Integrand size = 24, antiderivative size = 107 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {x^2}{a b d \log (F)}-\frac {x^2}{b d \left (a+b F^{c+d x}\right ) \log (F)}-\frac {2 x \log \left (1+\frac {b F^{c+d x}}{a}\right )}{a b d^2 \log ^2(F)}-\frac {2 \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \log ^3(F)} \]
x^2/a/b/d/ln(F)-x^2/b/d/(a+b*F^(d*x+c))/ln(F)-2*x*ln(1+b*F^(d*x+c)/a)/a/b/ d^2/ln(F)^2-2*polylog(2,-b*F^(d*x+c)/a)/a/b/d^3/ln(F)^3
Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.96 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\frac {d x \log (F) \left (b d F^{c+d x} x \log (F)-2 \left (a+b F^{c+d x}\right ) \log \left (1+\frac {b F^{c+d x}}{a}\right )\right )-2 \left (a+b F^{c+d x}\right ) \operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{a b d^3 \left (a+b F^{c+d x}\right ) \log ^3(F)} \]
(d*x*Log[F]*(b*d*F^(c + d*x)*x*Log[F] - 2*(a + b*F^(c + d*x))*Log[1 + (b*F ^(c + d*x))/a]) - 2*(a + b*F^(c + d*x))*PolyLog[2, -((b*F^(c + d*x))/a)])/ (a*b*d^3*(a + b*F^(c + d*x))*Log[F]^3)
Time = 0.46 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2621, 2615, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {x^2 F^{c+d x}}{\left (a+b F^{c+d x}\right )^2} \, dx\) |
\(\Big \downarrow \) 2621 |
\(\displaystyle \frac {2 \int \frac {x}{b F^{c+d x}+a}dx}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \int \frac {F^{c+d x} x}{b F^{c+d x}+a}dx}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {\int \log \left (\frac {b F^{c+d x}}{a}+1\right )dx}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}-\frac {\int F^{-c-d x} \log \left (\frac {b F^{c+d x}}{a}+1\right )dF^{c+d x}}{b d^2 \log ^2(F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {2 \left (\frac {x^2}{2 a}-\frac {b \left (\frac {\operatorname {PolyLog}\left (2,-\frac {b F^{c+d x}}{a}\right )}{b d^2 \log ^2(F)}+\frac {x \log \left (\frac {b F^{c+d x}}{a}+1\right )}{b d \log (F)}\right )}{a}\right )}{b d \log (F)}-\frac {x^2}{b d \log (F) \left (a+b F^{c+d x}\right )}\) |
-(x^2/(b*d*(a + b*F^(c + d*x))*Log[F])) + (2*(x^2/(2*a) - (b*((x*Log[1 + ( b*F^(c + d*x))/a])/(b*d*Log[F]) + PolyLog[2, -((b*F^(c + d*x))/a)]/(b*d^2* Log[F]^2)))/a))/(b*d*Log[F])
3.1.83.3.1 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[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Leaf count of result is larger than twice the leaf count of optimal. \(230\) vs. \(2(107)=214\).
