Integrand size = 23, antiderivative size = 296 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {b (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a e^2 (1+n)} \]
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Time = 0.31 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {974, 67, 844, 70} \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=-\frac {c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )}-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{a^2 (n+1) \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}+\frac {b (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {f x}{e}+1\right )}{a^2 e (n+1)}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a e^2 (n+1)} \]
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Rule 67
Rule 70
Rule 844
Rule 974
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(e+f x)^n}{a x^2}-\frac {b (e+f x)^n}{a^2 x}+\frac {\left (b^2-a c+b c x\right ) (e+f x)^n}{a^2 \left (a+b x+c x^2\right )}\right ) \, dx \\ & = \frac {\int \frac {\left (b^2-a c+b c x\right ) (e+f x)^n}{a+b x+c x^2} \, dx}{a^2}+\frac {\int \frac {(e+f x)^n}{x^2} \, dx}{a}-\frac {b \int \frac {(e+f x)^n}{x} \, dx}{a^2} \\ & = \frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac {f x}{e}\right )}{a e^2 (1+n)}+\frac {\int \left (\frac {\left (b c+\frac {c \left (b^2-2 a c\right )}{\sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (b c-\frac {c \left (b^2-2 a c\right )}{\sqrt {b^2-4 a c}}\right ) (e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{a^2} \\ & = \frac {b (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac {f x}{e}\right )}{a e^2 (1+n)}+\frac {\left (c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e+f x)^n}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{a^2}+\frac {\left (c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(e+f x)^n}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{a^2} \\ & = -\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) (e+f x)^{1+n} \, _2F_1\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} \, _2F_1\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} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 e (1+n)}+\frac {f (e+f x)^{1+n} \, _2F_1\left (2,1+n;2+n;1+\frac {f x}{e}\right )}{a e^2 (1+n)} \\ \end{align*}
Time = 0.40 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.83 \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {c \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e+\left (-b+\sqrt {b^2-4 a c}\right ) f}-\frac {c \left (b+\frac {-b^2+2 a c}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}+\frac {b \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{e}+\frac {a f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{e^2}\right )}{a^2 (1+n)} \]
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\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (c \,x^{2}+b x +a \right )}d x\]
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\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
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\[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (c x^{2} + b x + a\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e+f x)^n}{x^2 \left (a+b x+c x^2\right )} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (c\,x^2+b\,x+a\right )} \,d x \]
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