Integrand size = 24, antiderivative size = 115 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)} \]
x^3*ln(1+b*F^(d*x+c)/a)/b/d/ln(F)+3*x^2*polylog(2,-b*F^(d*x+c)/a)/b/d^2/ln (F)^2-6*x*polylog(3,-b*F^(d*x+c)/a)/b/d^3/ln(F)^3+6*polylog(4,-b*F^(d*x+c) /a)/b/d^4/ln(F)^4
Time = 0.16 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {x^3 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{b d \log (F)}+\frac {3 x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{b d^4 \log ^4(F)} \]
(x^3*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) + (3*x^2*PolyLog[2, -((b*F^( c + d*x))/a)])/(b*d^2*Log[F]^2) - (6*x*PolyLog[3, -((b*F^(c + d*x))/a)])/( b*d^3*Log[F]^3) + (6*PolyLog[4, -((b*F^(c + d*x))/a)])/(b*d^4*Log[F]^4)
Time = 0.56 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2620, 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.
\(\displaystyle \int \frac {x^3 F^{c+d x}}{a+b F^{c+d x}} \, dx\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {3 \int x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )dx}{b d \log (F)}\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {3 \left (\frac {2 \int x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )dx}{d \log (F)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}-\frac {\int \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )dx}{d \log (F)}\right )}{d \log (F)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}-\frac {\int F^{-c-d x} \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )dF^{c+d x}}{d^2 \log ^2(F)}\right )}{d \log (F)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {x^3 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {3 \left (\frac {2 \left (\frac {x \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}-\frac {\operatorname {PolyLog}\left (4,-\frac {b F^{c+d x}}{a}\right )}{d^2 \log ^2(F)}\right )}{d \log (F)}-\frac {x^2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\) |
(x^3*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) - (3*(-((x^2*PolyLog[2, -((b *F^(c + d*x))/a)])/(d*Log[F])) + (2*((x*PolyLog[3, -((b*F^(c + d*x))/a)])/ (d*Log[F]) - PolyLog[4, -((b*F^(c + d*x))/a)]/(d^2*Log[F]^2)))/(d*Log[F])) )/(b*d*Log[F])
3.1.76.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
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]
Time = 0.06 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.96
method | result | size |
risch | \(\frac {6 \,\operatorname {Li}_{4}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{d^{4} \ln \left (F \right )^{4} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c^{3}}{d^{4} \ln \left (F \right ) b}-\frac {c^{3} \ln \left (F^{c} F^{d x} b +a \right )}{d^{4} \ln \left (F \right ) b}+\frac {c^{3} \ln \left (F^{d x} F^{c}\right )}{d^{4} \ln \left (F \right ) b}-\frac {c^{3} x}{d^{3} b}+\frac {\ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x^{3}}{d \ln \left (F \right ) b}+\frac {3 \,\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x^{2}}{d^{2} \ln \left (F \right )^{2} b}-\frac {6 \,\operatorname {Li}_{3}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x}{d^{3} \ln \left (F \right )^{3} b}-\frac {3 c^{4}}{4 d^{4} b}\) | \(225\) |
6/d^4/ln(F)^4/b*polylog(4,-b*F^(d*x)*F^c/a)+1/d^4/ln(F)/b*ln(1+b*F^(d*x)*F ^c/a)*c^3-1/d^4/ln(F)/b*c^3*ln(F^c*F^(d*x)*b+a)+1/d^4/ln(F)/b*c^3*ln(F^(d* x)*F^c)-1/d^3/b*c^3*x+1/d/ln(F)/b*ln(1+b*F^(d*x)*F^c/a)*x^3+3/d^2/ln(F)^2/ b*polylog(2,-b*F^(d*x)*F^c/a)*x^2-6/d^3/ln(F)^3/b*polylog(3,-b*F^(d*x)*F^c /a)*x-3/4/d^4/b*c^4
Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.17 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) \log \left (F\right )^{2} - c^{3} \log \left (F^{d x + c} b + a\right ) \log \left (F\right )^{3} + {\left (d^{3} x^{3} + c^{3}\right )} \log \left (F\right )^{3} \log \left (\frac {F^{d x + c} b + a}{a}\right ) - 6 \, d x \log \left (F\right ) {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right ) + 6 \, {\rm polylog}\left (4, -\frac {F^{d x + c} b}{a}\right )}{b d^{4} \log \left (F\right )^{4}} \]
(3*d^2*x^2*dilog(-(F^(d*x + c)*b + a)/a + 1)*log(F)^2 - c^3*log(F^(d*x + c )*b + a)*log(F)^3 + (d^3*x^3 + c^3)*log(F)^3*log((F^(d*x + c)*b + a)/a) - 6*d*x*log(F)*polylog(3, -F^(d*x + c)*b/a) + 6*polylog(4, -F^(d*x + c)*b/a) )/(b*d^4*log(F)^4)
\[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int \frac {F^{c + d x} x^{3}}{F^{c + d x} b + a}\, dx \]
Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.92 \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\frac {d^{3} x^{3} \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right )^{3} + 3 \, d^{2} x^{2} {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F\right )^{2} - 6 \, d x \log \left (F\right ) {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a}) + 6 \, {\rm Li}_{4}(-\frac {F^{d x} F^{c} b}{a})}{b d^{4} \log \left (F\right )^{4}} \]
(d^3*x^3*log(F^(d*x)*F^c*b/a + 1)*log(F)^3 + 3*d^2*x^2*dilog(-F^(d*x)*F^c* b/a)*log(F)^2 - 6*d*x*log(F)*polylog(3, -F^(d*x)*F^c*b/a) + 6*polylog(4, - F^(d*x)*F^c*b/a))/(b*d^4*log(F)^4)
\[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int { \frac {F^{d x + c} x^{3}}{F^{d x + c} b + a} \,d x } \]
Timed out. \[ \int \frac {F^{c+d x} x^3}{a+b F^{c+d x}} \, dx=\int \frac {F^{c+d\,x}\,x^3}{a+F^{c+d\,x}\,b} \,d x \]