Time = 0.04 (sec) , antiderivative size = 231, normalized size of antiderivative = 2.16
method | result | size |
risch | \(-\frac {x^{2}}{b d \left (a +b \,F^{d x +c}\right ) \ln \left (F \right )}+\frac {x^{2}}{a b d \ln \left (F \right )}+\frac {2 c x}{b \ln \left (F \right ) d^{2} a}+\frac {c^{2}}{b \ln \left (F \right ) d^{3} a}-\frac {2 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) x}{b \ln \left (F \right )^{2} d^{2} a}-\frac {2 \ln \left (1+\frac {b \,F^{d x} F^{c}}{a}\right ) c}{b \ln \left (F \right )^{2} d^{3} a}-\frac {2 \,\operatorname {Li}_{2}\left (-\frac {b \,F^{d x} F^{c}}{a}\right )}{b \ln \left (F \right )^{3} d^{3} a}+\frac {2 c \ln \left (F^{c} F^{d x} b +a \right )}{b \ln \left (F \right )^{2} d^{3} a}-\frac {2 c \ln \left (F^{d x} F^{c}\right )}{b \ln \left (F \right )^{2} d^{3} a}\) | \(231\) |
-x^2/b/d/(a+b*F^(d*x+c))/ln(F)+x^2/a/b/d/ln(F)+2/b/ln(F)/d^2/a*c*x+1/b/ln( F)/d^3/a*c^2-2/b/ln(F)^2/d^2/a*ln(1+b*F^(d*x)*F^c/a)*x-2/b/ln(F)^2/d^3/a*l n(1+b*F^(d*x)*F^c/a)*c-2/b/ln(F)^3/d^3/a*polylog(2,-b*F^(d*x)*F^c/a)+2/b/l n(F)^2/d^3*c/a*ln(F^c*F^(d*x)*b+a)-2/b/ln(F)^2/d^3*c/a*ln(F^(d*x)*F^c)
Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.74 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {a c^{2} \log \left (F\right )^{2} - {\left (b d^{2} x^{2} - b c^{2}\right )} F^{d x + c} \log \left (F\right )^{2} + 2 \, {\left (F^{d x + c} b + a\right )} {\rm Li}_2\left (-\frac {F^{d x + c} b + a}{a} + 1\right ) - 2 \, {\left (F^{d x + c} b c \log \left (F\right ) + a c \log \left (F\right )\right )} \log \left (F^{d x + c} b + a\right ) + 2 \, {\left ({\left (b d x + b c\right )} F^{d x + c} \log \left (F\right ) + {\left (a d x + a c\right )} \log \left (F\right )\right )} \log \left (\frac {F^{d x + c} b + a}{a}\right )}{F^{d x + c} a b^{2} d^{3} \log \left (F\right )^{3} + a^{2} b d^{3} \log \left (F\right )^{3}} \]
-(a*c^2*log(F)^2 - (b*d^2*x^2 - b*c^2)*F^(d*x + c)*log(F)^2 + 2*(F^(d*x + c)*b + a)*dilog(-(F^(d*x + c)*b + a)/a + 1) - 2*(F^(d*x + c)*b*c*log(F) + a*c*log(F))*log(F^(d*x + c)*b + a) + 2*((b*d*x + b*c)*F^(d*x + c)*log(F) + (a*d*x + a*c)*log(F))*log((F^(d*x + c)*b + a)/a))/(F^(d*x + c)*a*b^2*d^3* log(F)^3 + a^2*b*d^3*log(F)^3)
\[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=- \frac {x^{2}}{F^{c + d x} b^{2} d \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {2 \int \frac {x}{a + b e^{c \log {\left (F \right )}} e^{d x \log {\left (F \right )}}}\, dx}{b d \log {\left (F \right )}} \]
-x**2/(F**(c + d*x)*b**2*d*log(F) + a*b*d*log(F)) + 2*Integral(x/(a + b*ex p(c*log(F))*exp(d*x*log(F))), x)/(b*d*log(F))
Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=-\frac {x^{2}}{F^{d x} F^{c} b^{2} d \log \left (F\right ) + a b d \log \left (F\right )} + \frac {x^{2}}{a b d \log \left (F\right )} - \frac {2 \, {\left (d x \log \left (\frac {F^{d x} F^{c} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{d x} F^{c} b}{a}\right )\right )}}{a b d^{3} \log \left (F\right )^{3}} \]
-x^2/(F^(d*x)*F^c*b^2*d*log(F) + a*b*d*log(F)) + x^2/(a*b*d*log(F)) - 2*(d *x*log(F^(d*x)*F^c*b/a + 1)*log(F) + dilog(-F^(d*x)*F^c*b/a))/(a*b*d^3*log (F)^3)
\[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\int { \frac {F^{d x + c} x^{2}}{{\left (F^{d x + c} b + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {F^{c+d x} x^2}{\left (a+b F^{c+d x}\right )^2} \, dx=\int \frac {F^{c+d\,x}\,x^2}{{\left (a+F^{c+d\,x}\,b\right )}^2} \,d x \]