\(\int (c+d x)^3 \tanh (e+f x) \, dx\) [1]

Optimal result

Integrand size = 14, antiderivative size = 117 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=-\frac {(c+d x)^4}{4 d}+\frac {(c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac {3 d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}-\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac {3 d^3 \operatorname {PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.32 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=\frac {1}{4} \left (\frac {2 (c+d x)^4}{d \left (1+e^{2 e}\right )}+\frac {4 (c+d x)^3 \log \left (1+e^{-2 (e+f x)}\right )}{f}-\frac {3 d \left (2 f^2 (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \operatorname {PolyLog}\left (3,-e^{-2 (e+f x)}\right )+d \operatorname {PolyLog}\left (4,-e^{-2 (e+f x)}\right )\right )\right )}{f^4}+x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \tanh (e)\right ) \] Input:

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

Output:

((2*(c + d*x)^4)/(d*(1 + E^(2*e))) + (4*(c + d*x)^3*Log[1 + E^(-2*(e + f*x 
))])/f - (3*d*(2*f^2*(c + d*x)^2*PolyLog[2, -E^(-2*(e + f*x))] + d*(2*f*(c 
 + d*x)*PolyLog[3, -E^(-2*(e + f*x))] + d*PolyLog[4, -E^(-2*(e + f*x))]))) 
/f^4 + x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3)*Tanh[e])/4
 

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.73 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.20, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 26, 4201, 2620, 3011, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4201

\(\displaystyle -i \left (2 i \int \frac {e^{2 (e+f x)} (c+d x)^3}{1+e^{2 (e+f x)}}dx-\frac {i (c+d x)^4}{4 d}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle -i \left (2 i \left (\frac {(c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{2 f}-\frac {3 d \int (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )dx}{2 f}\right )-\frac {i (c+d x)^4}{4 d}\right )\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

(-I)*(((-1/4*I)*(c + d*x)^4)/d + (2*I)*(((c + d*x)^3*Log[1 + E^(2*(e + f*x 
))])/(2*f) - (3*d*(-1/2*((c + d*x)^2*PolyLog[2, -E^(2*(e + f*x))])/f + (d* 
(((c + d*x)*PolyLog[3, -E^(2*(e + f*x))])/(2*f) - (d*PolyLog[4, -E^(2*(e + 
 f*x))])/(4*f^2)))/f))/(2*f)))
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4201

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x 
_Symbol] :> Simp[(-I)*((c + d*x)^(m + 1)/(d*(m + 1))), x] + Simp[2*I   Int[ 
(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x], x] 
 /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(109)=218\).

Time = 0.83 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.44

method	result	size
risch	\(-\frac {6 c^{2} d e x}{f}-\frac {3 d^{3} e^{4}}{2 f^{4}}+\frac {c^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right )}{f}-\frac {2 c^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f}-d^{2} c \,x^{3}-\frac {3 d \,c^{2} x^{2}}{2}+c^{3} x -\frac {3 c^{2} d \,e^{2}}{f^{2}}+\frac {4 c \,d^{2} e^{3}}{f^{3}}+\frac {d^{3} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{3}}{f}+\frac {3 d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{2 f^{2}}-\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right ) x}{2 f^{3}}+\frac {2 d^{3} e^{3} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{4}}+\frac {6 c \,d^{2} e^{2} x}{f^{2}}-\frac {2 d^{3} e^{3} x}{f^{3}}+\frac {3 c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{2}}-\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 f x +2 e}\right )}{2 f^{3}}+\frac {3 c \,d^{2} \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x^{2}}{f}+\frac {3 c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 f x +2 e}\right ) x}{f^{2}}-\frac {6 c \,d^{2} e^{2} \ln \left ({\mathrm e}^{f x +e}\right )}{f^{3}}+\frac {6 c^{2} d e \ln \left ({\mathrm e}^{f x +e}\right )}{f^{2}}+\frac {3 c^{2} d \ln \left (1+{\mathrm e}^{2 f x +2 e}\right ) x}{f}-\frac {d^{3} x^{4}}{4}+\frac {c^{4}}{4 d}+\frac {3 d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 f x +2 e}\right )}{4 f^{4}}\)	\(402\)

Input:

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

Output:

-6/f*c^2*d*e*x-3/2/f^4*d^3*e^4+1/f*c^3*ln(1+exp(2*f*x+2*e))-2/f*c^3*ln(exp 
(f*x+e))-d^2*c*x^3-3/2*d*c^2*x^2+c^3*x-3/f^2*c^2*d*e^2+4/f^3*c*d^2*e^3+1/f 
*d^3*ln(1+exp(2*f*x+2*e))*x^3+3/2/f^2*d^3*polylog(2,-exp(2*f*x+2*e))*x^2-3 
/2/f^3*d^3*polylog(3,-exp(2*f*x+2*e))*x+2/f^4*d^3*e^3*ln(exp(f*x+e))+6/f^2 
*c*d^2*e^2*x-2/f^3*d^3*e^3*x+3/2/f^2*c^2*d*polylog(2,-exp(2*f*x+2*e))-3/2/ 
f^3*c*d^2*polylog(3,-exp(2*f*x+2*e))+3/f*c*d^2*ln(1+exp(2*f*x+2*e))*x^2+3/ 
f^2*c*d^2*polylog(2,-exp(2*f*x+2*e))*x-6/f^3*c*d^2*e^2*ln(exp(f*x+e))+6/f^ 
2*c^2*d*e*ln(exp(f*x+e))+3/f*c^2*d*ln(1+exp(2*f*x+2*e))*x-1/4*d^3*x^4+1/4/ 
d*c^4+3/4*d^3*polylog(4,-exp(2*f*x+2*e))/f^4
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 531, normalized size of antiderivative = 4.54 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=-\frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x - 24 \, d^{3} {\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 24 \, d^{3} {\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} {\rm Li}_2\left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} {\rm Li}_2\left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 4 \, {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) + 4 \, {\left (d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2} - c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right ) + 1\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + d^{3} e^{3} - 3 \, c d^{2} e^{2} f + 3 \, c^{2} d e f^{2}\right )} \log \left (-i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right ) + 1\right ) + 24 \, {\left (d^{3} f x + c d^{2} f\right )} {\rm polylog}\left (3, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \, {\left (d^{3} f x + c d^{2} f\right )} {\rm polylog}\left (3, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right )}{4 \, f^{4}} \] Input:

