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

Optimal. Leaf size=137 \[ -\frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 b d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac{3 b d^3 \text{PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}+\frac{a (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{b (c+d x)^4}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.26511, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {3722, 3718, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 b d (c+d x)^2 \text{PolyLog}\left (2,-e^{2 (e+f x)}\right )}{2 f^2}+\frac{3 b d^3 \text{PolyLog}\left (4,-e^{2 (e+f x)}\right )}{4 f^4}+\frac{a (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (e^{2 (e+f x)}+1\right )}{f}-\frac{b (c+d x)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3722

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]

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 6609

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] - Dist[(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]

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]

Rubi steps

\begin{align*} \int (c+d x)^3 (a+b \tanh (e+f x)) \, dx &=\int \left (a (c+d x)^3+b (c+d x)^3 \tanh (e+f x)\right ) \, dx\\ &=\frac{a (c+d x)^4}{4 d}+b \int (c+d x)^3 \tanh (e+f x) \, dx\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+(2 b) \int \frac{e^{2 (e+f x)} (c+d x)^3}{1+e^{2 (e+f x)}} \, dx\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}-\frac{(3 b d) \int (c+d x)^2 \log \left (1+e^{2 (e+f x)}\right ) \, dx}{f}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{\left (3 b d^2\right ) \int (c+d x) \text{Li}_2\left (-e^{2 (e+f x)}\right ) \, dx}{f^2}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 b d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{\left (3 b d^3\right ) \int \text{Li}_3\left (-e^{2 (e+f x)}\right ) \, dx}{2 f^3}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 b d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{\left (3 b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 (e+f x)}\right )}{4 f^4}\\ &=\frac{a (c+d x)^4}{4 d}-\frac{b (c+d x)^4}{4 d}+\frac{b (c+d x)^3 \log \left (1+e^{2 (e+f x)}\right )}{f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-e^{2 (e+f x)}\right )}{2 f^2}-\frac{3 b d^2 (c+d x) \text{Li}_3\left (-e^{2 (e+f x)}\right )}{2 f^3}+\frac{3 b d^3 \text{Li}_4\left (-e^{2 (e+f x)}\right )}{4 f^4}\\ \end{align*}

Mathematica [A]  time = 2.09144, size = 161, normalized size = 1.18 \[ \frac{1}{4} \left (b \left (-\frac{3 d \left (2 f^2 (c+d x)^2 \text{PolyLog}\left (2,-e^{-2 (e+f x)}\right )+d \left (2 f (c+d x) \text{PolyLog}\left (3,-e^{-2 (e+f x)}\right )+d \text{PolyLog}\left (4,-e^{-2 (e+f x)}\right )\right )\right )}{f^4}+\frac{4 (c+d x)^3 \log \left (e^{-2 (e+f x)}+1\right )}{f}+\frac{2 (c+d x)^4}{d \left (e^{2 e}+1\right )}\right )+x \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right ) (a+b \tanh (e))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*((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)*(a + b*Tanh[e]))/4

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Maple [B]  time = 0.063, size = 452, normalized size = 3.3 \begin{align*} -3\,{\frac{b{c}^{2}d{e}^{2}}{{f}^{2}}}+4\,{\frac{bc{d}^{2}{e}^{3}}{{f}^{3}}}-2\,{\frac{b{d}^{3}{e}^{3}x}{{f}^{3}}}+{\frac{b{d}^{3}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{3}}{f}}+{\frac{3\,b{d}^{3}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ){x}^{2}}{2\,{f}^{2}}}-{\frac{3\,b{d}^{3}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{2\,{f}^{3}}}+2\,{\frac{b{d}^{3}{e}^{3}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{4}}}+{\frac{3\,b{c}^{2}d{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{2}}}-{\frac{3\,bc{d}^{2}{\it polylog} \left ( 3,-{{\rm e}^{2\,fx+2\,e}} \right ) }{2\,{f}^{3}}}+6\,{\frac{bc{d}^{2}{e}^{2}x}{{f}^{2}}}+{\frac{a{d}^{3}{x}^{4}}{4}}+ac{d}^{2}{x}^{3}-bc{d}^{2}{x}^{3}-6\,{\frac{b{c}^{2}dex}{f}}+3\,{\frac{bc{d}^{2}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ){x}^{2}}{f}}+3\,{\frac{bc{d}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,fx+2\,e}} \right ) x}{{f}^{2}}}+3\,{\frac{b{c}^{2}d\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) x}{f}}+{\frac{3\,a{c}^{2}d{x}^{2}}{2}}-{\frac{3\,b{c}^{2}d{x}^{2}}{2}}-{\frac{3\,b{d}^{3}{e}^{4}}{2\,{f}^{4}}}+{\frac{b{c}^{3}\ln \left ({{\rm e}^{2\,fx+2\,e}}+1 \right ) }{f}}-2\,{\frac{b{c}^{3}\ln \left ({{\rm e}^{fx+e}} \right ) }{f}}+6\,{\frac{b{c}^{2}de\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{2}}}-6\,{\frac{bc{d}^{2}{e}^{2}\ln \left ({{\rm e}^{fx+e}} \right ) }{{f}^{3}}}+{\frac{3\,b{d}^{3}{\it polylog} \left ( 4,-{{\rm e}^{2\,fx+2\,e}} \right ) }{4\,{f}^{4}}}-{\frac{b{d}^{3}{x}^{4}}{4}}+{c}^{3}ax+b{c}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.42192, size = 410, normalized size = 2.99 \begin{align*} \frac{1}{4} \, a d^{3} x^{4} + \frac{1}{4} \, b d^{3} x^{4} + a c d^{2} x^{3} + b c d^{2} x^{3} + \frac{3}{2} \, a c^{2} d x^{2} + \frac{3}{2} \, b c^{2} d x^{2} + a c^{3} x + \frac{b c^{3} \log \left (\cosh \left (f x + e\right )\right )}{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 )} b 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 )} b 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 )} b d^{3}}{3 \, f^{4}} - \frac{b d^{3} f^{4} x^{4} + 4 \, b c d^{2} f^{4} x^{3} + 6 \, b c^{2} d f^{4} x^{2}}{2 \, f^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C]  time = 2.6661, size = 1438, normalized size = 10.5 \begin{align*} \frac{{\left (a - b\right )} d^{3} f^{4} x^{4} + 4 \,{\left (a - b\right )} c d^{2} f^{4} x^{3} + 6 \,{\left (a - b\right )} c^{2} d f^{4} x^{2} + 4 \,{\left (a - b\right )} c^{3} f^{4} x + 24 \, b d^{3}{\rm polylog}\left (4, i \, \cosh \left (f x + e\right ) + i \, \sinh \left (f x + e\right )\right ) + 24 \, b d^{3}{\rm polylog}\left (4, -i \, \cosh \left (f x + e\right ) - i \, \sinh \left (f x + e\right )\right ) + 12 \,{\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b 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 (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b 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 (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) + i\right ) - 4 \,{\left (b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b c^{2} d e f^{2} - b c^{3} f^{3}\right )} \log \left (\cosh \left (f x + e\right ) + \sinh \left (f x + e\right ) - i\right ) + 4 \,{\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b 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 (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + b d^{3} e^{3} - 3 \, b c d^{2} e^{2} f + 3 \, b 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 (b d^{3} f x + b 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 (b d^{3} f x + b 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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