Integrand size = 20, antiderivative size = 98 \[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}+\frac {2 x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {2 \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)} \]
-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*polylog(4,-e*(f^(c*(b*x+a)))^n)/b^3/c^3/n ^3/ln(f)^3
Time = 0.01 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}+\frac {2 x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {2 \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)} \]
-((x^2*PolyLog[2, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f])) + (2*x*PolyLo g[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)
Time = 0.46 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3011, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \log \left (e \left (f^{c (a+b x)}\right )^n+1\right ) \, dx\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {2 \int x \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )dx}{b c n \log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\int \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )dx}{b c n \log (f)}\right )}{b c n \log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\int f^{-c (a+b x)} \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )df^{c (a+b x)}}{b^2 c^2 n \log ^2(f)}\right )}{b c n \log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}\right )}{b c n \log (f)}-\frac {x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
-((x^2*PolyLog[2, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f])) + (2*((x*Poly Log[3, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f]) - PolyLog[4, -(e*(f^(c*(a + b*x)))^n)]/(b^2*c^2*n^2*Log[f]^2)))/(b*c*n*Log[f])
3.2.19.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
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] - Simp[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]
Leaf count of result is larger than twice the leaf count of optimal. \(429\) vs. \(2(98)=196\).
Time = 1.06 (sec) , antiderivative size = 430, normalized size of antiderivative = 4.39
method | result | size |
risch | \(\frac {x^{3} \ln \left (1+e \left (f^{c \left (b x +a \right )}\right )^{n}\right )}{3}+\frac {2 \,\operatorname {Li}_{3}\left (-f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n^{2}}-\frac {2 \,\operatorname {Li}_{2}\left (-f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n}+\frac {\operatorname {Li}_{2}\left (-f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{c^{3} b^{3} \ln \left (f \right )^{3} n}-\frac {\ln \left (1+f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x^{3}}{3}-\frac {\operatorname {dilog}\left (1+f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) x^{2}}{c b \ln \left (f \right ) n}+\frac {2 \operatorname {dilog}\left (1+f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right ) x}{c^{2} b^{2} \ln \left (f \right )^{2} n}-\frac {\operatorname {dilog}\left (1+f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right ) \ln \left (f^{c \left (b x +a \right )}\right )^{2}}{c^{3} b^{3} \ln \left (f \right )^{3} n}-\frac {2 \,\operatorname {Li}_{4}\left (-f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right )}{c^{3} b^{3} \ln \left (f \right )^{3} n^{3}}\) | \(430\) |
1/3*x^3*ln(1+e*(f^(c*(b*x+a)))^n)+2/c^2/b^2/ln(f)^2/n^2*polylog(3,-f^(x*b* c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x-2/c^2/b^2/ln(f)^2/n*polylog(2,-f^ (x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))*x+1/c^3/b^3/ ln(f)^3/n*polylog(2,-f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f^(c *(b*x+a)))^2-1/3*ln(1+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x^3-1/ c/b/ln(f)/n*dilog(1+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x^2+2/c^ 2/b^2/ln(f)^2/n*dilog(1+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*ln(f ^(c*(b*x+a)))*x-1/c^3/b^3/ln(f)^3/n*dilog(1+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c *(b*x+a)))^n*e)*ln(f^(c*(b*x+a)))^2-2/c^3/b^3/ln(f)^3/n^3*polylog(4,-f^(x* b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)
Time = 0.34 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.95 \[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \left (f\right )^{2} - 2 \, b c n x \log \left (f\right ) {\rm polylog}\left (3, -e f^{b c n x + a c n}\right ) + 2 \, {\rm polylog}\left (4, -e f^{b c n x + a c n}\right )}{b^{3} c^{3} n^{3} \log \left (f\right )^{3}} \]
-(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)
\[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^{2} \log {\left (e \left (f^{a c + b c x}\right )^{n} + 1 \right )}\, dx \]
Time = 0.24 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.56 \[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{3} \, x^{3} \log \left (e f^{{\left (b x + a\right )} c n} + 1\right ) - \frac {b^{3} c^{3} n^{3} x^{3} \log \left (e f^{b c n x} f^{a c n} + 1\right ) \log \left (f\right )^{3} + 3 \, b^{2} c^{2} n^{2} x^{2} {\rm Li}_2\left (-e f^{b c n x} f^{a c n}\right ) \log \left (f\right )^{2} - 6 \, b c n x \log \left (f\right ) {\rm Li}_{3}(-e f^{b c n x} f^{a c n}) + 6 \, {\rm Li}_{4}(-e f^{b c n x} f^{a c n})}{3 \, b^{3} c^{3} n^{3} \log \left (f\right )^{3}} \]
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)
\[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { x^{2} \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + 1\right ) \,d x } \]
Timed out. \[ \int x^2 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^2\,\ln \left (e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n+1\right ) \,d x \]