Integrand size = 25, antiderivative size = 222 \[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=-\frac {b^3 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f) (1+n)}+\frac {d^3 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{c^2 (b c-a d) (d e-c f) (1+n)}+\frac {(b c+a d) (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )}{a^2 c^2 e (1+n)}+\frac {f (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a c e^2 (1+n)} \]
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Time = 0.11 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {186, 67, 70} \[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=-\frac {b^3 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b (e+f x)}{b e-a f}\right )}{a^2 (n+1) (b c-a d) (b e-a f)}+\frac {(a d+b c) (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {f x}{e}+1\right )}{a^2 c^2 e (n+1)}+\frac {d^3 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {d (e+f x)}{d e-c f}\right )}{c^2 (n+1) (b c-a d) (d e-c f)}+\frac {f (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (2,n+1,n+2,\frac {f x}{e}+1\right )}{a c e^2 (n+1)} \]
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Rule 67
Rule 70
Rule 186
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(e+f x)^n}{a c x^2}+\frac {(-b c-a d) (e+f x)^n}{a^2 c^2 x}-\frac {b^3 (e+f x)^n}{a^2 (-b c+a d) (a+b x)}-\frac {d^3 (e+f x)^n}{c^2 (b c-a d) (c+d x)}\right ) \, dx \\ & = \frac {\int \frac {(e+f x)^n}{x^2} \, dx}{a c}+\frac {b^3 \int \frac {(e+f x)^n}{a+b x} \, dx}{a^2 (b c-a d)}-\frac {d^3 \int \frac {(e+f x)^n}{c+d x} \, dx}{c^2 (b c-a d)}-\frac {(b c+a d) \int \frac {(e+f x)^n}{x} \, dx}{a^2 c^2} \\ & = -\frac {b^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f) (1+n)}+\frac {d^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {d (e+f x)}{d e-c f}\right )}{c^2 (b c-a d) (d e-c f) (1+n)}+\frac {(b c+a d) (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {f x}{e}\right )}{a^2 c^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 c e^2 (1+n)} \\ \end{align*}
Time = 0.31 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.80 \[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\frac {(e+f x)^{1+n} \left (-\frac {b^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{a^2 (b c-a d) (b e-a f)}+\frac {-\frac {d^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{(b c-a d) (-d e+c f)}+\frac {(b c+a d) e \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {f x}{e}\right )+a c f \operatorname {Hypergeometric2F1}\left (2,1+n,2+n,1+\frac {f x}{e}\right )}{a^2 e^2}}{c^2}\right )}{1+n} \]
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\[\int \frac {\left (f x +e \right )^{n}}{x^{2} \left (b x +a \right ) \left (d x +c \right )}d x\]
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\[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x + a\right )} {\left (d x + c\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\text {Timed out} \]
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\[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x + a\right )} {\left (d x + c\right )} x^{2}} \,d x } \]
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\[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n}}{{\left (b x + a\right )} {\left (d x + c\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e+f x)^n}{x^2 (a+b x) (c+d x)} \, dx=\int \frac {{\left (e+f\,x\right )}^n}{x^2\,\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]
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