c)^4/d-1/2*a*b*(d*x+c)^4/d+1/4*b^2*(d*x+c)^4 
/d+3*b^2*d*(d*x+c)^2*ln(1+exp(2*f*x+2*e))/f^2+2*a*b*(d*x+c)^3*ln(1+exp(2*f 
*x+2*e))/f+3*b^2*d^2*(d*x+c)*polylog(2,-exp(2*f*x+2*e))/f^3+3*a*b*d*(d*x+c 
)^2*polylog(2,-exp(2*f*x+2*e))/f^2-3/2*b^2*d^3*polylog(3,-exp(2*f*x+2*e))/ 
f^4-3*a*b*d^2*(d*x+c)*polylog(3,-exp(2*f*x+2*e))/f^3+3/2*a*b*d^3*polylog(4 
,-exp(2*f*x+2*e))/f^4-b^2*(d*x+c)^3*tanh(f*x+e)/f
 

Mathematica [A] (verified)

Time = 2.78 (sec) , antiderivative size = 544, normalized size of antiderivative = 1.96 \[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\frac {-16 b c^2 (3 b d+2 a c f) x+\frac {16 b^2 (c+d x)^3}{1+e^{2 e}}+\frac {8 a b f (c+d x)^4}{d \left (1+e^{2 e}\right )}+\frac {48 b c d (b d+a c f) x \log \left (1+e^{-2 (e+f x)}\right )}{f}+\frac {24 b d^2 (b d+2 a c f) x^2 \log \left (1+e^{-2 (e+f x)}\right )}{f}+16 a b d^3 x^3 \log \left (1+e^{-2 (e+f x)}\right )+\frac {8 b c^2 (3 b d+2 a c f) \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac {24 b c d (b d+a c f) \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^2}-\frac {24 b d^2 (b d+2 a c f) x \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f^2}-\frac {24 a b d^3 x^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )}{f}-\frac {12 b d^2 (b d+2 a c f) \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f^3}-\frac {24 a b d^3 x \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )}{f^2}-\frac {12 a b d^3 \operatorname {PolyLog}\left (4,-e^{-2 (e+f x)}\right )}{f^3}+\text {sech}(e) \text {sech}(e+f x) \left (\left (a^2+b^2\right ) f x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \cosh (f x)+\left (a^2+b^2\right ) f x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \cosh (2 e+f x)-2 b \left (4 b (c+d x)^3+a f x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )\right ) \sinh (f x)+2 a b f x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \sinh (2 e+f x)\right )}{8 f} \] Input:

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

Output:

(-16*b*c^2*(3*b*d + 2*a*c*f)*x + (16*b^2*(c + d*x)^3)/(1 + E^(2*e)) + (8*a 
*b*f*(c + d*x)^4)/(d*(1 + E^(2*e))) + (48*b*c*d*(b*d + a*c*f)*x*Log[1 + E^ 
(-2*(e + f*x))])/f + (24*b*d^2*(b*d + 2*a*c*f)*x^2*Log[1 + E^(-2*(e + f*x) 
)])/f + 16*a*b*d^3*x^3*Log[1 + E^(-2*(e + f*x))] + (8*b*c^2*(3*b*d + 2*a*c 
*f)*Log[1 + E^(2*(e + f*x))])/f - (24*b*c*d*(b*d + a*c*f)*PolyLog[2, -E^(- 
2*(e + f*x))])/f^2 - (24*b*d^2*(b*d + 2*a*c*f)*x*PolyLog[2, -E^(-2*(e + f* 
x))])/f^2 - (24*a*b*d^3*x^2*PolyLog[2, -E^(-2*(e + f*x))])/f - (12*b*d^2*( 
b*d + 2*a*c*f)*PolyLog[3, -E^(-2*(e + f*x))])/f^3 - (24*a*b*d^3*x*PolyLog[ 
3, -E^(-2*(e + f*x))])/f^2 - (12*a*b*d^3*PolyLog[4, -E^(-2*(e + f*x))])/f^ 
3 + Sech[e]*Sech[e + f*x]*((a^2 + b^2)*f*x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^ 
2 + d^3*x^3)*Cosh[f*x] + (a^2 + b^2)*f*x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 
+ d^3*x^3)*Cosh[2*e + f*x] - 2*b*(4*b*(c + d*x)^3 + a*f*x*(4*c^3 + 6*c^2*d 
*x + 4*c*d^2*x^2 + d^3*x^3))*Sinh[f*x] + 2*a*b*f*x*(4*c^3 + 6*c^2*d*x + 4* 
c*d^2*x^2 + d^3*x^3)*Sinh[2*e + f*x]))/(8*f)
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4205, 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.

