\(\int x^3 \log (d+e (f^{c (a+b x)})^n) \, dx\) [123]

Optimal result

Integrand size = 20, antiderivative size = 193 \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \operatorname {PolyLog}\left (4,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \operatorname {PolyLog}\left (5,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^4 c^4 n^4 \log ^4(f)} \]

[Out]

1/4*x^4*ln(d+e*(f^(c*(b*x+a)))^n)-1/4*x^4*ln(1+e*(f^(c*(b*x+a)))^n/d)-x^3*polylog(2,-e*(f^(c*(b*x+a)))^n/d)/b/
c/n/ln(f)+3*x^2*polylog(3,-e*(f^(c*(b*x+a)))^n/d)/b^2/c^2/n^2/ln(f)^2-6*x*polylog(4,-e*(f^(c*(b*x+a)))^n/d)/b^
3/c^3/n^3/ln(f)^3+6*polylog(5,-e*(f^(c*(b*x+a)))^n/d)/b^4/c^4/n^4/ln(f)^4

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2612, 2611, 6744, 2320, 6724} \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {6 \operatorname {PolyLog}\left (5,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^4 c^4 n^4 \log ^4(f)}-\frac {6 x \operatorname {PolyLog}\left (4,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {1}{4} x^4 \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-\frac {1}{4} x^4 \log \left (\frac {e \left (f^{c (a+b x)}\right )^n}{d}+1\right ) \]

[In]

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

[Out]

(x^4*Log[d + e*(f^(c*(a + b*x)))^n])/4 - (x^4*Log[1 + (e*(f^(c*(a + b*x)))^n)/d])/4 - (x^3*PolyLog[2, -((e*(f^
(c*(a + b*x)))^n)/d)])/(b*c*n*Log[f]) + (3*x^2*PolyLog[3, -((e*(f^(c*(a + b*x)))^n)/d)])/(b^2*c^2*n^2*Log[f]^2
) - (6*x*PolyLog[4, -((e*(f^(c*(a + b*x)))^n)/d)])/(b^3*c^3*n^3*Log[f]^3) + (6*PolyLog[5, -((e*(f^(c*(a + b*x)
))^n)/d)])/(b^4*c^4*n^4*Log[f]^4)

Rule 2320

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 2611

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 2612

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

Rule 6724

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 6744

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )+\int x^3 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx \\ & = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 \int x^2 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx}{b c n \log (f)} \\ & = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 \int x \text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx}{b^2 c^2 n^2 \log ^2(f)} \\ & = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \text {Li}_4\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \int \text {Li}_4\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right ) \, dx}{b^3 c^3 n^3 \log ^3(f)} \\ & = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \text {Li}_4\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \text {Subst}\left (\int \frac {\text {Li}_4\left (-\frac {e x^n}{d}\right )}{x} \, dx,x,f^{c (a+b x)}\right )}{b^4 c^4 n^3 \log ^4(f)} \\ & = \frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \text {Li}_2\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \text {Li}_3\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \text {Li}_4\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \text {Li}_5\left (-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^4 c^4 n^4 \log ^4(f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00 \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{4} x^4 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right )-\frac {1}{4} x^4 \log \left (1+\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )-\frac {x^3 \operatorname {PolyLog}\left (2,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \operatorname {PolyLog}\left (4,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \operatorname {PolyLog}\left (5,-\frac {e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^4 c^4 n^4 \log ^4(f)} \]

[In]

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

[Out]

(x^4*Log[d + e*(f^(c*(a + b*x)))^n])/4 - (x^4*Log[1 + (e*(f^(c*(a + b*x)))^n)/d])/4 - (x^3*PolyLog[2, -((e*(f^
(c*(a + b*x)))^n)/d)])/(b*c*n*Log[f]) + (3*x^2*PolyLog[3, -((e*(f^(c*(a + b*x)))^n)/d)])/(b^2*c^2*n^2*Log[f]^2
) - (6*x*PolyLog[4, -((e*(f^(c*(a + b*x)))^n)/d)])/(b^3*c^3*n^3*Log[f]^3) + (6*PolyLog[5, -((e*(f^(c*(a + b*x)
))^n)/d)])/(b^4*c^4*n^4*Log[f]^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1275\) vs. \(2(189)=378\).

Time = 2.12 (sec) , antiderivative size = 1276, normalized size of antiderivative = 6.61

method	result	size
risch	\(\text {Expression too large to display}\)	\(1276\)

[In]

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

[Out]