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

Output:

-1/4*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2 + 4*c^3*f^4*x - 24*d 
^3*polylog(4, I*cosh(f*x + e) + I*sinh(f*x + e)) - 24*d^3*polylog(4, -I*co 
sh(f*x + e) - I*sinh(f*x + e)) - 12*(d^3*f^2*x^2 + 2*c*d^2*f^2*x + c^2*d*f 
^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) - 12*(d^3*f^2*x^2 + 2*c*d^2*f 
^2*x + c^2*d*f^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 4*(d^3*e^3 - 
 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*log(cosh(f*x + e) + sinh(f*x + e 
) + I) + 4*(d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 - c^3*f^3)*log(cosh(f* 
x + e) + sinh(f*x + e) - I) - 4*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f 
^3*x + d^3*e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2)*log(I*cosh(f*x + e) + I*si 
nh(f*x + e) + 1) - 4*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + d^3* 
e^3 - 3*c*d^2*e^2*f + 3*c^2*d*e*f^2)*log(-I*cosh(f*x + e) - I*sinh(f*x + e 
) + 1) + 24*(d^3*f*x + c*d^2*f)*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + 
e)) + 24*(d^3*f*x + c*d^2*f)*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e) 
))/f^4
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (108) = 216\).

Time = 0.19 (sec) , antiderivative size = 286, normalized size of antiderivative = 2.44 \[ \int (c+d x)^3 \tanh (e+f x) \, dx=\frac {1}{4} \, d^{3} x^{4} + c d^{2} x^{3} + \frac {3}{2} \, c^{2} d x^{2} + \frac {c^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}{2 \, f} + \frac {c^{3} \log \left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}{2 \, f} + \frac {3 \, {\left (2 \, f x \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right )\right )} c^{2} d}{2 \, f^{2}} + \frac {3 \, {\left (2 \, f^{2} x^{2} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 2 \, f x {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} c d^{2}}{2 \, f^{3}} + \frac {{\left (4 \, f^{3} x^{3} \log \left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right ) + 6 \, f^{2} x^{2} {\rm Li}_2\left (-e^{\left (2 \, f x + 2 \, e\right )}\right ) - 6 \, f x {\rm Li}_{3}(-e^{\left (2 \, f x + 2 \, e\right )}) + 3 \, {\rm Li}_{4}(-e^{\left (2 \, f x + 2 \, e\right )})\right )} d^{3}}{3 \, f^{4}} - \frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2}}{2 \, f^{4}} \] Input:

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

Output:

1/4*d^3*x^4 + c*d^2*x^3 + 3/2*c^2*d*x^2 + 1/2*c^3*log(e^(2*f*x + 2*e) + 1) 
/f + 1/2*c^3*log(e^(-2*f*x - 2*e) + 1)/f + 3/2*(2*f*x*log(e^(2*f*x + 2*e) 
+ 1) + dilog(-e^(2*f*x + 2*e)))*c^2*d/f^2 + 3/2*(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)))* 
c*d^2/f^3 + 1/3*(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)))*d^3/f^4 - 1/2*(d^3*f^4*x^4 + 4*c*d^2*f^4*x^3 + 6*c^2*d*f^4*x^2) 
/f^4
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \tanh (e+f x) \, dx=\frac {8 e^{2 e} \left (\int \frac {e^{2 f x} x^{3}}{e^{2 f x +2 e}+1}d x \right ) d^{3} f +24 e^{2 e} \left (\int \frac {e^{2 f x} x^{2}}{e^{2 f x +2 e}+1}d x \right ) c \,d^{2} f +24 e^{2 e} \left (\int \frac {e^{2 f x} x}{e^{2 f x +2 e}+1}d x \right ) c^{2} d f +4 \,\mathrm {log}\left (e^{2 f x +2 e}+1\right ) c^{3}-4 c^{3} f x -6 c^{2} d f \,x^{2}-4 c \,d^{2} f \,x^{3}-d^{3} f \,x^{4}}{4 f} \] Input:

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

Output:

(8*e**(2*e)*int((e**(2*f*x)*x**3)/(e**(2*e + 2*f*x) + 1),x)*d**3*f + 24*e* 
*(2*e)*int((e**(2*f*x)*x**2)/(e**(2*e + 2*f*x) + 1),x)*c*d**2*f + 24*e**(2 
*e)*int((e**(2*f*x)*x)/(e**(2*e + 2*f*x) + 1),x)*c**2*d*f + 4*log(e**(2*e 
+ 2*f*x) + 1)*c**3 - 4*c**3*f*x - 6*c**2*d*f*x**2 - 4*c*d**2*f*x**3 - d**3 
*f*x**4)/(4*f)