3.119 $$\int x^2 \log (1+e (f^{c (a+b x)})^n) \, dx$$

Optimal. Leaf size=98 $\frac{2 x \text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{2 \text{PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}-\frac{x^2 \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

[Out]

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

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Rubi [A]  time = 0.0615725, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, Rules used = {2531, 6609, 2282, 6589} $\frac{2 x \text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{2 \text{PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}-\frac{x^2 \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2*Log[1 + e*(f^(c*(a + b*x)))^n],x]

[Out]

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

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

Mathematica [A]  time = 0.0054603, size = 98, normalized size = 1. $\frac{2 x \text{PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac{2 \text{PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}-\frac{x^2 \text{PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2*Log[1 + e*(f^(c*(a + b*x)))^n],x]

[Out]

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

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Maple [B]  time = 0.049, size = 462, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*ln(1+e*(f^(c*(b*x+a)))^n),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{12} \, b c n x^{4} \log \left (f\right ) + b c n \int \frac{x^{3}}{3 \,{\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}{3} \, x^{3} \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^2*log(1+e*(f^(c*(b*x+a)))^n),x, algorithm="maxima")

[Out]

-1/12*b*c*n*x^4*log(f) + b*c*n*integrate(1/3*x^3/(e*(f^(b*c*x))^n*(f^(a*c))^n + 1), x)*log(f) + 1/3*x^3*log(e*
(f^(b*c*x))^n*(f^(a*c))^n + 1)

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Fricas [C]  time = 2.15546, size = 236, normalized size = 2.41 \begin{align*} -\frac{b^{2} c^{2} n^{2} x^{2}{\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \left (f\right )^{2} - 2 \, b c n x \log \left (f\right ){\rm polylog}\left (3, -e f^{b c n x + a c n}\right ) + 2 \,{\rm polylog}\left (4, -e f^{b c n x + a c n}\right )}{b^{3} c^{3} n^{3} \log \left (f\right )^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(1+e*(f^(c*(b*x+a)))^n),x, algorithm="fricas")

[Out]

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

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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^{3} 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}{3} + \frac{x^{3} \log{\left (e \left (f^{c \left (a + b x\right )}\right )^{n} + 1 \right )}}{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \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^2*log(1+e*(f^(c*(b*x+a)))^n),x, algorithm="giac")

[Out]

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