Integrand size = 20, antiderivative size = 132 \[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \operatorname {PolyLog}\left (5,-e \left (f^{c (a+b x)}\right )^n\right )}{b^4 c^4 n^4 \log ^4(f)} \]
-x^3*polylog(2,-e*(f^(c*(b*x+a)))^n)/b/c/n/ln(f)+3*x^2*polylog(3,-e*(f^(c* (b*x+a)))^n)/b^2/c^2/n^2/ln(f)^2-6*x*polylog(4,-e*(f^(c*(b*x+a)))^n)/b^3/c ^3/n^3/ln(f)^3+6*polylog(5,-e*(f^(c*(b*x+a)))^n)/b^4/c^4/n^4/ln(f)^4
Time = 0.01 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00 \[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}+\frac {3 x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}-\frac {6 x \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b^3 c^3 n^3 \log ^3(f)}+\frac {6 \operatorname {PolyLog}\left (5,-e \left (f^{c (a+b x)}\right )^n\right )}{b^4 c^4 n^4 \log ^4(f)} \]
-((x^3*PolyLog[2, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f])) + (3*x^2*Poly Log[3, -(e*(f^(c*(a + b*x)))^n)])/(b^2*c^2*n^2*Log[f]^2) - (6*x*PolyLog[4, -(e*(f^(c*(a + b*x)))^n)])/(b^3*c^3*n^3*Log[f]^3) + (6*PolyLog[5, -(e*(f^ (c*(a + b*x)))^n)])/(b^4*c^4*n^4*Log[f]^4)
Time = 0.65 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3011, 7163, 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.
| \(\displaystyle \int x^3 \log \left (e \left (f^{c (a+b x)}\right )^n+1\right ) \, dx\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {3 \int x^2 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )dx}{b c n \log (f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {2 \int x \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )dx}{b c n \log (f)}\right )}{b c n \log (f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\int \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )dx}{b c n \log (f)}\right )}{b c n \log (f)}\right )}{b c n \log (f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\int f^{-c (a+b x)} \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )df^{c (a+b x)}}{b^2 c^2 n \log ^2(f)}\right )}{b c n \log (f)}\right )}{b c n \log (f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {3 \left (\frac {x^2 \operatorname {PolyLog}\left (3,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {2 \left (\frac {x \operatorname {PolyLog}\left (4,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}-\frac {\operatorname {PolyLog}\left (5,-e \left (f^{c (a+b x)}\right )^n\right )}{b^2 c^2 n^2 \log ^2(f)}\right )}{b c n \log (f)}\right )}{b c n \log (f)}-\frac {x^3 \operatorname {PolyLog}\left (2,-e \left (f^{c (a+b x)}\right )^n\right )}{b c n \log (f)}\) |
-((x^3*PolyLog[2, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f])) + (3*((x^2*Po lyLog[3, -(e*(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f]) - (2*((x*PolyLog[4, -(e *(f^(c*(a + b*x)))^n)])/(b*c*n*Log[f]) - PolyLog[5, -(e*(f^(c*(a + b*x)))^ n)]/(b^2*c^2*n^2*Log[f]^2)))/(b*c*n*Log[f])))/(b*c*n*Log[f])
3.2.18.3.1 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. \(600\) vs. \(2(132)=264\).
