3.68 \(\int \frac{(c+d x)^3}{a+b \tanh (e+f x)} \, dx\)

Optimal. Leaf size=212 \[ \frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^3 \left (a^2-b^2\right )}+\frac{3 b d (c+d x)^2 \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^2 \left (a^2-b^2\right )}+\frac{3 b d^3 \text{PolyLog}\left (4,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{4 f^4 \left (a^2-b^2\right )}-\frac{b (c+d x)^3 \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f \left (a^2-b^2\right )}+\frac{(c+d x)^4}{4 d (a+b)} \]

[Out]

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

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Rubi [A]  time = 0.340666, antiderivative size = 212, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3732, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 b d^2 (c+d x) \text{PolyLog}\left (3,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^3 \left (a^2-b^2\right )}+\frac{3 b d (c+d x)^2 \text{PolyLog}\left (2,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 f^2 \left (a^2-b^2\right )}+\frac{3 b d^3 \text{PolyLog}\left (4,-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{4 f^4 \left (a^2-b^2\right )}-\frac{b (c+d x)^3 \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f \left (a^2-b^2\right )}+\frac{(c+d x)^4}{4 d (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3732

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[((c + d*x)^m*E^Simp[2*I*(e + f*x), x])/((a + I*b)^2 + (a^2 + b^2)*E^S
imp[2*I*(e + f*x), x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0] && 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 \frac{(c+d x)^3}{a+b \tanh (e+f x)} \, dx &=\frac{(c+d x)^4}{4 (a+b) d}+(2 b) \int \frac{e^{-2 (e+f x)} (c+d x)^3}{(a+b)^2+\left (a^2-b^2\right ) e^{-2 (e+f x)}} \, dx\\ &=\frac{(c+d x)^4}{4 (a+b) d}-\frac{b (c+d x)^3 \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{(3 b d) \int (c+d x)^2 \log \left (1+\frac{\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f}\\ &=\frac{(c+d x)^4}{4 (a+b) d}-\frac{b (c+d x)^3 \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2}-\frac{\left (3 b d^2\right ) \int (c+d x) \text{Li}_2\left (-\frac{\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{\left (a^2-b^2\right ) f^2}\\ &=\frac{(c+d x)^4}{4 (a+b) d}-\frac{b (c+d x)^3 \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3}-\frac{\left (3 b d^3\right ) \int \text{Li}_3\left (-\frac{\left (a^2-b^2\right ) e^{-2 (e+f x)}}{(a+b)^2}\right ) \, dx}{2 \left (a^2-b^2\right ) f^3}\\ &=\frac{(c+d x)^4}{4 (a+b) d}-\frac{b (c+d x)^3 \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3}+\frac{\left (3 b d^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{(-a+b) x}{a+b}\right )}{x} \, dx,x,e^{-2 (e+f x)}\right )}{4 \left (a^2-b^2\right ) f^4}\\ &=\frac{(c+d x)^4}{4 (a+b) d}-\frac{b (c+d x)^3 \log \left (1+\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{\left (a^2-b^2\right ) f}+\frac{3 b d (c+d x)^2 \text{Li}_2\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^2}+\frac{3 b d^2 (c+d x) \text{Li}_3\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{2 \left (a^2-b^2\right ) f^3}+\frac{3 b d^3 \text{Li}_4\left (-\frac{(a-b) e^{-2 (e+f x)}}{a+b}\right )}{4 \left (a^2-b^2\right ) f^4}\\ \end{align*}

Mathematica [A]  time = 3.41063, size = 239, normalized size = 1.13 \[ \frac{b \left (\frac{3 d \left (2 f^2 (c+d x)^2 \text{PolyLog}\left (2,\frac{(b-a) e^{-2 (e+f x)}}{a+b}\right )+d \left (2 f (c+d x) \text{PolyLog}\left (3,\frac{(b-a) e^{-2 (e+f x)}}{a+b}\right )+d \text{PolyLog}\left (4,\frac{(b-a) e^{-2 (e+f x)}}{a+b}\right )\right )\right )}{f^4 (a-b)}-\frac{4 (c+d x)^3 \log \left (\frac{(a-b) e^{-2 (e+f x)}}{a+b}+1\right )}{f (a-b)}-\frac{2 (c+d x)^4}{d \left (a \left (e^{2 e}+1\right )+b \left (e^{2 e}-1\right )\right )}\right )}{4 (a+b)}+\frac{x \cosh (e) \left (6 c^2 d x+4 c^3+4 c d^2 x^2+d^3 x^3\right )}{4 (a \cosh (e)+b \sinh (e))} \]

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

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Maple [B]  time = 0.203, size = 1143, normalized size = 5.4 \begin{align*} \text{result too large to display} \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]

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

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Maxima [B]  time = 1.68302, size = 713, normalized size = 3.36 \begin{align*} -\frac{3 \,{\left (2 \, f x \log \left (\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) +{\rm Li}_2\left (-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right )\right )} b c^{2} d}{2 \,{\left (a^{2} f^{2} - b^{2} f^{2}\right )}} - \frac{3 \,{\left (2 \, f^{2} x^{2} \log \left (\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + 2 \, f x{\rm Li}_2\left (-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right ) -{\rm Li}_{3}(-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b})\right )} b c d^{2}}{2 \,{\left (a^{2} f^{3} - b^{2} f^{3}\right )}} - \frac{{\left (4 \, f^{3} x^{3} \log \left (\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b} + 1\right ) + 6 \, f^{2} x^{2}{\rm Li}_2\left (-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}\right ) - 6 \, f x{\rm Li}_{3}(-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b}) + 3 \,{\rm Li}_{4}(-\frac{{\left (a e^{\left (2 \, e\right )} + b e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{a - b})\right )} b d^{3}}{3 \,{\left (a^{2} f^{4} - b^{2} f^{4}\right )}} - c^{3}{\left (\frac{b \log \left (-{\left (a - b\right )} e^{\left (-2 \, f x - 2 \, e\right )} - a - b\right )}{{\left (a^{2} - b^{2}\right )} f} - \frac{f x + e}{{\left (a + b\right )} f}\right )} + \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 \,{\left (a^{2} f^{4} - b^{2} f^{4}\right )}} + \frac{d^{3} x^{4} + 4 \, c d^{2} x^{3} + 6 \, c^{2} d x^{2}}{4 \,{\left (a + b\right )}} \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]

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

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Fricas [C]  time = 2.45542, size = 1821, normalized size = 8.59 \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/(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, sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*x + e))) - 24*b*d^3*polylog(4, -sqrt(-(a + b)/(a - b
))*(cosh(f*x + e) + 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(sqrt(-(a + b)/(
a - b))*(cosh(f*x + e) + 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(-sqrt(-(a
+ b)/(a - b))*(cosh(f*x + e) + 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(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) + 2*(a - b)*sqrt(-(a + b)/(a - b))) + 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(2*(a + b)*cosh(f*x + e) + 2*(a + b)*sinh(f*x + e) - 2*(a - b)
*sqrt(-(a + b)/(a - b))) - 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(sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + 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(-sqrt(-(a + b)/(a - b))*(
cosh(f*x + e) + sinh(f*x + e)) + 1) + 24*(b*d^3*f*x + b*c*d^2*f)*polylog(3, sqrt(-(a + b)/(a - b))*(cosh(f*x +
e) + sinh(f*x + e))) + 24*(b*d^3*f*x + b*c*d^2*f)*polylog(3, -sqrt(-(a + b)/(a - b))*(cosh(f*x + e) + sinh(f*
x + e))))/((a^2 - b^2)*f^4)

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

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