Integrand size = 15, antiderivative size = 229 \[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\frac {a^2 f^{\frac {c}{a+b x}} (a+b x)}{b^3}-\frac {a f^{\frac {c}{a+b x}} (a+b x)^2}{b^3}+\frac {f^{\frac {c}{a+b x}} (a+b x)^3}{3 b^3}-\frac {a c f^{\frac {c}{a+b x}} (a+b x) \log (f)}{b^3}+\frac {c f^{\frac {c}{a+b x}} (a+b x)^2 \log (f)}{6 b^3}-\frac {a^2 c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log (f)}{b^3}+\frac {c^2 f^{\frac {c}{a+b x}} (a+b x) \log ^2(f)}{6 b^3}+\frac {a c^2 \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log ^2(f)}{b^3}-\frac {c^3 \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log ^3(f)}{6 b^3} \]
a^2*f^(c/(b*x+a))*(b*x+a)/b^3-a*f^(c/(b*x+a))*(b*x+a)^2/b^3+1/3*f^(c/(b*x+ a))*(b*x+a)^3/b^3-a*c*f^(c/(b*x+a))*(b*x+a)*ln(f)/b^3+1/6*c*f^(c/(b*x+a))* (b*x+a)^2*ln(f)/b^3-a^2*c*Ei(c*ln(f)/(b*x+a))*ln(f)/b^3+1/6*c^2*f^(c/(b*x+ a))*(b*x+a)*ln(f)^2/b^3+a*c^2*Ei(c*ln(f)/(b*x+a))*ln(f)^2/b^3-1/6*c^3*Ei(c *ln(f)/(b*x+a))*ln(f)^3/b^3
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.56 \[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\frac {a f^{\frac {c}{a+b x}} \left (2 a^2-5 a c \log (f)+c^2 \log ^2(f)\right )}{6 b^3}+\frac {-c \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right ) \log (f) \left (6 a^2-6 a c \log (f)+c^2 \log ^2(f)\right )+b f^{\frac {c}{a+b x}} x \left (2 b^2 x^2+(-4 a c+b c x) \log (f)+c^2 \log ^2(f)\right )}{6 b^3} \]
(a*f^(c/(a + b*x))*(2*a^2 - 5*a*c*Log[f] + c^2*Log[f]^2))/(6*b^3) + (-(c*E xpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]*(6*a^2 - 6*a*c*Log[f] + c^2*Log[ f]^2)) + b*f^(c/(a + b*x))*x*(2*b^2*x^2 + (-4*a*c + b*c*x)*Log[f] + c^2*Lo g[f]^2))/(6*b^3)
Time = 0.44 (sec) , antiderivative size = 229, 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.
\(\displaystyle \int x^2 f^{\frac {c}{a+b x}} \, dx\) |
\(\Big \downarrow \) 2656 |
\(\displaystyle \int \left (\frac {a^2 f^{\frac {c}{a+b x}}}{b^2}+\frac {(a+b x)^2 f^{\frac {c}{a+b x}}}{b^2}-\frac {2 a (a+b x) f^{\frac {c}{a+b x}}}{b^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 c \log (f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{b^3}+\frac {a^2 (a+b x) f^{\frac {c}{a+b x}}}{b^3}-\frac {c^3 \log ^3(f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{6 b^3}+\frac {a c^2 \log ^2(f) \operatorname {ExpIntegralEi}\left (\frac {c \log (f)}{a+b x}\right )}{b^3}+\frac {c^2 \log ^2(f) (a+b x) f^{\frac {c}{a+b x}}}{6 b^3}+\frac {(a+b x)^3 f^{\frac {c}{a+b x}}}{3 b^3}-\frac {a (a+b x)^2 f^{\frac {c}{a+b x}}}{b^3}+\frac {c \log (f) (a+b x)^2 f^{\frac {c}{a+b x}}}{6 b^3}-\frac {a c \log (f) (a+b x) f^{\frac {c}{a+b x}}}{b^3}\) |
(a^2*f^(c/(a + b*x))*(a + b*x))/b^3 - (a*f^(c/(a + b*x))*(a + b*x)^2)/b^3 + (f^(c/(a + b*x))*(a + b*x)^3)/(3*b^3) - (a*c*f^(c/(a + b*x))*(a + b*x)*L og[f])/b^3 + (c*f^(c/(a + b*x))*(a + b*x)^2*Log[f])/(6*b^3) - (a^2*c*ExpIn tegralEi[(c*Log[f])/(a + b*x)]*Log[f])/b^3 + (c^2*f^(c/(a + b*x))*(a + b*x )*Log[f]^2)/(6*b^3) + (a*c^2*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]^2) /b^3 - (c^3*ExpIntegralEi[(c*Log[f])/(a + b*x)]*Log[f]^3)/(6*b^3)
3.3.18.3.1 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 = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.99
method | result | size |
