Integrand size = 15, antiderivative size = 269 \[ \int f^{\frac {c}{a+b x}} x^3 \, dx=-\frac {a^3 f^{\frac {c}{a+b x}} (a+b x)}{b^4}+\frac {3 a^2 f^{\frac {c}{a+b x}} (a+b x)^2}{2 b^4}-\frac {a f^{\frac {c}{a+b x}} (a+b x)^3}{b^4}+\frac {3 a^2 c f^{\frac {c}{a+b x}} (a+b x) \log (f)}{2 b^4}-\frac {a c f^{\frac {c}{a+b x}} (a+b x)^2 \log (f)}{2 b^4}+\frac {a^3 c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log (f)}{b^4}-\frac {a c^2 f^{\frac {c}{a+b x}} (a+b x) \log ^2(f)}{2 b^4}-\frac {3 a^2 c^2 \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log ^2(f)}{2 b^4}+\frac {a c^3 \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log ^3(f)}{2 b^4}+\frac {c^4 \Gamma \left (-4,-\frac {c \log (f)}{a+b x}\right ) \log ^4(f)}{b^4} \]
-a^3*f^(c/(b*x+a))*(b*x+a)/b^4+3/2*a^2*f^(c/(b*x+a))*(b*x+a)^2/b^4-a*f^(c/ (b*x+a))*(b*x+a)^3/b^4+3/2*a^2*c*f^(c/(b*x+a))*(b*x+a)*ln(f)/b^4-1/2*a*c*f ^(c/(b*x+a))*(b*x+a)^2*ln(f)/b^4+a^3*c*Ei(c*ln(f)/(b*x+a))*ln(f)/b^4-1/2*a *c^2*f^(c/(b*x+a))*(b*x+a)*ln(f)^2/b^4-3/2*a^2*c^2*Ei(c*ln(f)/(b*x+a))*ln( f)^2/b^4+1/2*a*c^3*Ei(c*ln(f)/(b*x+a))*ln(f)^3/b^4+(b*x+a)^4*Ei(5,-c*ln(f) /(b*x+a))/b^4
Time = 0.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.67 \[ \int f^{\frac {c}{a+b x}} x^3 \, dx=-\frac {a f^{\frac {c}{a+b x}} \left (6 a^3-26 a^2 c \log (f)+11 a c^2 \log ^2(f)-c^3 \log ^3(f)\right )}{24 b^4}+\frac {c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log (f) \left (24 a^3-36 a^2 c \log (f)+12 a c^2 \log ^2(f)-c^3 \log ^3(f)\right )+b f^{\frac {c}{a+b x}} x \left (6 b^3 x^3+2 c \left (9 a^2-3 a b x+b^2 x^2\right ) \log (f)+c^2 (-10 a+b x) \log ^2(f)+c^3 \log ^3(f)\right )}{24 b^4} \]
-1/24*(a*f^(c/(a + b*x))*(6*a^3 - 26*a^2*c*Log[f] + 11*a*c^2*Log[f]^2 - c^ 3*Log[f]^3))/b^4 + (c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]*(24*a^3 - 36*a^2*c*Log[f] + 12*a*c^2*Log[f]^2 - c^3*Log[f]^3) + b*f^(c/(a + b*x))*x *(6*b^3*x^3 + 2*c*(9*a^2 - 3*a*b*x + b^2*x^2)*Log[f] + c^2*(-10*a + b*x)*L og[f]^2 + c^3*Log[f]^3))/(24*b^4)
Time = 0.51 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2656, 2009}
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.
