Integrand size = 24, antiderivative size = 140 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {x^3}{a b d \log (F)}-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {3 x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {6 x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)}+\frac {6 \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a b d^4 \log ^4(F)} \]
x^3/a/b/d/ln(F)-x^3/b/d/(a+b*F^(d*x+c))/ln(F)-3*x^2*ln(1+b*F^(d*x+c)/a)/a/ b/d^2/ln(F)^2-6*x*polylog(2,-b*F^(d*x+c)/a)/a/b/d^3/ln(F)^3+6*polylog(3,-b *F^(d*x+c)/a)/a/b/d^4/ln(F)^4
Time = 0.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.98 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^3}{b d \left (a+b F^{c+d x}\right ) \log (F)}+\frac {3 \left (\frac {x^3}{3 a}-\frac {x^2 \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a d \log (F)}-\frac {2 x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a d^2 \log ^2(F)}+\frac {2 \operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{a d^3 \log ^3(F)}\right )}{b d \log (F)} \]
-(x^3/(b*d*(a + b*F^(c + d*x))*Log[F])) + (3*(x^3/(3*a) - (x^2*Log[1 + (b* F^(c + d*x))/a])/(a*d*Log[F]) - (2*x*PolyLog[2, -((b*F^(c + d*x))/a)])/(a* d^2*Log[F]^2) + (2*PolyLog[3, -((b*F^(c + d*x))/a)])/(a*d^3*Log[F]^3)))/(b *d*Log[F])
Time = 0.65 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2621, 2615, 2620, 3011, 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.
| \(\displaystyle \int \frac {x^3 F^{c+d x}}{\left (a+b F^{c+d x}\right )^2} \, dx\) |
| \(\Big \downarrow \) 2621 |
| \(\displaystyle \frac {3 \int \frac {x^2}{b F^{c+d x}+a}dx}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {3 \left (\frac {x^3}{3 a}-\frac {b \int \frac {F^{c+d x} x^2}{b F^{c+d x}+a}dx}{a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {2 \int x \log \left (\frac {b F^{c+d x}}{a}+1\right )dx}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {2 \left (\frac {\int \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )dx}{d \log (F)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {2 \left (\frac {\int F^{-c-d x} \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )dF^{c+d x}}{d^2 \log ^2(F)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {3 \left (\frac {x^3}{3 a}-\frac {b \left (\frac {x^2 \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {2 \left (\frac {\operatorname {PolyLog}\left (3,-\frac {b F^{c+d x}}{a}\right )}{d^2 \log ^2(F)}-\frac {x \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{d \log (F)}\right )}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^3}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
-(x^3/(b*d*(a + b*F^(c + d*x))*Log[F])) + (3*(x^3/(3*a) - (b*((x^2*Log[1 + (b*F^(c + d*x))/a])/(b*d*Log[F]) - (2*(-((x*PolyLog[2, -((b*F^(c + d*x))/ a)])/(d*Log[F])) + PolyLog[3, -((b*F^(c + d*x))/a)]/(d^2*Log[F]^2)))/(b*d* Log[F])))/a))/(b*d*Log[F])
3.1.82.3.1 Defintions of rubi rules used
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x _))))^(n_.)), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(a*d*(m + 1)), x] - Simp[ b/a Int[(c + d*x)^m*((F^(g*(e + f*x)))^n/(a + b*(F^(g*(e + f*x)))^n)), x] , x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
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[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*( (e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*((a + b*(F^(g*(e + f*x)))^n)^(p + 1)/(b*f*g*n*(p + 1)*Log [F])), x] - Simp[d*(m/(b*f*g*n*(p + 1)*Log[F])) Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]
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]
Time = 0.05 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.96
| method | result | size |
