Integrand size = 23, antiderivative size = 296 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {b (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a e^2 (1+n)} \]
b*(f*x+e)^(1+n)*hypergeom([1, 1+n],[2+n],1+f*x/e)/a^2/e/(1+n)+f*(f*x+e)^(1 +n)*hypergeom([2, 1+n],[2+n],1+f*x/e)/a/e^2/(1+n)-c*(f*x+e)^(1+n)*hypergeo m([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))/a^2/(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))))*(b+(2*a*c-b^2)/(-4*a*c+b^2)^(1/2))/a^2/(1+n)/(2*c*e-f*(b+(-4*a*c+b^ 2)^(1/2)))
Time = 0.40 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.83 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}-\frac {c \left (b+\frac {-b^2+2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}+\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{e}+\frac {a f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{e^2}\right )}{a^2 (1+n)} \]
((e + f*x)^(1 + n)*(-((c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*Hypergeomet ric2F1[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*(b + (-b^2 + 2*a*c)/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) + (b*Hyper geometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e])/e + (a*f*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/e^2))/(a^2*(1 + n))
Time = 0.57 (sec) , antiderivative size = 296, 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.
\(\displaystyle \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle \int \left (\frac {\left (-a c+b^2+b c x\right ) (e+f x)^n}{a^2 \left (a+b x+c x^2\right )}-\frac {b (e+f x)^n}{a^2 x}+\frac {(e+f x)^n}{a x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {b (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {f x}{e}+1\right )}{a^2 e (n+1)}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a e^2 (n+1)}\) |
-((c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(e + f*x)^(1 + n)*Hypergeometri c2F1[1, 1 + n, 2 + n, (2*c*(e + f*x))/(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)] )/(a^2*(2*c*e - (b - Sqrt[b^2 - 4*a*c])*f)*(1 + n))) - (c*(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)])/(a^2*(2*c*e - (b + Sqrt[b^2 - 4*a*c])*f)*(1 + n)) + (b*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (f*x)/e])/(a^2*e*(1 + n)) + (f*(e + f*x)^(1 + n)*Hyperg eometric2F1[2, 1 + n, 2 + n, 1 + (f*x)/e])/(a*e^2*(1 + n))
3.6.47.3.1 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]
\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (c \,x^{2}+b x +a \right )}d x\]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (c\,x^2+b\,x+a\right )} \,d x \]