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

Optimal. Leaf size=98 \[ -\frac {2 \text {Li}_4\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {2 x \text {Li}_3\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x^2 \text {Li}_2\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 98, normalized size = 1.00 \[ -\frac {2 \text {Li}_4\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {2 x \text {Li}_3\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x^2 \text {Li}_2\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)} \]

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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fricas [C]  time = 0.80, size = 93, normalized size = 0.95 \[ -\frac {b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \relax (f)^{2} - 2 \, b c n x \log \relax (f) {\rm polylog}\left (3, -e f^{b c n x + a c n}\right ) + 2 \, {\rm polylog}\left (4, -e f^{b c n x + a c n}\right )}{b^{3} c^{3} n^{3} \log \relax (f)^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + 1\right )\,{d x} \]

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)

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maple [B]  time = 0.35, size = 430, normalized size = 4.39 \[ \frac {x^{3} \ln \left (e \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{3}-\frac {x^{3} \ln \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{3}-\frac {x^{2} \dilog \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{b c n \ln \relax (f )}+\frac {2 x \dilog \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right ) \ln \left (f^{\left (b x +a \right ) c}\right )}{b^{2} c^{2} n \ln \relax (f )^{2}}-\frac {2 x \polylog \left (2, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right ) \ln \left (f^{\left (b x +a \right ) c}\right )}{b^{2} c^{2} n \ln \relax (f )^{2}}+\frac {2 x \polylog \left (3, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right )}{b^{2} c^{2} n^{2} \ln \relax (f )^{2}}-\frac {\dilog \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right ) \ln \left (f^{\left (b x +a \right ) c}\right )^{2}}{b^{3} c^{3} n \ln \relax (f )^{3}}+\frac {\polylog \left (2, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right ) \ln \left (f^{\left (b x +a \right ) c}\right )^{2}}{b^{3} c^{3} n \ln \relax (f )^{3}}-\frac {2 \polylog \left (4, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right )}{b^{3} c^{3} n^{3} \ln \relax (f )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.81, size = 153, normalized size = 1.56 \[ \frac {1}{3} \, x^{3} \log \left (e f^{{\left (b x + a\right )} c n} + 1\right ) - \frac {b^{3} c^{3} n^{3} x^{3} \log \left (e f^{b c n x} f^{a c n} + 1\right ) \log \relax (f)^{3} + 3 \, b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-e f^{b c n x} f^{a c n}\right ) \log \relax (f)^{2} - 6 \, b c n x \log \relax (f) {\rm Li}_{3}(-e f^{b c n x} f^{a c n}) + 6 \, {\rm Li}_{4}(-e f^{b c n x} f^{a c n})}{3 \, b^{3} c^{3} n^{3} \log \relax (f)^{3}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\ln \left (e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n+1\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {b c e n e^{a c n \log {\relax (f )}} \log {\relax (f )} \int \frac {x^{3} e^{b c n x \log {\relax (f )}}}{e e^{a c n \log {\relax (f )}} e^{b c n x \log {\relax (f )}} + 1}\, dx}{3} + \frac {x^{3} \log {\left (e \left (f^{c \left (a + b x\right )}\right )^{n} + 1 \right )}}{3} \]

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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