Integrand size = 23, antiderivative size = 237 \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\frac {(e+f x)^{1+n}}{c f (1+n)}+\frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {\left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]
(f*x+e)^(1+n)/c/f/(1+n)+(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],2*c*(f*x+e) /(2*c*e-f*(b-(-4*a*c+b^2)^(1/2))))*(b+(2*a*c-b^2)/(-4*a*c+b^2)^(1/2))/c/(1 +n)/(2*c*e-f*(b-(-4*a*c+b^2)^(1/2)))+(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n ],2*c*(f*x+e)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2))))*(b+(-2*a*c+b^2)/(-4*a*c+b^ 2)^(1/2))/c/(1+n)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2)))
Time = 0.40 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.22 \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {2 (e+f x)^{1+n} \left (2 \sqrt {b^2-4 a c} \left (c e^2+f (-b e+a f)\right )+f \left (-b^2 e+2 a c e+b \sqrt {b^2-4 a c} e+a b f-a \sqrt {b^2-4 a c} f\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )+f \left (b^2 e-2 a c e+b \sqrt {b^2-4 a c} e-a b f-a \sqrt {b^2-4 a c} f\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )\right )}{\sqrt {b^2-4 a c} f \left (2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f\right ) \left (-2 c e+\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]
(-2*(e + f*x)^(1 + n)*(2*Sqrt[b^2 - 4*a*c]*(c*e^2 + f*(-(b*e) + a*f)) + f* (-(b^2*e) + 2*a*c*e + b*Sqrt[b^2 - 4*a*c]*e + a*b*f - a*Sqrt[b^2 - 4*a*c]* f)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e + (-b + Sqrt[ b^2 - 4*a*c])*f)] + f*(b^2*e - 2*a*c*e + b*Sqrt[b^2 - 4*a*c]*e - a*b*f - a *Sqrt[b^2 - 4*a*c]*f)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/( 2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)]))/(Sqrt[b^2 - 4*a*c]*f*(2*c*e + (-b + Sqrt[b^2 - 4*a*c])*f)*(-2*c*e + (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n))
Time = 0.48 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {1200, 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.
\(\displaystyle \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {\left (\frac {b^2-2 a c}{c \sqrt {b^2-4 a c}}-\frac {b}{c}\right ) (e+f x)^n}{-\sqrt {b^2-4 a c}+b+2 c x}+\frac {\left (-\frac {b^2-2 a c}{c \sqrt {b^2-4 a c}}-\frac {b}{c}\right ) (e+f x)^n}{\sqrt {b^2-4 a c}+b+2 c x}+\frac {(e+f x)^n}{c}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {\left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{c (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {(e+f x)^{n+1}}{c f (n+1)}\) |
(e + f*x)^(1 + n)/(c*f*(1 + n)) + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*( e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)])/(c*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + ((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*x)^(1 + n)*Hypergeom etric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b + Sqrt[b^2 - 4*a*c]) *f)])/(c*(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n))
3.6.43.3.1 Defintions of rubi rules used
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
\[\int \frac {x^{2} \left (f x +e \right )^{n}}{c \,x^{2}+b x +a}d x\]
\[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{2}}{c x^{2} + b x + a} \,d x } \]
\[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x^{2} \left (e + f x\right )^{n}}{a + b x + c x^{2}}\, dx \]
\[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{2}}{c x^{2} + b x + a} \,d x } \]
\[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{2}}{c x^{2} + b x + a} \,d x } \]
Timed out. \[ \int \frac {x^2 (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x^2\,{\left (e+f\,x\right )}^n}{c\,x^2+b\,x+a} \,d x \]