1/4*x^4*ln(d+e*(f^(c*(b*x+a)))^n)-3/4/c^4/b^4/ln(f)^4*ln(1+e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*ln(
f^(c*(b*x+a)))^4+1/c^4/b^4/ln(f)^4*ln((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x+a)))^4-
1/4*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x^4-1/4/c^4/b^4/ln(f)^4*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(
f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))^4+6/c^4/b^4/ln(f)^4/n^4*polylog(5,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x
+a)))^n/d)-1/c^4/b^4/ln(f)^4/n*polylog(2,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*ln(f^(c*(b*x+a)))^3-
3/c^2/b^2/ln(f)^2/n*polylog(2,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*ln(f^(c*(b*x+a)))*x^2+3/c^3/b^3
/ln(f)^3/n*polylog(2,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*ln(f^(c*(b*x+a)))^2*x+3/c^2/b^2/ln(f)^2/
n*dilog((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x+a)))*x^2-3/c^3/b^3/ln(f)^3/n*dilog((d
+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x+a)))^2*x-6/c^3/b^3/ln(f)^3/n^3*polylog(4,-e*f^(
x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*x+1/c/b/ln(f)*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln
(f^(c*(b*x+a)))*x^3-3/2/c^2/b^2/ln(f)^2*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))^2
*x^2+1/c^3/b^3/ln(f)^3*ln(d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))^3*x-3/2/c^2/b^2/ln
(f)^2*ln(1+e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*x^2*ln(f^(c*(b*x+a)))^2+2/c^3/b^3/ln(f)^3*ln(1+e*f^
(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*x*ln(f^(c*(b*x+a)))^3+3/c^2/b^2/ln(f)^2*ln((d+f^(x*b*c*n)*f^(-x*b*
c*n)*(f^(c*(b*x+a)))^n*e)/d)*x^2*ln(f^(c*(b*x+a)))^2-3/c^3/b^3/ln(f)^3*ln((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b
*x+a)))^n*e)/d)*x*ln(f^(c*(b*x+a)))^3-1/c/b/ln(f)*ln((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*x^3*l
n(f^(c*(b*x+a)))+3/c^2/b^2/ln(f)^2/n^2*polylog(3,-e*f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n/d)*x^2-1/c/b/ln
(f)/n*dilog((d+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*x^3+1/c^4/b^4/ln(f)^4/n*dilog((d+f^(x*b*c*n)*f
^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)/d)*ln(f^(c*(b*x+a)))^3

Fricas [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.27 \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {4 \, b^{3} c^{3} n^{3} x^{3} {\rm Li}_2\left (-\frac {e f^{b c n x + a c n} + d}{d} + 1\right ) \log \left (f\right )^{3} - 12 \, b^{2} c^{2} n^{2} x^{2} \log \left (f\right )^{2} {\rm polylog}\left (3, -\frac {e f^{b c n x + a c n}}{d}\right ) - {\left (b^{4} c^{4} n^{4} x^{4} - a^{4} c^{4} n^{4}\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \left (f\right )^{4} + {\left (b^{4} c^{4} n^{4} x^{4} - a^{4} c^{4} n^{4}\right )} \log \left (f\right )^{4} \log \left (\frac {e f^{b c n x + a c n} + d}{d}\right ) + 24 \, b c n x \log \left (f\right ) {\rm polylog}\left (4, -\frac {e f^{b c n x + a c n}}{d}\right ) - 24 \, {\rm polylog}\left (5, -\frac {e f^{b c n x + a c n}}{d}\right )}{4 \, b^{4} c^{4} n^{4} \log \left (f\right )^{4}} \]

[In]

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

[Out]

-1/4*(4*b^3*c^3*n^3*x^3*dilog(-(e*f^(b*c*n*x + a*c*n) + d)/d + 1)*log(f)^3 - 12*b^2*c^2*n^2*x^2*log(f)^2*polyl
og(3, -e*f^(b*c*n*x + a*c*n)/d) - (b^4*c^4*n^4*x^4 - a^4*c^4*n^4)*log(e*f^(b*c*n*x + a*c*n) + d)*log(f)^4 + (b
^4*c^4*n^4*x^4 - a^4*c^4*n^4)*log(f)^4*log((e*f^(b*c*n*x + a*c*n) + d)/d) + 24*b*c*n*x*log(f)*polylog(4, -e*f^
(b*c*n*x + a*c*n)/d) - 24*polylog(5, -e*f^(b*c*n*x + a*c*n)/d))/(b^4*c^4*n^4*log(f)^4)

Sympy [F]

\[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^{3} \log {\left (d + e \left (f^{a c + b c x}\right )^{n} \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.06 \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (e f^{{\left (b x + a\right )} c n} + d\right ) - \frac {b^{4} c^{4} n^{4} x^{4} \log \left (\frac {e f^{b c n x} f^{a c n}}{d} + 1\right ) \log \left (f\right )^{4} + 4 \, b^{3} c^{3} n^{3} x^{3} {\rm Li}_2\left (-\frac {e f^{b c n x} f^{a c n}}{d}\right ) \log \left (f\right )^{3} - 12 \, b^{2} c^{2} n^{2} x^{2} \log \left (f\right )^{2} {\rm Li}_{3}(-\frac {e f^{b c n x} f^{a c n}}{d}) + 24 \, b c n x \log \left (f\right ) {\rm Li}_{4}(-\frac {e f^{b c n x} f^{a c n}}{d}) - 24 \, {\rm Li}_{5}(-\frac {e f^{b c n x} f^{a c n}}{d})}{4 \, b^{4} c^{4} n^{4} \log \left (f\right )^{4}} \]

[In]

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

[Out]

1/4*x^4*log(e*f^((b*x + a)*c*n) + d) - 1/4*(b^4*c^4*n^4*x^4*log(e*f^(b*c*n*x)*f^(a*c*n)/d + 1)*log(f)^4 + 4*b^
3*c^3*n^3*x^3*dilog(-e*f^(b*c*n*x)*f^(a*c*n)/d)*log(f)^3 - 12*b^2*c^2*n^2*x^2*log(f)^2*polylog(3, -e*f^(b*c*n*
x)*f^(a*c*n)/d) + 24*b*c*n*x*log(f)*polylog(4, -e*f^(b*c*n*x)*f^(a*c*n)/d) - 24*polylog(5, -e*f^(b*c*n*x)*f^(a
*c*n)/d))/(b^4*c^4*n^4*log(f)^4)

Giac [F]

\[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { x^{3} \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + d\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^3 \log \left (d+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^3\,\ln \left (d+e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n\right ) \,d x \]

[In]

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

[Out]

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