3.2.19 \(\int x^2 \log (1+e (f^{c (a+b x)})^n) \, dx\) [119]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.04, 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 = {2611, 6744, 2320, 6724} \begin {gather*} -\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 {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 {x^2 \text {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)} \end {gather*}

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 2320

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 2611

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 6724

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 6744

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 \text {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 = 0.01, size = 98, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(429\) vs. \(2(98)=196\).
time = 0.02, size = 430, normalized size = 4.39

method	result	size
risch	\(\frac {x^{3} \ln \left (1+e \left (f^{c \left (b x +a \right )}\right )^{n}\right )}{3}-\frac {\ln \left (1+e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) x^{3}}{3}-\frac {2 \polylog \left (4, -e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right )}{c^{3} b^{3} \ln \left (f \right )^{3} n^{3}}-\frac {\dilog \left (1+e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) x^{2}}{c b \ln \left (f \right ) n}+\frac {2 \dilog \left (1+e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (f^{c \left (b x +a \right )}\right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n}-\frac {\dilog \left (1+e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{c^{3} b^{3} \ln \left (f \right )^{3} n}-\frac {2 \polylog \left (2, -e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (f^{c \left (b x +a \right )}\right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n}+\frac {\polylog \left (2, -e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{c^{3} b^{3} \ln \left (f \right )^{3} n}+\frac {2 \polylog \left (3, -e \,f^{b c n x} f^{-b c n x} \left (f^{c \left (b x +a \right )}\right )^{n}\right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n^{2}}\)	\(430\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 162, normalized size = 1.65 \begin {gather*} \frac {1}{3} \, x^{3} \log \left (f^{{\left (b x + a\right )} c n} e + 1\right ) - \frac {b^{3} c^{3} n^{3} x^{3} \log \left (f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )} + 1\right ) \log \left (f\right )^{3} + 3 \, b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )}\right ) \log \left (f\right )^{2} - 6 \, b c n x \log \left (f\right ) {\rm Li}_{3}(-f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )}) + 6 \, {\rm Li}_{4}(-f^{a c n} e^{\left (b c n x \log \left (f\right ) + 1\right )})}{3 \, b^{3} c^{3} n^{3} \log \left (f\right )^{3}} \end {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.38, size = 96, normalized size = 0.98 \begin {gather*} -\frac {b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-f^{b c n x + a c n} e\right ) \log \left (f\right )^{2} - 2 \, b c n x \log \left (f\right ) {\rm polylog}\left (3, -f^{b c n x + a c n} e\right ) + 2 \, {\rm polylog}\left (4, -f^{b c n x + a c n} e\right )}{b^{3} c^{3} n^{3} \log \left (f\right )^{3}} \end {gather*}

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((f^((b*x + a)*c))^n*e + 1), x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________