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

Optimal. Leaf size=119 \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}-\frac{3 d^3 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^4}+\frac{3 d (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{(c+d x)^3 \tanh (e+f x)}{f}-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.209542, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3720, 3718, 2190, 2531, 2282, 6589, 32} \[ \frac{3 d^2 (c+d x) \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{f^3}-\frac{3 d^3 \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^4}+\frac{3 d (c+d x)^2 \log \left (e^{2 (e+f x)}+1\right )}{f^2}-\frac{(c+d x)^3 \tanh (e+f x)}{f}-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^3*Tanh[e + f*x]^2,x]

[Out]

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

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3718

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] + Dist[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 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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 2531

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] + Dist[(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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin{align*} \int (c+d x)^3 \tanh ^2(e+f x) \, dx &=-\frac{(c+d x)^3 \tanh (e+f x)}{f}+\frac{(3 d) \int (c+d x)^2 \tanh (e+f x) \, dx}{f}+\int (c+d x)^3 \, dx\\ &=-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d}-\frac{(c+d x)^3 \tanh (e+f x)}{f}+\frac{(6 d) \int \frac{e^{2 (e+f x)} (c+d x)^2}{1+e^{2 (e+f x)}} \, dx}{f}\\ &=-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}-\frac{(c+d x)^3 \tanh (e+f x)}{f}-\frac{\left (6 d^2\right ) \int (c+d x) \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^3}-\frac{(c+d x)^3 \tanh (e+f x)}{f}-\frac{\left (3 d^3\right ) \int \text{Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^3}\\ &=-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^3}-\frac{(c+d x)^3 \tanh (e+f x)}{f}-\frac{\left (3 d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{2 f^4}\\ &=-\frac{(c+d x)^3}{f}+\frac{(c+d x)^4}{4 d}+\frac{3 d (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right )}{f^2}+\frac{3 d^2 (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right )}{f^3}-\frac{3 d^3 \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^4}-\frac{(c+d x)^3 \tanh (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 2.02132, size = 179, normalized size = 1.5 \[ \frac{1}{4} \left (\frac{2 d e^{2 e} \left (-\frac{3 d \left (e^{-2 e}+1\right ) \left (2 f (c+d x) \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )\right )}{f^3}+\frac{6 \left (e^{-2 e}+1\right ) (c+d x)^2 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac{4 e^{-2 e} (c+d x)^3}{d}\right )}{\left (e^{2 e}+1\right ) f}+x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )-\frac{4 \text{sech}(e) (c+d x)^3 \sinh (f x) \text{sech}(e+f x)}{f}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.041, size = 328, normalized size = 2.8 \begin{align*}{\frac{{d}^{3}{x}^{4}}{4}}+c{d}^{2}{x}^{3}+{\frac{3\,{c}^{2}d{x}^{2}}{2}}+{c}^{3}x+2\,{\frac{{d}^{3}{x}^{3}+3\,c{d}^{2}{x}^{2}+3\,{c}^{2}dx+{c}^{3}}{f \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }}+3\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{{f}^{2}}}-6\,{\frac{{c}^{2}d\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-6\,{\frac{{d}^{3}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{4}}}-2\,{\frac{{d}^{3}{x}^{3}}{f}}+6\,{\frac{{d}^{3}{e}^{2}x}{{f}^{3}}}+4\,{\frac{{d}^{3}{e}^{3}}{{f}^{4}}}+3\,{\frac{{d}^{3}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{{f}^{2}}}+3\,{\frac{{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{3}}}-{\frac{3\,{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{4}}}+12\,{\frac{c{d}^{2}e\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}-6\,{\frac{c{d}^{2}{x}^{2}}{f}}-12\,{\frac{c{d}^{2}ex}{{f}^{2}}}-6\,{\frac{c{d}^{2}{e}^{2}}{{f}^{3}}}+6\,{\frac{c{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{{f}^{2}}}+3\,{\frac{c{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{{f}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.82238, size = 463, normalized size = 3.89 \begin{align*} c^{3}{\left (x + \frac{e}{f} - \frac{2}{f{\left (e^{\left (-2 \, f x - 2 \, e\right )} + 1\right )}}\right )} - \frac{3}{2} \, c^{2} d{\left (\frac{2 \, x e^{\left (2 \, f x + 2 \, e\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac{f x^{2} +{\left (f x^{2} e^{\left (2 \, e\right )} - 2 \, x e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f e^{\left (2 \, f x + 2 \, e\right )} + f} - \frac{2 \, \log \left ({\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )} e^{\left (-2 \, e\right )}\right )}{f^{2}}\right )} + \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 d^{2}}{f^{3}} + \frac{d^{3} f x^{4} + 24 \, c d^{2} x^{2} + 4 \,{\left (c d^{2} f + 2 \, d^{3}\right )} x^{3} +{\left (d^{3} f x^{4} e^{\left (2 \, e\right )} + 4 \, c d^{2} f x^{3} e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{4 \,{\left (f e^{\left (2 \, f x + 2 \, e\right )} + f\right )}} + \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 )} d^{3}}{2 \, f^{4}} - \frac{2 \,{\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2}\right )}}{f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(x + e/f - 2/(f*(e^(-2*f*x - 2*e) + 1))) - 3/2*c^2*d*(2*x*e^(2*f*x + 2*e)/(f*e^(2*f*x + 2*e) + f) - (f*x^2
+ (f*x^2*e^(2*e) - 2*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) + 3*(2*f*x*log(e^(2*f*x + 2*e) + 1) + dilog(-e^(2*f*x + 2*e)))*c*d^2/f^3 + 1/4*(d^3*f*x^4 + 24*c*d^2*x^2 +
4*(c*d^2*f + 2*d^3)*x^3 + (d^3*f*x^4*e^(2*e) + 4*c*d^2*f*x^3*e^(2*e))*e^(2*f*x))/(f*e^(2*f*x + 2*e) + f) + 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)))*d^3/f^4 -
2*(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2)/f^4

