Integrand size = 21, antiderivative size = 198 \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{\left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{\left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]
-(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))))*(1-b/(-4*a*c+b^2)^(1/2))/(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))))*(1+b/(-4*a*c+b^2)^(1/2))/(1+n)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2) ))
Time = 0.21 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.92 \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}-\frac {\left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{1+n} \]
((e + f*x)^(1 + n)*(-(((1 - b/Sqrt[b^2 - 4*a*c])*Hypergeometric2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e + (-b + Sqrt[b^2 - 4*a*c])*f)])/(2*c*e + (-b + Sqrt[b^2 - 4*a*c])*f)) - ((1 + b/Sqrt[b^2 - 4*a*c])*Hypergeometric2F 1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(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 = 0.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
\(\displaystyle \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^n}{-\sqrt {b^2-4 a c}+b+2 c x}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (e+f x)^n}{\sqrt {b^2-4 a c}+b+2 c x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}\) |
-(((1 - b/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)])/((2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n))) - ((1 + b/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 + S qrt[b^2 - 4*a*c])*f)])/((2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n))
3.6.44.3.1 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]
\[\int \frac {x \left (f x +e \right )^{n}}{c \,x^{2}+b x +a}d x\]
\[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x}{c x^{2} + b x + a} \,d x } \]
\[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x \left (e + f x\right )^{n}}{a + b x + c x^{2}}\, dx \]
\[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x}{c x^{2} + b x + a} \,d x } \]
\[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x}{c x^{2} + b x + a} \,d x } \]
Timed out. \[ \int \frac {x (e+f x)^n}{a+b x+c x^2} \, dx=\int \frac {x\,{\left (e+f\,x\right )}^n}{c\,x^2+b\,x+a} \,d x \]