risch | \(\frac {f^{\frac {c}{b x +a}} \ln \left (f \right )^{2} c^{2} x}{6 b^{2}}+\frac {f^{\frac {c}{b x +a}} \ln \left (f \right ) c \,x^{2}}{6 b}+\frac {f^{\frac {c}{b x +a}} x^{3}}{3}+\frac {\operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) c^{3} \ln \left (f \right )^{3}}{6 b^{3}}+\frac {f^{\frac {c}{b x +a}} \ln \left (f \right )^{2} a \,c^{2}}{6 b^{3}}-\frac {2 f^{\frac {c}{b x +a}} \ln \left (f \right ) a c x}{3 b^{2}}-\frac {\ln \left (f \right )^{2} \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) a \,c^{2}}{b^{3}}-\frac {5 f^{\frac {c}{b x +a}} \ln \left (f \right ) a^{2} c}{6 b^{3}}+\frac {\ln \left (f \right ) \operatorname {Ei}_{1}\left (-\frac {c \ln \left (f \right )}{b x +a}\right ) a^{2} c}{b^{3}}+\frac {f^{\frac {c}{b x +a}} a^{3}}{3 b^{3}}\) | \(227\) |
1/6/b^2*f^(c/(b*x+a))*ln(f)^2*c^2*x+1/6/b*f^(c/(b*x+a))*ln(f)*c*x^2+1/3*f^ (c/(b*x+a))*x^3+1/6/b^3*Ei(1,-c*ln(f)/(b*x+a))*c^3*ln(f)^3+1/6/b^3*f^(c/(b *x+a))*ln(f)^2*a*c^2-2/3/b^2*f^(c/(b*x+a))*ln(f)*a*c*x-1/b^3*ln(f)^2*Ei(1, -c*ln(f)/(b*x+a))*a*c^2-5/6/b^3*f^(c/(b*x+a))*ln(f)*a^2*c+1/b^3*ln(f)*Ei(1 ,-c*ln(f)/(b*x+a))*a^2*c+1/3/b^3*f^(c/(b*x+a))*a^3
Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.50 \[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\frac {{\left (2 \, b^{3} x^{3} + 2 \, a^{3} + {\left (b c^{2} x + a c^{2}\right )} \log \left (f\right )^{2} + {\left (b^{2} c x^{2} - 4 \, a b c x - 5 \, a^{2} c\right )} \log \left (f\right )\right )} f^{\frac {c}{b x + a}} - {\left (c^{3} \log \left (f\right )^{3} - 6 \, a c^{2} \log \left (f\right )^{2} + 6 \, a^{2} c \log \left (f\right )\right )} {\rm Ei}\left (\frac {c \log \left (f\right )}{b x + a}\right )}{6 \, b^{3}} \]
1/6*((2*b^3*x^3 + 2*a^3 + (b*c^2*x + a*c^2)*log(f)^2 + (b^2*c*x^2 - 4*a*b* c*x - 5*a^2*c)*log(f))*f^(c/(b*x + a)) - (c^3*log(f)^3 - 6*a*c^2*log(f)^2 + 6*a^2*c*log(f))*Ei(c*log(f)/(b*x + a)))/b^3
\[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\int f^{\frac {c}{a + b x}} x^{2}\, dx \]
\[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\int { f^{\frac {c}{b x + a}} x^{2} \,d x } \]
1/6*(2*b^2*x^3 + b*c*x^2*log(f) + (c^2*log(f)^2 - 4*a*c*log(f))*x)*f^(c/(b *x + a))/b^2 + integrate(-1/6*(a^2*c^2*log(f)^2 - 4*a^3*c*log(f) - (b*c^3* log(f)^3 - 6*a*b*c^2*log(f)^2 + 6*a^2*b*c*log(f))*x)*f^(c/(b*x + a))/(b^4* x^2 + 2*a*b^3*x + a^2*b^2), x)
\[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\int { f^{\frac {c}{b x + a}} x^{2} \,d x } \]
Time = 0.57 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.91 \[ \int f^{\frac {c}{a+b x}} x^2 \, dx=\frac {\frac {b\,f^{\frac {c}{a+b\,x}}\,x^4}{3}+f^{\frac {c}{a+b\,x}}\,x^3\,\left (\frac {a}{3}+\frac {c\,\ln \left (f\right )}{6}\right )+\frac {f^{\frac {c}{a+b\,x}}\,x\,\left (2\,a^3-9\,a^2\,c\,\ln \left (f\right )+2\,a\,c^2\,{\ln \left (f\right )}^2\right )}{6\,b^2}+\frac {f^{\frac {c}{a+b\,x}}\,x^2\,\left (c^2\,{\ln \left (f\right )}^2-3\,a\,c\,\ln \left (f\right )\right )}{6\,b}+\frac {a^2\,f^{\frac {c}{a+b\,x}}\,\left (2\,a^2-5\,a\,c\,\ln \left (f\right )+c^2\,{\ln \left (f\right )}^2\right )}{6\,b^3}}{a+b\,x}-\frac {\mathrm {ei}\left (\frac {c\,\ln \left (f\right )}{a+b\,x}\right )\,\left (6\,a^2\,c\,\ln \left (f\right )-6\,a\,c^2\,{\ln \left (f\right )}^2+c^3\,{\ln \left (f\right )}^3\right )}{6\,b^3} \]
((b*f^(c/(a + b*x))*x^4)/3 + f^(c/(a + b*x))*x^3*(a/3 + (c*log(f))/6) + (f ^(c/(a + b*x))*x*(2*a^3 - 9*a^2*c*log(f) + 2*a*c^2*log(f)^2))/(6*b^2) + (f ^(c/(a + b*x))*x^2*(c^2*log(f)^2 - 3*a*c*log(f)))/(6*b) + (a^2*f^(c/(a + b *x))*(c^2*log(f)^2 + 2*a^2 - 5*a*c*log(f)))/(6*b^3))/(a + b*x) - (ei((c*lo g(f))/(a + b*x))*(c^3*log(f)^3 + 6*a^2*c*log(f) - 6*a*c^2*log(f)^2))/(6*b^ 3)