________________________________________________________________________________________

Fricas [C]  time = 1.90288, size = 3497, normalized size = 29.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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 - 8*d^3*e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f
^2 + 8*c^3*f^3 + (d^3*f^4*x^4 - 8*d^3*e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f^2 + 4*(c*d^2*f^4 - 2*d^3*f^3)*x^3 +
6*(c^2*d*f^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4 - 6*c^2*d*f^3)*x)*cosh(f*x + e)^2 + 2*(d^3*f^4*x^4 - 8*d^3*e^3 +
24*c*d^2*e^2*f - 24*c^2*d*e*f^2 + 4*(c*d^2*f^4 - 2*d^3*f^3)*x^3 + 6*(c^2*d*f^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4
- 6*c^2*d*f^3)*x)*cosh(f*x + e)*sinh(f*x + e) + (d^3*f^4*x^4 - 8*d^3*e^3 + 24*c*d^2*e^2*f - 24*c^2*d*e*f^2 +
4*(c*d^2*f^4 - 2*d^3*f^3)*x^3 + 6*(c^2*d*f^4 - 4*c*d^2*f^3)*x^2 + 4*(c^3*f^4 - 6*c^2*d*f^3)*x)*sinh(f*x + e)^2
+ 24*(d^3*f*x + c*d^2*f + (d^3*f*x + c*d^2*f)*cosh(f*x + e)^2 + 2*(d^3*f*x + c*d^2*f)*cosh(f*x + e)*sinh(f*x
+ e) + (d^3*f*x + c*d^2*f)*sinh(f*x + e)^2)*dilog(I*cosh(f*x + e) + I*sinh(f*x + e)) + 24*(d^3*f*x + c*d^2*f +
(d^3*f*x + c*d^2*f)*cosh(f*x + e)^2 + 2*(d^3*f*x + c*d^2*f)*cosh(f*x + e)*sinh(f*x + e) + (d^3*f*x + c*d^2*f)
*sinh(f*x + e)^2)*dilog(-I*cosh(f*x + e) - I*sinh(f*x + e)) + 12*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 + (d^3*e^2
- 2*c*d^2*e*f + c^2*d*f^2)*cosh(f*x + e)^2 + 2*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cosh(f*x + e)*sinh(f*x + e
) + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*sinh(f*x + e)^2)*log(cosh(f*x + e) + sinh(f*x + e) + I) + 12*(d^3*e^2
- 2*c*d^2*e*f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cosh(f*x + e)^2 + 2*(d^3*e^2 - 2*c*d^2*e*f + c
^2*d*f^2)*cosh(f*x + e)*sinh(f*x + e) + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*sinh(f*x + e)^2)*log(cosh(f*x + e)
+ sinh(f*x + e) - I) + 12*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f + (d^3*f^2*x^2 + 2*c*d^2*f^2*x
- d^3*e^2 + 2*c*d^2*e*f)*cosh(f*x + e)^2 + 2*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cosh(f*x +
e)*sinh(f*x + e) + (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sinh(f*x + e)^2)*log(I*cosh(f*x + e)
+ I*sinh(f*x + e) + 1) + 12*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f + (d^3*f^2*x^2 + 2*c*d^2*f^2
*x - d^3*e^2 + 2*c*d^2*e*f)*cosh(f*x + e)^2 + 2*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cosh(f*x
+ e)*sinh(f*x + e) + (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sinh(f*x + e)^2)*log(-I*cosh(f*x +
e) - I*sinh(f*x + e) + 1) - 24*(d^3*cosh(f*x + e)^2 + 2*d^3*cosh(f*x + e)*sinh(f*x + e) + d^3*sinh(f*x + e)^2
+ d^3)*polylog(3, I*cosh(f*x + e) + I*sinh(f*x + e)) - 24*(d^3*cosh(f*x + e)^2 + 2*d^3*cosh(f*x + e)*sinh(f*x
+ e) + d^3*sinh(f*x + e)^2 + d^3)*polylog(3, -I*cosh(f*x + e) - I*sinh(f*x + e)))/(f^4*cosh(f*x + e)^2 + 2*f^
4*cosh(f*x + e)*sinh(f*x + e) + f^4*sinh(f*x + e)^2 + f^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{3} \tanh ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{3} \tanh \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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