### 3.125 $$\int x \log (d+e (f^{c (a+b x)})^n) \, dx$$

Optimal. Leaf size=118 $\frac{\text{PolyLog}\left (3,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac{1}{2} x^2 \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-\frac{1}{2} x^2 \log \left (\frac{e \left (f^{c (a+b x)}\right )^n}{d}+1\right )$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0650263, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.222, Rules used = {2532, 2531, 2282, 6589} $\frac{\text{PolyLog}\left (3,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac{1}{2} x^2 \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-\frac{1}{2} x^2 \log \left (\frac{e \left (f^{c (a+b x)}\right )^n}{d}+1\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2532

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*(F^(c*(a + b*
x)))^n)/d], x] - Simp[((f + g*x)^(m + 1)*Log[1 + (e*(F^(c*(a + b*x)))^n)/d])/(g*(m + 1)), x]) /; FreeQ[{F, a,
b, c, d, e, f, g, n}, x] && GtQ[m, 0] && NeQ[d, 1]

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]

Rubi steps

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

Mathematica [A]  time = 0.0054364, size = 118, normalized size = 1. $\frac{\text{PolyLog}\left (3,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-\frac{e \left (f^{c (a+b x)}\right )^n}{d}\right )}{b c n \log (f)}+\frac{1}{2} x^2 \log \left (e \left (f^{c (a+b x)}\right )^n+d\right )-\frac{1}{2} x^2 \log \left (\frac{e \left (f^{c (a+b x)}\right )^n}{d}+1\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.059, size = 598, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{6} \, b c n x^{3} \log \left (f\right ) + b c d n \int \frac{x^{2}}{2 \,{\left (e{\left (f^{b c x}\right )}^{n}{\left (f^{a c}\right )}^{n} + d\right )}}\,{d x} \log \left (f\right ) + \frac{1}{2} \, x^{2} \log \left (e{\left (f^{b c x}\right )}^{n}{\left (f^{a c}\right )}^{n} + d\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b*c*n*x^3*log(f) + b*c*d*n*integrate(1/2*x^2/(e*(f^(b*c*x))^n*(f^(a*c))^n + d), x)*log(f) + 1/2*x^2*log(e
*(f^(b*c*x))^n*(f^(a*c))^n + d)

________________________________________________________________________________________

Fricas [C]  time = 2.08905, size = 382, normalized size = 3.24 \begin{align*} -\frac{2 \, b c n x{\rm Li}_2\left (-\frac{e f^{b c n x + a c n} + d}{d} + 1\right ) \log \left (f\right ) -{\left (b^{2} c^{2} n^{2} x^{2} - a^{2} c^{2} n^{2}\right )} \log \left (e f^{b c n x + a c n} + d\right ) \log \left (f\right )^{2} +{\left (b^{2} c^{2} n^{2} x^{2} - a^{2} c^{2} n^{2}\right )} \log \left (f\right )^{2} \log \left (\frac{e f^{b c n x + a c n} + d}{d}\right ) - 2 \,{\rm polylog}\left (3, -\frac{e f^{b c n x + a c n}}{d}\right )}{2 \, b^{2} c^{2} n^{2} \log \left (f\right )^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*c*e*n*exp(a*c*n*log(f))*log(f)*Integral(x**2*exp(b*c*n*x*log(f))/(d + e*exp(a*c*n*log(f))*exp(b*c*n*x*log(f
))), x)/2 + x**2*log(d + e*(f**(c*(a + b*x)))**n)/2

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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