\(\displaystyle \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^3 (a-i b \tan (i e+i f x))^2dx\)

\(\Big \downarrow \) 4205

\(\displaystyle \int \left (a^2 (c+d x)^3+2 a b (c+d x)^3 \tanh (e+f x)+b^2 (c+d x)^3 \tanh ^2(e+f x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^2 (c+d x)^4}{4 d}-\frac {3 a b d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{f^3}+\frac {3 a b d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^2}+\frac {2 a b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac {a b (c+d x)^4}{2 d}+\frac {3 a b d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{2 f^4}+\frac {3 b^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}+\frac {3 b^2 d (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac {b^2 (c+d x)^3 \tanh (e+f x)}{f}-\frac {b^2 (c+d x)^3}{f}+\frac {b^2 (c+d x)^4}{4 d}-\frac {3 b^2 d^3 \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^4}\)

Input:

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

Output:

-((b^2*(c + d*x)^3)/f) + (a^2*(c + d*x)^4)/(4*d) - (a*b*(c + d*x)^4)/(2*d) 
 + (b^2*(c + d*x)^4)/(4*d) + (3*b^2*d*(c + d*x)^2*Log[1 + E^(2*(e + f*x))] 
)/f^2 + (2*a*b*(c + d*x)^3*Log[1 + E^(2*(e + f*x))])/f + (3*b^2*d^2*(c + d 
*x)*PolyLog[2, -E^(2*(e + f*x))])/f^3 + (3*a*b*d*(c + d*x)^2*PolyLog[2, -E 
^(2*(e + f*x))])/f^2 - (3*b^2*d^3*PolyLog[3, -E^(2*(e + f*x))])/(2*f^4) - 
(3*a*b*d^2*(c + d*x)*PolyLog[3, -E^(2*(e + f*x))])/f^3 + (3*a*b*d^3*PolyLo 
g[4, -E^(2*(e + f*x))])/(2*f^4) - (b^2*(c + d*x)^3*Tanh[e + f*x])/f
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4205

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(904\) vs. \(2(267)=534\).

Time = 1.34 (sec) , antiderivative size = 905, normalized size of antiderivative = 3.27

method	result	size
risch	\(\frac {12 b \,d^{2} c a \,e^{2} x}{f^{2}}-\frac {12 b a \,c^{2} d e x}{f}+\frac {12 b e a \,c^{2} d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}-\frac {12 b \,e^{2} d^{2} c a \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {2 b^{2} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{f \left (1+{\mathrm e}^{2 f x +2 e}\right )}+\frac {6 b \,d^{2} c a \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {6 b \,d^{2} c a \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {6 b a \,c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}-\frac {3 b^{2} d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{4}}+\frac {3 a b \,d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{4}}-2 d^{2} a b c \,x^{3}-3 d a b \,c^{2} x^{2}+2 a b \,c^{3} x +\frac {d^{3} a^{2} x^{4}}{4}+\frac {d^{3} b^{2} x^{4}}{4}+\frac {a^{2} c^{4}}{4 d}+\frac {b^{2} c^{4}}{4 d}-\frac {d^{3} a b \,x^{4}}{2}+d^{2} a^{2} c \,x^{3}+d^{2} b^{2} c \,x^{3}+\frac {3 d \,a^{2} c^{2} x^{2}}{2}+\frac {3 d \,b^{2} c^{2} x^{2}}{2}+a^{2} c^{3} x +b^{2} c^{3} x +\frac {a b \,c^{4}}{2 d}+\frac {6 b^{2} d^{3} e^{2} x}{f^{3}}+\frac {3 b^{2} c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}-\frac {6 b^{2} e^{2} d^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {3 b^{2} d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f^{2}}+\frac {3 b^{2} d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{3}}-\frac {4 b a \,c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}+\frac {2 b a \,c^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {6 b^{2} c^{2} d \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {3 b^{2} c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {2 b^{2} d^{3} x^{3}}{f}+\frac {4 b^{2} d^{3} e^{3}}{f^{4}}-\frac {3 b \,d^{3} a \,e^{4}}{f^{4}}-\frac {6 b^{2} c \,d^{2} x^{2}}{f}-\frac {6 b^{2} c \,d^{2} e^{2}}{f^{3}}-\frac {12 b^{2} c \,d^{2} e x}{f^{2}}+\frac {12 b^{2} e c \,d^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {6 b^{2} c \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}+\frac {3 b \,d^{3} a \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f^{2}}-\frac {3 b \,d^{3} a \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{3}}+\frac {3 b a \,c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{f^{2}}-\frac {3 b \,d^{2} c a \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{f^{3}}+\frac {4 b \,e^{3} d^{3} a \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {2 b \,d^{3} a \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{3}}{f}-\frac {6 b a \,c^{2} d \,e^{2}}{f^{2}}+\frac {8 b \,d^{2} c a \,e^{3}}{f^{3}}-\frac {4 b \,d^{3} a \,e^{3} x}{f^{3}}\)	\(905\)

