Integrand size = 21, antiderivative size = 559 \[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=-\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}+\frac {c e \left (3 a e^2 (2+p)-c d^2 \left (18+11 p+2 p^2\right )\right ) (d+e x)^{-3-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^3 (2+p) (3+2 p) (5+2 p)}+\frac {c^2 d e (3+p) \left (a e^2 (8+5 p)-c d^2 \left (8+7 p+2 p^2\right )\right ) (d+e x)^{-2 (1+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^4 (1+p) (2+p) (3+2 p) (5+2 p)}-\frac {c d e (4+p) (d+e x)^{-2 (2+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^2 (2+p) (5+2 p)}-\frac {c^2 \left (3 a^2 e^4-6 a c d^2 e^2 (5+2 p)+c^2 d^4 \left (15+16 p+4 p^2\right )\right ) \left (\sqrt {-a}-\sqrt {c} x\right ) \left (-\frac {\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (\sqrt {-a}+\sqrt {c} x\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^4 (1+2 p) (3+2 p) (5+2 p)} \]
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Time = 0.50 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {759, 851, 821, 741} \[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=-\frac {c^2 \left (\sqrt {-a}-\sqrt {c} x\right ) \left (a+c x^2\right )^p (d+e x)^{-2 p-1} \left (3 a^2 e^4-6 a c d^2 e^2 (2 p+5)+c^2 d^4 \left (4 p^2+16 p+15\right )\right ) \left (-\frac {\left (\sqrt {-a}+\sqrt {c} x\right ) \left (\sqrt {-a} e+\sqrt {c} d\right )}{\left (\sqrt {-a}-\sqrt {c} x\right ) \left (\sqrt {c} d-\sqrt {-a} e\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{(2 p+1) (2 p+3) (2 p+5) \left (\sqrt {-a} e+\sqrt {c} d\right ) \left (a e^2+c d^2\right )^4}+\frac {c^2 d e (p+3) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+1)} \left (a e^2 (5 p+8)-c d^2 \left (2 p^2+7 p+8\right )\right )}{(p+1) (p+2) (2 p+3) (2 p+5) \left (a e^2+c d^2\right )^4}+\frac {c e \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-3} \left (3 a e^2 (p+2)-c d^2 \left (2 p^2+11 p+18\right )\right )}{(p+2) (2 p+3) (2 p+5) \left (a e^2+c d^2\right )^3}-\frac {e \left (a+c x^2\right )^{p+1} (d+e x)^{-2 p-5}}{(2 p+5) \left (a e^2+c d^2\right )}-\frac {c d e (p+4) \left (a+c x^2\right )^{p+1} (d+e x)^{-2 (p+2)}}{(p+2) (2 p+5) \left (a e^2+c d^2\right )^2} \]
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Rule 741
Rule 759
Rule 821
Rule 851
Rubi steps \begin{align*} \text {integral}& = -\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}-\frac {c \int (d+e x)^{-5-2 p} (-d (5+2 p)+3 e x) \left (a+c x^2\right )^p \, dx}{\left (c d^2+a e^2\right ) (5+2 p)} \\ & = -\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}-\frac {c d e (4+p) (d+e x)^{-2 (2+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^2 (2+p) (5+2 p)}+\frac {c \int (d+e x)^{-4-2 p} \left (-2 (2+p) \left (3 a e^2-c d^2 (5+2 p)\right )-4 c d e (4+p) x\right ) \left (a+c x^2\right )^p \, dx}{2 \left (c d^2+a e^2\right )^2 (2+p) (5+2 p)} \\ & = -\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}+\frac {c e \left (3 a e^2 (2+p)-c d^2 \left (18+11 p+2 p^2\right )\right ) (d+e x)^{-3-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^3 (2+p) (3+2 p) (5+2 p)}-\frac {c d e (4+p) (d+e x)^{-2 (2+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^2 (2+p) (5+2 p)}-\frac {c \int (d+e x)^{-3-2 p} \left (2 c d (3+2 p) \left (a e^2 (14+5 p)-c d^2 \left (10+9 p+2 p^2\right )\right )-2 c e \left (3 a e^2 (2+p)-c d^2 \left (18+11 p+2 p^2\right )\right ) x\right ) \left (a+c x^2\right )^p \, dx}{2 \left (c d^2+a e^2\right )^3 (2+p) (3+2 p) (5+2 p)} \\ & = -\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}+\frac {c e \left (3 a e^2 (2+p)-c d^2 \left (18+11 p+2 p^2\right )\right ) (d+e x)^{-3-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^3 (2+p) (3+2 p) (5+2 p)}+\frac {c^2 d e (3+p) \left (a e^2 (8+5 p)-c d^2 \left (8+7 p+2 p^2\right )\right ) (d+e x)^{-2 (1+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^4 (1+p) (2+p) (3+2 p) (5+2 p)}-\frac {c d e (4+p) (d+e x)^{-2 (2+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^2 (2+p) (5+2 p)}+\frac {\left (c^2 \left (3 a^2 e^4-6 a c d^2 e^2 (5+2 p)+c^2 d^4 \left (15+16 p+4 p^2\right )\right )\right ) \int (d+e x)^{-2-2 p} \left (a+c x^2\right )^p \, dx}{\left (c d^2+a e^2\right )^4 (3+2 p) (5+2 p)} \\ & = -\frac {e (d+e x)^{-5-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right ) (5+2 p)}+\frac {c e \left (3 a e^2 (2+p)-c d^2 \left (18+11 p+2 p^2\right )\right ) (d+e x)^{-3-2 p} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^3 (2+p) (3+2 p) (5+2 p)}+\frac {c^2 d e (3+p) \left (a e^2 (8+5 p)-c d^2 \left (8+7 p+2 p^2\right )\right ) (d+e x)^{-2 (1+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^4 (1+p) (2+p) (3+2 p) (5+2 p)}-\frac {c d e (4+p) (d+e x)^{-2 (2+p)} \left (a+c x^2\right )^{1+p}}{\left (c d^2+a e^2\right )^2 (2+p) (5+2 p)}-\frac {c^2 \left (3 a^2 e^4-6 a c d^2 e^2 (5+2 p)+c^2 d^4 \left (15+16 p+4 p^2\right )\right ) \left (\sqrt {-a}-\sqrt {c} x\right ) \left (-\frac {\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (\sqrt {-a}+\sqrt {c} x\right )}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+c x^2\right )^p \, _2F_1\left (-1-2 p,-p;-2 p;\frac {2 \sqrt {-a} \sqrt {c} (d+e x)}{\left (\sqrt {c} d-\sqrt {-a} e\right ) \left (\sqrt {-a}-\sqrt {c} x\right )}\right )}{\left (\sqrt {c} d+\sqrt {-a} e\right ) \left (c d^2+a e^2\right )^4 (1+2 p) (3+2 p) (5+2 p)} \\ \end{align*}
\[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx \]
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\[\int \left (e x +d \right )^{-6-2 p} \left (c \,x^{2}+a \right )^{p}d x\]
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\[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\int { {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 6} \,d x } \]
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Timed out. \[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\text {Timed out} \]
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\[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\int { {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 6} \,d x } \]
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\[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\int { {\left (c x^{2} + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 6} \,d x } \]
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Timed out. \[ \int (d+e x)^{-6-2 p} \left (a+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+6}} \,d x \]
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