Integrand size = 20, antiderivative size = 489 \[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\frac {c (d-e x) (d+e x)^{1+n}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\sqrt {c} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {\sqrt {c} e \left (\sqrt {c} d+\sqrt {-a} e\right ) n (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}+\frac {\sqrt {c} (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}-\frac {\sqrt {c} e \left (\sqrt {-a} \sqrt {c} d+a e\right ) n (d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {(d+e x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,1+\frac {e x}{d}\right )}{a^2 d (1+n)} \]
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Time = 0.41 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {975, 67, 837, 845, 70} \[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=-\frac {\sqrt {c} e n \left (\sqrt {-a} \sqrt {c} d+a e\right ) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right )}+\frac {\sqrt {c} (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 (n+1) \left (\sqrt {-a} e+\sqrt {c} d\right )}-\frac {(d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {e x}{d}+1\right )}{a^2 d (n+1)}+\frac {\sqrt {c} e n \left (\sqrt {-a} e+\sqrt {c} d\right ) (d+e x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 (-a)^{3/2} (n+1) \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (a e^2+c d^2\right )}+\frac {c (d-e x) (d+e x)^{n+1}}{2 a \left (a+c x^2\right ) \left (a e^2+c d^2\right )} \]
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Rule 67
Rule 70
Rule 837
Rule 845
Rule 975
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(d+e x)^n}{a^2 x}-\frac {c x (d+e x)^n}{a \left (a+c x^2\right )^2}-\frac {c x (d+e x)^n}{a^2 \left (a+c x^2\right )}\right ) \, dx \\ & = \frac {\int \frac {(d+e x)^n}{x} \, dx}{a^2}-\frac {c \int \frac {x (d+e x)^n}{a+c x^2} \, dx}{a^2}-\frac {c \int \frac {x (d+e x)^n}{\left (a+c x^2\right )^2} \, dx}{a} \\ & = \frac {c (d-e x) (d+e x)^{1+n}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {e x}{d}\right )}{a^2 d (1+n)}-\frac {c \int \left (-\frac {(d+e x)^n}{2 \sqrt {c} \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {(d+e x)^n}{2 \sqrt {c} \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{a^2}+\frac {\int \frac {(d+e x)^n \left (-a c d e n+a c e^2 n x\right )}{a+c x^2} \, dx}{2 a^2 \left (c d^2+a e^2\right )} \\ & = \frac {c (d-e x) (d+e x)^{1+n}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {e x}{d}\right )}{a^2 d (1+n)}+\frac {\sqrt {c} \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{2 a^2}-\frac {\sqrt {c} \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{2 a^2}+\frac {\int \left (\frac {\left (-\sqrt {-a} a c d e n-a^2 \sqrt {c} e^2 n\right ) (d+e x)^n}{2 a \left (\sqrt {-a}-\sqrt {c} x\right )}+\frac {\left (-\sqrt {-a} a c d e n+a^2 \sqrt {c} e^2 n\right ) (d+e x)^n}{2 a \left (\sqrt {-a}+\sqrt {c} x\right )}\right ) \, dx}{2 a^2 \left (c d^2+a e^2\right )} \\ & = \frac {c (d-e x) (d+e x)^{1+n}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\sqrt {c} (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {\sqrt {c} (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {e x}{d}\right )}{a^2 d (1+n)}-\frac {\left (\sqrt {c} e \left (\sqrt {-a} \sqrt {c} d-a e\right ) n\right ) \int \frac {(d+e x)^n}{\sqrt {-a}+\sqrt {c} x} \, dx}{4 a^2 \left (c d^2+a e^2\right )}-\frac {\left (\sqrt {c} e \left (\sqrt {-a} \sqrt {c} d+a e\right ) n\right ) \int \frac {(d+e x)^n}{\sqrt {-a}-\sqrt {c} x} \, dx}{4 a^2 \left (c d^2+a e^2\right )} \\ & = \frac {c (d-e x) (d+e x)^{1+n}}{2 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )}+\frac {\sqrt {c} (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {\sqrt {c} e \left (\sqrt {-a} \sqrt {c} d-a e\right ) n (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{4 a^2 \left (\sqrt {c} d-\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}+\frac {\sqrt {c} (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{2 a^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}-\frac {\sqrt {c} e \left (\sqrt {-a} \sqrt {c} d+a e\right ) n (d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{4 a^2 \left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right ) (1+n)}-\frac {(d+e x)^{1+n} \, _2F_1\left (1,1+n;2+n;1+\frac {e x}{d}\right )}{a^2 d (1+n)} \\ \end{align*}
Time = 0.59 (sec) , antiderivative size = 391, normalized size of antiderivative = 0.80 \[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\frac {(d+e x)^{1+n} \left (\frac {2 a c (d-e x)}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}-\frac {4 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d+e x}{d}\right )}{d+d n}+\frac {2 \sqrt {c} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) (1+n)}+\frac {2 \sqrt {c} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) (1+n)}+\frac {\sqrt {c} e n \left (\left (\sqrt {-a} c d^2-2 a \sqrt {c} d e+(-a)^{3/2} e^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d-\sqrt {-a} e}\right )+\left (-\sqrt {-a} c d^2-2 a \sqrt {c} d e+\sqrt {-a} a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {\sqrt {c} (d+e x)}{\sqrt {c} d+\sqrt {-a} e}\right )\right )}{\left (c d^2+a e^2\right )^2 (1+n)}\right )}{4 a^2} \]
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\[\int \frac {\left (e x +d \right )^{n}}{x \left (c \,x^{2}+a \right )^{2}}d x\]
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\[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n}}{{\left (c x^{2} + a\right )}^{2} x} \,d x } \]
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\[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\int \frac {\left (d + e x\right )^{n}}{x \left (a + c x^{2}\right )^{2}}\, dx \]
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\[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n}}{{\left (c x^{2} + a\right )}^{2} x} \,d x } \]
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\[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\int { \frac {{\left (e x + d\right )}^{n}}{{\left (c x^{2} + a\right )}^{2} x} \,d x } \]
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Timed out. \[ \int \frac {(d+e x)^n}{x \left (a+c x^2\right )^2} \, dx=\int \frac {{\left (d+e\,x\right )}^n}{x\,{\left (c\,x^2+a\right )}^2} \,d x \]
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