| \(\displaystyle \int x^3 f^{\frac {c}{a+b x}} \, dx\) |
| \(\Big \downarrow \) 2656 |
| \(\displaystyle \int \left (-\frac {a^3 f^{\frac {c}{a+b x}}}{b^3}+\frac {3 a^2 (a+b x) f^{\frac {c}{a+b x}}}{b^3}+\frac {(a+b x)^3 f^{\frac {c}{a+b x}}}{b^3}-\frac {3 a (a+b x)^2 f^{\frac {c}{a+b x}}}{b^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 c \log (f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{b^4}-\frac {a^3 (a+b x) f^{\frac {c}{a+b x}}}{b^4}-\frac {3 a^2 c^2 \log ^2(f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{2 b^4}+\frac {3 a^2 (a+b x)^2 f^{\frac {c}{a+b x}}}{2 b^4}+\frac {3 a^2 c \log (f) (a+b x) f^{\frac {c}{a+b x}}}{2 b^4}+\frac {c^4 \log ^4(f) \Gamma \left (-4,-\frac {c \log (f)}{a+b x}\right )}{b^4}+\frac {a c^3 \log ^3(f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{2 b^4}-\frac {a c^2 \log ^2(f) (a+b x) f^{\frac {c}{a+b x}}}{2 b^4}-\frac {a (a+b x)^3 f^{\frac {c}{a+b x}}}{b^4}-\frac {a c \log (f) (a+b x)^2 f^{\frac {c}{a+b x}}}{2 b^4}\) |
-((a^3*f^(c/(a + b*x))*(a + b*x))/b^4) + (3*a^2*f^(c/(a + b*x))*(a + b*x)^ 2)/(2*b^4) - (a*f^(c/(a + b*x))*(a + b*x)^3)/b^4 + (3*a^2*c*f^(c/(a + b*x) )*(a + b*x)*Log[f])/(2*b^4) - (a*c*f^(c/(a + b*x))*(a + b*x)^2*Log[f])/(2* b^4) + (a^3*c*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^4 - (a*c^2*f^( c/(a + b*x))*(a + b*x)*Log[f]^2)/(2*b^4) - (3*a^2*c^2*ExpIntegralEi[(c*Log [f])/(a + b*x)]*Log[f]^2)/(2*b^4) + (a*c^3*ExpIntegralEi[(c*Log[f])/(a + b *x)]*Log[f]^3)/(2*b^4) + (c^4*Gamma[-4, -((c*Log[f])/(a + b*x))]*Log[f]^4) /b^4
3.3.17.3.1 Defintions of rubi rules used
Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(Px_), x_Symbol] :> Int[ ExpandLinearProduct[F^(a + b*(c + d*x)^n), Px, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[Px, x]
Time = 0.14 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.33
| method | result | size |
| risch | \(\frac {f^{\frac {c}{b x +a}} \ln \left (f \right )^{3} c^{3} x}{24 b^{3}}+\frac {f^{\frac {c}{b x +a}} \ln \left (f \right )^{2} c^{2} x^{2}}{24 b^{2}}+\frac {f^{\frac {c}{b x +a}} \ln \left (f \right ) c \,x^{3}}{12 b}+\frac {f^{\frac {c}{b x +a}} x^{4}}{4}+\frac {\operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) c^{4} \ln \left (f \right )^{4}}{24 b^{4}}+\frac {f^{\frac {c}{b x +a}} \ln \left (f \right )^{3} a \,c^{3}}{24 b^{4}}-\frac {5 f^{\frac {c}{b x +a}} \ln \left (f \right )^{2} a \,c^{2} x}{12 b^{3}}-\frac {f^{\frac {c}{b x +a}} \ln \left (f \right ) a c \,x^{2}}{4 b^{2}}-\frac {\ln \left (f \right )^{3} \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) a \,c^{3}}{2 b^{4}}-\frac {11 f^{\frac {c}{b x +a}} \ln \left (f \right )^{2} a^{2} c^{2}}{24 b^{4}}+\frac {3 f^{\frac {c}{b x +a}} \ln \left (f \right ) a^{2} c x}{4 b^{3}}+\frac {3 \ln \left (f \right )^{2} \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) a^{2} c^{2}}{2 b^{4}}+\frac {13 f^{\frac {c}{b x +a}} \ln \left (f \right ) a^{3} c}{12 b^{4}}-\frac {\ln \left (f \right ) \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) a^{3} c}{b^{4}}-\frac {f^{\frac {c}{b x +a}} a^{4}}{4 b^{4}}\) | \(359\) |
