### 3.120 $$\int x \log (1+e (f^{c (a+b x)})^n) \, dx$$

Optimal. Leaf size=63 $\frac{\text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

[Out]

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

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Rubi [A]  time = 0.0382002, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {2531, 2282, 6589} $\frac{\text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Log[1 + e*(f^(c*(a + b*x)))^n],x]

[Out]

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

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

Mathematica [A]  time = 0.0047085, size = 63, normalized size = 1. $\frac{\text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{x \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Log[1 + e*(f^(c*(a + b*x)))^n],x]

[Out]

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

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Maple [B]  time = 0.043, size = 282, normalized size = 4.5 \begin{align*}{\frac{{x}^{2}\ln \left ( 1+e \left ({f}^{c \left ( bx+a \right ) } \right ) ^{n} \right ) }{2}}-{\frac{\ln \left ( 1+e{f}^{bcnx}{{\rm e}^{-n \left ( \ln \left ( f \right ) bcx-\ln \left ({f}^{c \left ( bx+a \right ) } \right ) \right ) }} \right ){x}^{2}}{2}}-{\frac{{\it polylog} \left ( 2,-e{f}^{bcnx}{{\rm e}^{-n \left ( \ln \left ( f \right ) bcx-\ln \left ({f}^{c \left ( bx+a \right ) } \right ) \right ) }} \right ) \ln \left ({f}^{c \left ( bx+a \right ) } \right ) }{{c}^{2}{b}^{2} \left ( \ln \left ( f \right ) \right ) ^{2}n}}+{\frac{{\it polylog} \left ( 3,-e{f}^{bcnx}{{\rm e}^{-n \left ( \ln \left ( f \right ) bcx-\ln \left ({f}^{c \left ( bx+a \right ) } \right ) \right ) }} \right ) }{{c}^{2}{b}^{2} \left ( \ln \left ( f \right ) \right ) ^{2}{n}^{2}}}-{\frac{{\it dilog} \left ( 1+e{f}^{bcnx}{{\rm e}^{-n \left ( \ln \left ( f \right ) bcx-\ln \left ({f}^{c \left ( bx+a \right ) } \right ) \right ) }} \right ) x}{ncb\ln \left ( f \right ) }}+{\frac{{\it dilog} \left ( 1+e{f}^{bcnx}{{\rm e}^{-n \left ( \ln \left ( f \right ) bcx-\ln \left ({f}^{c \left ( bx+a \right ) } \right ) \right ) }} \right ) \ln \left ({f}^{c \left ( bx+a \right ) } \right ) }{{c}^{2}{b}^{2} \left ( \ln \left ( f \right ) \right ) ^{2}n}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(1+e*(f^(c*(b*x+a)))^n),x)

[Out]

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

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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 n \int \frac{x^{2}}{2 \,{\left (e{\left (f^{b c x}\right )}^{n}{\left (f^{a c}\right )}^{n} + 1\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} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(1+e*(f^(c*(b*x+a)))^n),x, algorithm="maxima")

[Out]

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

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Fricas [C]  time = 2.21496, size = 147, normalized size = 2.33 \begin{align*} -\frac{b c n x{\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \left (f\right ) -{\rm polylog}\left (3, -e f^{b c n x + a c n}\right )}{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(1+e*(f^(c*(b*x+a)))^n),x, algorithm="fricas")

[Out]

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

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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 )}}}{e e^{a c n \log{\left (f \right )}} e^{b c n x \log{\left (f \right )}} + 1}\, dx}{2} + \frac{x^{2} \log{\left (e \left (f^{c \left (a + b x\right )}\right )^{n} + 1 \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(1+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))/(e*exp(a*c*n*log(f))*exp(b*c*n*x*log(f)) +
1), x)/2 + x**2*log(e*(f**(c*(a + b*x)))**n + 1)/2

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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} + 1\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(1+e*(f^(c*(b*x+a)))^n),x, algorithm="giac")

[Out]

integrate(x*log(e*(f^((b*x + a)*c))^n + 1), x)