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

Optimal. Leaf size=63 \[ \frac {\text {Li}_3\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x \text {Li}_2\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, 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 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]

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*}

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Mathematica [A]  time = 0.00, size = 63, normalized size = 1.00 \[ \frac {\text {Li}_3\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {x \text {Li}_2\left (-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)} \]

Antiderivative was successfully verified.

[In]

Integrate[x*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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fricas [C]  time = 1.32, size = 58, normalized size = 0.92 \[ -\frac {b c n x {\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \relax (f) - {\rm polylog}\left (3, -e f^{b c n x + a c n}\right )}{b^{2} c^{2} n^{2} \log \relax (f)^{2}} \]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*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)

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maple [B]  time = 0.35, size = 262, normalized size = 4.16 \[ \frac {x^{2} \ln \left (e \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{2}-\frac {x^{2} \ln \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{2}-\frac {x \dilog \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right )}{b c n \ln \relax (f )}+\frac {\dilog \left (e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}+1\right ) \ln \left (f^{\left (b x +a \right ) c}\right )}{b^{2} c^{2} n \ln \relax (f )^{2}}-\frac {\polylog \left (2, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right ) \ln \left (f^{\left (b x +a \right ) c}\right )}{b^{2} c^{2} n \ln \relax (f )^{2}}+\frac {\polylog \left (3, -e \,f^{b c n x} f^{-b c n x} \left (f^{\left (b x +a \right ) c}\right )^{n}\right )}{b^{2} c^{2} n^{2} \ln \relax (f )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 117, normalized size = 1.86 \[ \frac {1}{2} \, x^{2} \log \left (e f^{{\left (b x + a\right )} c n} + 1\right ) - \frac {b^{2} c^{2} n^{2} x^{2} \log \left (e f^{b c n x} f^{a c n} + 1\right ) \log \relax (f)^{2} + 2 \, b c n x {\rm Li}_2\left (-e f^{b c n x} f^{a c n}\right ) \log \relax (f) - 2 \, {\rm Li}_{3}(-e f^{b c n x} f^{a c n})}{2 \, b^{2} c^{2} n^{2} \log \relax (f)^{2}} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x\,\ln \left (e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n+1\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x*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 \[ - \frac {b c e n e^{a c n \log {\relax (f )}} \log {\relax (f )} \int \frac {x^{2} e^{b c n x \log {\relax (f )}}}{e e^{a c n \log {\relax (f )}} e^{b c n x \log {\relax (f )}} + 1}\, dx}{2} + \frac {x^{2} \log {\left (e \left (f^{c \left (a + b x\right )}\right )^{n} + 1 \right )}}{2} \]

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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