Input:

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

Output:

12/f^2*b*d^2*c*a*e^2*x-12/f*b*a*c^2*d*e*x+12/f^2*b*e*a*c^2*d*ln(exp(f*x+e) 
)-12/f^3*b*e^2*d^2*c*a*ln(exp(f*x+e))+2/f*b^2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d 
*x+c^3)/(1+exp(2*f*x+2*e))+6/f*b*d^2*c*a*ln(1+exp(2*f*x+2*e))*x^2+6/f^2*b* 
d^2*c*a*polylog(2,-exp(2*f*x+2*e))*x+6/f*b*a*c^2*d*ln(1+exp(2*f*x+2*e))*x- 
3/2*b^2*d^3*polylog(3,-exp(2*f*x+2*e))/f^4+3/2*a*b*d^3*polylog(4,-exp(2*f* 
x+2*e))/f^4-2*d^2*a*b*c*x^3-3*d*a*b*c^2*x^2+2*a*b*c^3*x+1/4*d^3*a^2*x^4+1/ 
4*d^3*b^2*x^4+1/4/d*a^2*c^4+1/4/d*b^2*c^4-1/2*d^3*a*b*x^4+d^2*a^2*c*x^3+d^ 
2*b^2*c*x^3+3/2*d*a^2*c^2*x^2+3/2*d*b^2*c^2*x^2+a^2*c^3*x+b^2*c^3*x+1/2/d* 
a*b*c^4+6/f^3*b^2*d^3*e^2*x+3/f^3*b^2*c*d^2*polylog(2,-exp(2*f*x+2*e))-6/f 
^4*b^2*e^2*d^3*ln(exp(f*x+e))+3/f^2*b^2*d^3*ln(1+exp(2*f*x+2*e))*x^2+3/f^3 
*b^2*d^3*polylog(2,-exp(2*f*x+2*e))*x-4/f*b*a*c^3*ln(exp(f*x+e))+2/f*b*a*c 
^3*ln(1+exp(2*f*x+2*e))-6/f^2*b^2*c^2*d*ln(exp(f*x+e))+3/f^2*b^2*c^2*d*ln( 
1+exp(2*f*x+2*e))-2/f*b^2*d^3*x^3+4/f^4*b^2*d^3*e^3-3/f^4*b*d^3*a*e^4-6/f* 
b^2*c*d^2*x^2-6/f^3*b^2*c*d^2*e^2-12/f^2*b^2*c*d^2*e*x+12/f^3*b^2*e*c*d^2* 
ln(exp(f*x+e))+6/f^2*b^2*c*d^2*ln(1+exp(2*f*x+2*e))*x+3/f^2*b*d^3*a*polylo 
g(2,-exp(2*f*x+2*e))*x^2-3/f^3*b*d^3*a*polylog(3,-exp(2*f*x+2*e))*x+3/f^2* 
b*a*c^2*d*polylog(2,-exp(2*f*x+2*e))-3/f^3*b*d^2*c*a*polylog(3,-exp(2*f*x+ 
2*e))+4/f^4*b*e^3*d^3*a*ln(exp(f*x+e))+2/f*b*d^3*a*ln(1+exp(2*f*x+2*e))*x^ 
3-6/f^2*b*a*c^2*d*e^2+8/f^3*b*d^2*c*a*e^3-4/f^3*b*d^3*a*e^3*x
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 3744, normalized size of antiderivative = 13.52 \[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\int \left (a + b \tanh {\left (e + f x \right )}\right )^{2} \left (c + d x\right )^{3}\, dx \] Input:

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

Output:

Integral((a + b*tanh(e + f*x))**2*(c + d*x)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 625 vs. \(2 (265) = 530\).

