Integrand size = 25, antiderivative size = 145 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {(c+d x)^3}{3 a d}-\frac {(c+d x)^2 \log \left (1+\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f g n \log (F)}-\frac {2 d (c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^2 g^2 n^2 \log ^2(F)}+\frac {2 d^2 \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{a f^3 g^3 n^3 \log ^3(F)} \]
1/3*(d*x+c)^3/a/d-(d*x+c)^2*ln(1+b*(F^(g*(f*x+e)))^n/a)/a/f/g/n/ln(F)-2*d* (d*x+c)*polylog(2,-b*(F^(g*(f*x+e)))^n/a)/a/f^2/g^2/n^2/ln(F)^2+2*d^2*poly log(3,-b*(F^(g*(f*x+e)))^n/a)/a/f^3/g^3/n^3/ln(F)^3
Time = 0.50 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.83 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\frac {-f^2 g^2 n^2 (c+d x)^2 \log ^2(F) \log \left (1+\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d f g n (c+d x) \log (F) \operatorname {PolyLog}\left (2,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )+2 d^2 \operatorname {PolyLog}\left (3,-\frac {a \left (F^{g (e+f x)}\right )^{-n}}{b}\right )}{a f^3 g^3 n^3 \log ^3(F)} \]
(-(f^2*g^2*n^2*(c + d*x)^2*Log[F]^2*Log[1 + a/(b*(F^(g*(e + f*x)))^n)]) + 2*d*f*g*n*(c + d*x)*Log[F]*PolyLog[2, -(a/(b*(F^(g*(e + f*x)))^n))] + 2*d^ 2*PolyLog[3, -(a/(b*(F^(g*(e + f*x)))^n))])/(a*f^3*g^3*n^3*Log[F]^3)
Time = 0.68 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {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.
| \(\displaystyle \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {(c+d x)^3}{3 a d}-\frac {b \int \frac {\left (F^{g (e+f x)}\right )^n (c+d x)^2}{b \left (F^{g (e+f x)}\right )^n+a}dx}{a}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {(c+d x)^3}{3 a d}-\frac {b \left (\frac {(c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {2 d \int (c+d x) \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )dx}{b f g n \log (F)}\right )}{a}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {(c+d x)^3}{3 a d}-\frac {b \left (\frac {(c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {2 d \left (\frac {d \int \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )dx}{f g n \log (F)}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {(c+d x)^3}{3 a d}-\frac {b \left (\frac {(c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {2 d \left (\frac {d \int F^{-g (e+f x)} \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )dF^{g (e+f x)}}{f^2 g^2 n \log ^2(F)}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {(c+d x)^3}{3 a d}-\frac {b \left (\frac {(c+d x)^2 \log \left (\frac {b \left (F^{g (e+f x)}\right )^n}{a}+1\right )}{b f g n \log (F)}-\frac {2 d \left (\frac {d \operatorname {PolyLog}\left (3,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f^2 g^2 n^2 \log ^2(F)}-\frac {(c+d x) \operatorname {PolyLog}\left (2,-\frac {b \left (F^{g (e+f x)}\right )^n}{a}\right )}{f g n \log (F)}\right )}{b f g n \log (F)}\right )}{a}\) |
(c + d*x)^3/(3*a*d) - (b*(((c + d*x)^2*Log[1 + (b*(F^(g*(e + f*x)))^n)/a]) /(b*f*g*n*Log[F]) - (2*d*(-(((c + d*x)*PolyLog[2, -((b*(F^(g*(e + f*x)))^n )/a)])/(f*g*n*Log[F])) + (d*PolyLog[3, -((b*(F^(g*(e + f*x)))^n)/a)])/(f^2 *g^2*n^2*Log[F]^2)))/(b*f*g*n*Log[F])))/a
3.1.47.3.1 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[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]
Leaf count of result is larger than twice the leaf count of optimal. \(1123\) vs. \(2(143)=286\).
Time = 0.30 (sec) , antiderivative size = 1124, normalized size of antiderivative = 7.75
-1/ln(F)/f/g/n*c^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)+1/ ln(F)/f/g/n*c^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)-2/ln(F)^2 /f^2/g^2/n*c*d/a*ln(F^(g*(f*x+e)))*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*( f*x+e)))^n/a)-2/ln(F)^2/f^2/g^2/n*d^2/a*ln(F^(g*(f*x+e)))*ln(1+b*F^(n*g*f* x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*x+2/ln(F)^2/f^2/g^2/n*c*d/a*ln((F^(g* (f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*ln(F^(g*(f*x+e)))+2/ln(F)/f/g/n* c*d/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)*x-2/ln(F)^2/f^2/g^2/n *c*d/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n)*ln(F^(g*(f*x+e)))+1/ ln(F)^2/f^2/g^2*d^2/a*ln(F^(g*(f*x+e)))^2*x-2/ln(F)^2/f^2/g^2/n^2*d^2/a*po lylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)*x-1/ln(F)^3/f^3/g ^3/n*d^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x)*b+a)*ln(F^(g*(f*x +e)))^2+1/ln(F)/f/g/n*d^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n) *x^2+1/ln(F)^3/f^3/g^3/n*d^2/a*ln(F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e))) ^n)*ln(F^(g*(f*x+e)))^2-2/3/ln(F)^3/f^3/g^3*d^2/a*ln(F^(g*(f*x+e)))^3+2/ln (F)^3/f^3/g^3/n^3*d^2/a*polylog(3,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e )))^n/a)-1/ln(F)/f/g/n*d^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*g*f*x)*F^(n*g*f*x) *b+a)*x^2+1/ln(F)^2/f^2/g^2*c*d/a*ln(F^(g*(f*x+e)))^2-2/ln(F)^2/f^2/g^2/n^ 2*c*d/a*polylog(2,-b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^(g*(f*x+e)))^n/a)+1/ln(F) ^3/f^3/g^3/n*d^2/a*ln(F^(g*(f*x+e)))^2*ln(1+b*F^(n*g*f*x)*F^(-n*g*f*x)*(F^ (g*(f*x+e)))^n/a)+2/ln(F)^2/f^2/g^2/n*d^2/a*ln((F^(g*(f*x+e)))^n*F^(-n*...