1/24/b^3*f^(c/(b*x+a))*ln(f)^3*c^3*x+1/24/b^2*f^(c/(b*x+a))*ln(f)^2*c^2*x^ 2+1/12/b*f^(c/(b*x+a))*ln(f)*c*x^3+1/4*f^(c/(b*x+a))*x^4+1/24/b^4*Ei(1,-c* ln(f)/(b*x+a))*c^4*ln(f)^4+1/24/b^4*f^(c/(b*x+a))*ln(f)^3*a*c^3-5/12/b^3*f ^(c/(b*x+a))*ln(f)^2*a*c^2*x-1/4/b^2*f^(c/(b*x+a))*ln(f)*a*c*x^2-1/2/b^4*l n(f)^3*Ei(1,-c*ln(f)/(b*x+a))*a*c^3-11/24/b^4*f^(c/(b*x+a))*ln(f)^2*a^2*c^ 2+3/4/b^3*f^(c/(b*x+a))*ln(f)*a^2*c*x+3/2/b^4*ln(f)^2*Ei(1,-c*ln(f)/(b*x+a ))*a^2*c^2+13/12/b^4*f^(c/(b*x+a))*ln(f)*a^3*c-1/b^4*ln(f)*Ei(1,-c*ln(f)/( b*x+a))*a^3*c-1/4/b^4*f^(c/(b*x+a))*a^4
Time = 0.31 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.64 \[ \int f^{\frac {c}{a+b x}} x^3 \, dx=\frac {{\left (6 \, b^{4} x^{4} - 6 \, a^{4} + {\left (b c^{3} x + a c^{3}\right )} \log \left (f\right )^{3} + {\left (b^{2} c^{2} x^{2} - 10 \, a b c^{2} x - 11 \, a^{2} c^{2}\right )} \log \left (f\right )^{2} + 2 \, {\left (b^{3} c x^{3} - 3 \, a b^{2} c x^{2} + 9 \, a^{2} b c x + 13 \, a^{3} c\right )} \log \left (f\right )\right )} f^{\frac {c}{b x + a}} - {\left (c^{4} \log \left (f\right )^{4} - 12 \, a c^{3} \log \left (f\right )^{3} + 36 \, a^{2} c^{2} \log \left (f\right )^{2} - 24 \, a^{3} c \log \left (f\right )\right )} {\rm Ei}\left (\frac {c \log \left (f\right )}{b x + a}\right )}{24 \, b^{4}} \]
1/24*((6*b^4*x^4 - 6*a^4 + (b*c^3*x + a*c^3)*log(f)^3 + (b^2*c^2*x^2 - 10* a*b*c^2*x - 11*a^2*c^2)*log(f)^2 + 2*(b^3*c*x^3 - 3*a*b^2*c*x^2 + 9*a^2*b* c*x + 13*a^3*c)*log(f))*f^(c/(b*x + a)) - (c^4*log(f)^4 - 12*a*c^3*log(f)^ 3 + 36*a^2*c^2*log(f)^2 - 24*a^3*c*log(f))*Ei(c*log(f)/(b*x + a)))/b^4
\[ \int f^{\frac {c}{a+b x}} x^3 \, dx=\int f^{\frac {c}{a + b x}} x^{3}\, dx \]
\[ \int f^{\frac {c}{a+b x}} x^3 \, dx=\int { f^{\frac {c}{b x + a}} x^{3} \,d x } \]
1/24*(6*b^3*x^4 + 2*b^2*c*x^3*log(f) + (b*c^2*log(f)^2 - 6*a*b*c*log(f))*x ^2 + (c^3*log(f)^3 - 10*a*c^2*log(f)^2 + 18*a^2*c*log(f))*x)*f^(c/(b*x + a ))/b^3 - integrate(1/24*(a^2*c^3*log(f)^3 - 10*a^3*c^2*log(f)^2 + 18*a^4*c *log(f) - (b*c^4*log(f)^4 - 12*a*b*c^3*log(f)^3 + 36*a^2*b*c^2*log(f)^2 - 24*a^3*b*c*log(f))*x)*f^(c/(b*x + a))/(b^5*x^2 + 2*a*b^4*x + a^2*b^3), x)
\[ \int f^{\frac {c}{a+b x}} x^3 \, dx=\int { f^{\frac {c}{b x + a}} x^{3} \,d x } \]
Timed out. \[ \int f^{\frac {c}{a+b x}} x^3 \, dx=\int f^{\frac {c}{a+b\,x}}\,x^3 \,d x \]