Time = 0.16 (sec) , antiderivative size = 625, normalized size of antiderivative = 2.26 \[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*a^2*d^3*x^4 + a^2*c*d^2*x^3 + 3/2*a^2*c^2*d*x^2 + b^2*c^3*(x + e/f - 2 
/(f*(e^(-2*f*x - 2*e) + 1))) + a^2*c^3*x + 3/2*b^2*c^2*d*((f*x^2 + (f*x^2* 
e^(2*e) - 4*x*e^(2*e))*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) + 2*log((e^(2*f* 
x + 2*e) + 1)*e^(-2*e))/f^2) + 2*a*b*c^3*log(cosh(f*x + e))/f + 2/3*(4*f^3 
*x^3*log(e^(2*f*x + 2*e) + 1) + 6*f^2*x^2*dilog(-e^(2*f*x + 2*e)) - 6*f*x* 
polylog(3, -e^(2*f*x + 2*e)) + 3*polylog(4, -e^(2*f*x + 2*e)))*a*b*d^3/f^4 
 + 1/4*((2*a*b*d^3*f + b^2*d^3*f)*x^4 + 4*(2*a*b*c*d^2*f + (c*d^2*f + 2*d^ 
3)*b^2)*x^3 + 12*(a*b*c^2*d*f + 2*b^2*c*d^2)*x^2 + (12*a*b*c^2*d*f*x^2*e^( 
2*e) + (2*a*b*d^3*f*e^(2*e) + b^2*d^3*f*e^(2*e))*x^4 + 4*(2*a*b*c*d^2*f*e^ 
(2*e) + b^2*c*d^2*f*e^(2*e))*x^3)*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) + 3*( 
a*b*c^2*d*f + b^2*c*d^2)*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x 
 + 2*e)))/f^3 + 3/2*(2*a*b*c*d^2*f + b^2*d^3)*(2*f^2*x^2*log(e^(2*f*x + 2* 
e) + 1) + 2*f*x*dilog(-e^(2*f*x + 2*e)) - polylog(3, -e^(2*f*x + 2*e)))/f^ 
4 - (a*b*d^3*f^4*x^4 + 2*(2*a*b*c*d^2*f + b^2*d^3)*f^3*x^3 + 6*(a*b*c^2*d* 
f^2 + b^2*c*d^2*f)*f^2*x^2)/f^4
 

Giac [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\int { {\left (d x + c\right )}^{3} {\left (b \tanh \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\int {\left (a+b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^2\,{\left (c+d\,x\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 (a+b \tanh (e+f x))^2 \, dx=\text {too large to display} \] Input:

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

Output:

( - 16*e**(2*e + 2*f*x)*int(x**3/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 
1),x)*a*b*d**3*f**4 - 48*e**(2*e + 2*f*x)*int(x**2/(e**(4*e + 4*f*x) + 2*e 
**(2*e + 2*f*x) + 1),x)*a*b*c*d**2*f**4 - 24*e**(2*e + 2*f*x)*int(x**2/(e* 
*(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x)*a*b*d**3*f**3 - 24*e**(2*e + 2 
*f*x)*int(x**2/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x)*b**2*d**3*f* 
*3 - 48*e**(2*e + 2*f*x)*int(x/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1) 
,x)*a*b*c**2*d*f**4 - 48*e**(2*e + 2*f*x)*int(x/(e**(4*e + 4*f*x) + 2*e**( 
2*e + 2*f*x) + 1),x)*a*b*c*d**2*f**3 - 24*e**(2*e + 2*f*x)*int(x/(e**(4*e 
+ 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x)*a*b*d**3*f**2 - 48*e**(2*e + 2*f*x)* 
int(x/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x)*b**2*c*d**2*f**3 - 24 
*e**(2*e + 2*f*x)*int(x/(e**(4*e + 4*f*x) + 2*e**(2*e + 2*f*x) + 1),x)*b** 
2*d**3*f**2 + 8*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*a*b*c**3*f**3 + 
 12*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*a*b*c**2*d*f**2 + 12*e**(2* 
e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*a*b*c*d**2*f + 6*e**(2*e + 2*f*x)*log 
(e**(2*e + 2*f*x) + 1)*a*b*d**3 + 12*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) 
 + 1)*b**2*c**2*d*f**2 + 12*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*b** 
2*c*d**2*f + 6*e**(2*e + 2*f*x)*log(e**(2*e + 2*f*x) + 1)*b**2*d**3 + 4*e* 
*(2*e + 2*f*x)*a**2*c**3*f**4*x + 6*e**(2*e + 2*f*x)*a**2*c**2*d*f**4*x**2 
 + 4*e**(2*e + 2*f*x)*a**2*c*d**2*f**4*x**3 + e**(2*e + 2*f*x)*a**2*d**3*f 
**4*x**4 - 8*e**(2*e + 2*f*x)*a*b*c**3*f**4*x + 12*e**(2*e + 2*f*x)*a*b...