| risch | \(-\frac {x^{3}}{b d \left (a +b \,F^{d x +c}\right ) \ln \left (F \right )}+\frac {x^{3}}{a b d \ln \left (F \right )}-\frac {3 c^{2} x}{b \ln \left (F \right ) d^{3} a}-\frac {2 c^{3}}{b \ln \left (F \right ) d^{4} a}-\frac {3 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x^{2}}{b \ln \left (F \right )^{2} d^{2} a}+\frac {3 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c^{2}}{b \ln \left (F \right )^{2} d^{4} a}-\frac {6 \,\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right ) x}{b \ln \left (F \right )^{3} d^{3} a}+\frac {6 \,\operatorname {Li}_{3}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{b \ln \left (F \right )^{4} d^{4} a}-\frac {3 c^{2} \ln \left (F^{c} F^{d x} b +a \right )}{b \ln \left (F \right )^{2} d^{4} a}+\frac {3 c^{2} \ln \left (F^{d x} F^{c}\right )}{b \ln \left (F \right )^{2} d^{4} a}\) | \(274\) |
-x^3/b/d/(a+b*F^(d*x+c))/ln(F)+x^3/a/b/d/ln(F)-3/b/ln(F)/d^3/a*c^2*x-2/b/l n(F)/d^4/a*c^3-3/b/ln(F)^2/d^2/a*ln(1+b*F^(d*x)*F^c/a)*x^2+3/b/ln(F)^2/d^4 /a*ln(1+b*F^(d*x)*F^c/a)*c^2-6/b/ln(F)^3/d^3/a*polylog(2,-b*F^(d*x)*F^c/a) *x+6/b/ln(F)^4/d^4/a*polylog(3,-b*F^(d*x)*F^c/a)-3/b/ln(F)^2/d^4*c^2/a*ln( F^c*F^(d*x)*b+a)+3/b/ln(F)^2/d^4*c^2/a*ln(F^(d*x)*F^c)
Time = 0.24 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.76 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {a c^{3} \log \left (F\right )^{3} + {\left (b d^{3} x^{3} + b c^{3}\right )} F^{d x + c} \log \left (F\right )^{3} - 6 \, {\left (F^{d x + c} b d x \log \left (F\right ) + a d x \log \left (F\right )\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 3 \, {\left (F^{d x + c} b c^{2} \log \left (F\right )^{2} + a c^{2} \log \left (F\right )^{2}\right )} \log \left (F^{d x + c} b + a\right ) - 3 \, {\left ({\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \left (F\right )^{2} + {\left (a d^{2} x^{2} - a c^{2}\right )} \log \left (F\right )^{2}\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right ) + 6 \, {\left (F^{d x + c} b + a\right )} {\rm polylog}\left (3, -\frac {F^{d x + c} b}{a}\right )}{F^{d x + c} a b^{2} d^{4} \log \left (F\right )^{4} + a^{2} b d^{4} \log \left (F\right )^{4}} \]
(a*c^3*log(F)^3 + (b*d^3*x^3 + b*c^3)*F^(d*x + c)*log(F)^3 - 6*(F^(d*x + c )*b*d*x*log(F) + a*d*x*log(F))*dilog(-(F^(d*x + c)*b + a)/a + 1) - 3*(F^(d *x + c)*b*c^2*log(F)^2 + a*c^2*log(F)^2)*log(F^(d*x + c)*b + a) - 3*((b*d^ 2*x^2 - b*c^2)*F^(d*x + c)*log(F)^2 + (a*d^2*x^2 - a*c^2)*log(F)^2)*log((F ^(d*x + c)*b + a)/a) + 6*(F^(d*x + c)*b + a)*polylog(3, -F^(d*x + c)*b/a)) /(F^(d*x + c)*a*b^2*d^4*log(F)^4 + a^2*b*d^4*log(F)^4)
\[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=- \frac {x^{3}}{F^{c + d x} b^{2} d \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {3 \int \frac {x^{2}}{a + b e^{c \log {\left (F \right )}} e^{d x \log {\left (F \right )}}}\, dx}{b d \log {\left (F \right )}} \]
-x**3/(F**(c + d*x)*b**2*d*log(F) + a*b*d*log(F)) + 3*Integral(x**2/(a + b *exp(c*log(F))*exp(d*x*log(F))), x)/(b*d*log(F))
Time = 0.21 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.92 \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^{3}}{F^{d x} F^{c} b^{2} d \log \left (F\right ) + a b d \log \left (F\right )} + \frac {x^{3}}{a b d \log \left (F\right )} - \frac {3 \, {\left (d^{2} x^{2} \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right )^{2} + 2 \, d x {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right ) \log \left (F\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{d x} F^{c} b}{a})\right )}}{a b d^{4} \log \left (F\right )^{4}} \]
-x^3/(F^(d*x)*F^c*b^2*d*log(F) + a*b*d*log(F)) + x^3/(a*b*d*log(F)) - 3*(d ^2*x^2*log(F^(d*x)*F^c*b/a + 1)*log(F)^2 + 2*d*x*dilog(-F^(d*x)*F^c*b/a)*l og(F) - 2*polylog(3, -F^(d*x)*F^c*b/a))/(a*b*d^4*log(F)^4)
\[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\int { \frac {F^{d x + c} x^{3}}{{\left (F^{d x + c} b + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {F^{c+d x} x^3}{\left (a+b F^{c+d x}\right )^2} \, dx=\int \frac {F^{c+d\,x}\,x^3}{{\left (a+F^{c+d\,x}\,b\right )}^2} \,d x \]