Time = 0.26 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.87 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=-\frac {3 \, {\left (d^{2} e^{2} - 2 \, c d e f + c^{2} f^{2}\right )} g^{2} n^{2} \log \left (F^{f g n x + e g n} b + a\right ) \log \left (F\right )^{2} - {\left (d^{2} f^{3} g^{3} n^{3} x^{3} + 3 \, c d f^{3} g^{3} n^{3} x^{2} + 3 \, c^{2} f^{3} g^{3} n^{3} x\right )} \log \left (F\right )^{3} + 3 \, {\left (d^{2} f^{2} g^{2} n^{2} x^{2} + 2 \, c d f^{2} g^{2} n^{2} x - {\left (d^{2} e^{2} - 2 \, c d e f\right )} g^{2} n^{2}\right )} \log \left (F\right )^{2} \log \left (\frac {F^{f g n x + e g n} b + a}{a}\right ) + 6 \, {\left (d^{2} f g n x + c d f g n\right )} {\rm Li}_2\left (-\frac {F^{f g n x + e g n} b + a}{a} + 1\right ) \log \left (F\right ) - 6 \, d^{2} {\rm polylog}\left (3, -\frac {F^{f g n x + e g n} b}{a}\right )}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \]
-1/3*(3*(d^2*e^2 - 2*c*d*e*f + c^2*f^2)*g^2*n^2*log(F^(f*g*n*x + e*g*n)*b + a)*log(F)^2 - (d^2*f^3*g^3*n^3*x^3 + 3*c*d*f^3*g^3*n^3*x^2 + 3*c^2*f^3*g ^3*n^3*x)*log(F)^3 + 3*(d^2*f^2*g^2*n^2*x^2 + 2*c*d*f^2*g^2*n^2*x - (d^2*e ^2 - 2*c*d*e*f)*g^2*n^2)*log(F)^2*log((F^(f*g*n*x + e*g*n)*b + a)/a) + 6*( d^2*f*g*n*x + c*d*f*g*n)*dilog(-(F^(f*g*n*x + e*g*n)*b + a)/a + 1)*log(F) - 6*d^2*polylog(3, -F^(f*g*n*x + e*g*n)*b/a))/(a*f^3*g^3*n^3*log(F)^3)
\[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {\left (c + d x\right )^{2}}{a + b \left (F^{e g + f g x}\right )^{n}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 303 vs. \(2 (142) = 284\).
Time = 0.25 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.09 \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=c^{2} {\left (\frac {f g n x + e g n}{a f g n} - \frac {\log \left (F^{f g n x + e g n} b + a\right )}{a f g n \log \left (F\right )}\right )} - \frac {2 \, {\left (f g n x \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right ) + {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right )\right )} c d}{a f^{2} g^{2} n^{2} \log \left (F\right )^{2}} - \frac {{\left (f^{2} g^{2} n^{2} x^{2} \log \left (\frac {F^{f g n x} F^{e g n} b}{a} + 1\right ) \log \left (F\right )^{2} + 2 \, f g n x {\rm Li}_2\left (-\frac {F^{f g n x} F^{e g n} b}{a}\right ) \log \left (F\right ) - 2 \, {\rm Li}_{3}(-\frac {F^{f g n x} F^{e g n} b}{a})\right )} d^{2}}{a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} + \frac {d^{2} f^{3} g^{3} n^{3} x^{3} \log \left (F\right )^{3} + 3 \, c d f^{3} g^{3} n^{3} x^{2} \log \left (F\right )^{3}}{3 \, a f^{3} g^{3} n^{3} \log \left (F\right )^{3}} \]
c^2*((f*g*n*x + e*g*n)/(a*f*g*n) - log(F^(f*g*n*x + e*g*n)*b + a)/(a*f*g*n *log(F))) - 2*(f*g*n*x*log(F^(f*g*n*x)*F^(e*g*n)*b/a + 1)*log(F) + dilog(- F^(f*g*n*x)*F^(e*g*n)*b/a))*c*d/(a*f^2*g^2*n^2*log(F)^2) - (f^2*g^2*n^2*x^ 2*log(F^(f*g*n*x)*F^(e*g*n)*b/a + 1)*log(F)^2 + 2*f*g*n*x*dilog(-F^(f*g*n* x)*F^(e*g*n)*b/a)*log(F) - 2*polylog(3, -F^(f*g*n*x)*F^(e*g*n)*b/a))*d^2/( a*f^3*g^3*n^3*log(F)^3) + 1/3*(d^2*f^3*g^3*n^3*x^3*log(F)^3 + 3*c*d*f^3*g^ 3*n^3*x^2*log(F)^3)/(a*f^3*g^3*n^3*log(F)^3)
\[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a} \,d x } \]
Timed out. \[ \int \frac {(c+d x)^2}{a+b \left (F^{g (e+f x)}\right )^n} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n} \,d x \]