Time = 2.09 (sec) , antiderivative size = 601, normalized size of antiderivative = 4.55
| method | result | size |
| risch | \(\frac {x^{4} \ln \left (1+e \left (f^{c \left (b x +a \right )}\right )^{n}\right )}{4}-\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 )^{3}}{c^{4} b^{4} \ln \left (f \right )^{4} 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 )^{3}}{c^{4} b^{4} \ln \left (f \right )^{4} n}+\frac {3 \,\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^{2}}{c^{2} b^{2} \ln \left (f \right )^{2} n^{2}}-\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^{4}}{4}-\frac {6 \,\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 ) x}{c^{3} b^{3} \ln \left (f \right )^{3} n^{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^{3}}{c b \ln \left (f \right ) n}+\frac {6 \,\operatorname {Li}_{5}\left (-f^{x b c n} f^{-x b c n} \left (f^{c \left (b x +a \right )}\right )^{n} e \right )}{c^{4} b^{4} \ln \left (f \right )^{4} n^{4}}+\frac {3 \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^{2}}{c^{2} b^{2} \ln \left (f \right )^{2} n}-\frac {3 \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} x}{c^{3} b^{3} \ln \left (f \right )^{3} n}+\frac {3 \,\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} x}{c^{3} b^{3} \ln \left (f \right )^{3} n}-\frac {3 \,\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^{2}}{c^{2} b^{2} \ln \left (f \right )^{2} n}\) | \(601\) |
1/4*x^4*ln(1+e*(f^(c*(b*x+a)))^n)-1/c^4/b^4/ln(f)^4/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)))^3+1/c^4/b^4/ln(f)^4 /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))) ^3+3/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^2-1/4*ln(1+f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)))^n*e)*x^4-6/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 )*x-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^ 3+6/c^4/b^4/ln(f)^4/n^4*polylog(5,-f^(x*b*c*n)*f^(-x*b*c*n)*(f^(c*(b*x+a)) )^n*e)+3/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^2-3/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*x+3/c^3/b^3/ln(f)^3/n*pol ylog(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* x-3/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^2
Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=-\frac {b^{3} c^{3} n^{3} x^{3} {\rm Li}_2\left (-e f^{b c n x + a c n}\right ) \log \left (f\right )^{3} - 3 \, b^{2} c^{2} n^{2} x^{2} \log \left (f\right )^{2} {\rm polylog}\left (3, -e f^{b c n x + a c n}\right ) + 6 \, b c n x \log \left (f\right ) {\rm polylog}\left (4, -e f^{b c n x + a c n}\right ) - 6 \, {\rm polylog}\left (5, -e f^{b c n x + a c n}\right )}{b^{4} c^{4} n^{4} \log \left (f\right )^{4}} \]
-(b^3*c^3*n^3*x^3*dilog(-e*f^(b*c*n*x + a*c*n))*log(f)^3 - 3*b^2*c^2*n^2*x ^2*log(f)^2*polylog(3, -e*f^(b*c*n*x + a*c*n)) + 6*b*c*n*x*log(f)*polylog( 4, -e*f^(b*c*n*x + a*c*n)) - 6*polylog(5, -e*f^(b*c*n*x + a*c*n)))/(b^4*c^ 4*n^4*log(f)^4)
\[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^{3} \log {\left (e \left (f^{a c + b c x}\right )^{n} + 1 \right )}\, dx \]
Time = 0.23 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.43 \[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\frac {1}{4} \, x^{4} \log \left (e f^{{\left (b x + a\right )} c n} + 1\right ) - \frac {b^{4} c^{4} n^{4} x^{4} \log \left (e f^{b c n x} f^{a c n} + 1\right ) \log \left (f\right )^{4} + 4 \, b^{3} c^{3} n^{3} x^{3} {\rm Li}_2\left (-e f^{b c n x} f^{a c n}\right ) \log \left (f\right )^{3} - 12 \, b^{2} c^{2} n^{2} x^{2} \log \left (f\right )^{2} {\rm Li}_{3}(-e f^{b c n x} f^{a c n}) + 24 \, b c n x \log \left (f\right ) {\rm Li}_{4}(-e f^{b c n x} f^{a c n}) - 24 \, {\rm Li}_{5}(-e f^{b c n x} f^{a c n})}{4 \, b^{4} c^{4} n^{4} \log \left (f\right )^{4}} \]
1/4*x^4*log(e*f^((b*x + a)*c*n) + 1) - 1/4*(b^4*c^4*n^4*x^4*log(e*f^(b*c*n *x)*f^(a*c*n) + 1)*log(f)^4 + 4*b^3*c^3*n^3*x^3*dilog(-e*f^(b*c*n*x)*f^(a* c*n))*log(f)^3 - 12*b^2*c^2*n^2*x^2*log(f)^2*polylog(3, -e*f^(b*c*n*x)*f^( a*c*n)) + 24*b*c*n*x*log(f)*polylog(4, -e*f^(b*c*n*x)*f^(a*c*n)) - 24*poly log(5, -e*f^(b*c*n*x)*f^(a*c*n)))/(b^4*c^4*n^4*log(f)^4)
\[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int { x^{3} \log \left (e {\left (f^{{\left (b x + a\right )} c}\right )}^{n} + 1\right ) \,d x } \]
Timed out. \[ \int x^3 \log \left (1+e \left (f^{c (a+b x)}\right )^n\right ) \, dx=\int x^3\,\ln \left (e\,{\left (f^{c\,\left (a+b\,x\right )}\right )}^n+1\right ) \,d x \]