Integrand size = 23, antiderivative size = 242 \[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {(e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{a e (1+n)} \]
-(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],1+f*x/e)/a/e/(1+n)+c*(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))/a/(1+n)/(2*c*e-f*(b-(-4*a*c+b^2)^(1/2)))+c*(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))/a/(1+n)/(2*c*e-f*(b+(-4*a*c+b^2)^(1/2)))
Time = 0.35 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.86 \[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\frac {(e+f x)^{1+n} \left (\frac {c \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{e}\right )}{a (1+n)} \]
((e + f*x)^(1 + n)*((c*(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) + (c*(1 - b/Sqrt[b^2 - 4*a*c])*Hypergeometric2 F1[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) - Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e]/e))/(a*(1 + n))
Time = 0.49 (sec) , antiderivative size = 242, 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.
\(\displaystyle \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {(-b-c x) (e+f x)^n}{a \left (a+b x+c x^2\right )}+\frac {(e+f x)^n}{a x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}+\frac {c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {(e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {f x}{e}+1\right )}{a e (n+1)}\) |
(c*(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)])/(a*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + (c*(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)])/(a*(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n)) - ((e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e *(1 + n))
3.6.46.3.1 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]
\[\int \frac {\left (f x +e \right )^{n}}{x \left (c \,x^{2}+b x +a \right )}d x\]
\[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x} \,d x } \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\int \frac {\left (e + f x\right )^{n}}{x \left (a + b x + c x^{2}\right )}\, dx \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x} \,d x } \]
\[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x \left (a+b x+c x^2\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x\,\left (c\,x^2+b\,x+a\right )} \,d x \]