Integrand size = 17, antiderivative size = 133 \[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\frac {d e (2+p) \left (a+c x^2\right )^{1+p}}{c (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+c x^2\right )^{1+p}}{c (3+2 p)}-\frac {\left (a e^2-c d^2 (3+2 p)\right ) x \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )}{c (3+2 p)} \]
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Time = 0.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.94, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {757, 655, 252, 251} \[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (d^2-\frac {a e^2}{2 c p+3 c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )+\frac {e (d+e x) \left (a+c x^2\right )^{p+1}}{c (2 p+3)}+\frac {d e (p+2) \left (a+c x^2\right )^{p+1}}{c (p+1) (2 p+3)} \]
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Rule 251
Rule 252
Rule 655
Rule 757
Rubi steps \begin{align*} \text {integral}& = \frac {e (d+e x) \left (a+c x^2\right )^{1+p}}{c (3+2 p)}+\frac {\int \left (-a e^2+c d^2 (3+2 p)+2 c d e (2+p) x\right ) \left (a+c x^2\right )^p \, dx}{c (3+2 p)} \\ & = \frac {d e (2+p) \left (a+c x^2\right )^{1+p}}{c (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+c x^2\right )^{1+p}}{c (3+2 p)}+\left (d^2-\frac {a e^2}{3 c+2 c p}\right ) \int \left (a+c x^2\right )^p \, dx \\ & = \frac {d e (2+p) \left (a+c x^2\right )^{1+p}}{c (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+c x^2\right )^{1+p}}{c (3+2 p)}+\left (\left (d^2-\frac {a e^2}{3 c+2 c p}\right ) \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {c x^2}{a}\right )^p \, dx \\ & = \frac {d e (2+p) \left (a+c x^2\right )^{1+p}}{c (1+p) (3+2 p)}+\frac {e (d+e x) \left (a+c x^2\right )^{1+p}}{c (3+2 p)}+\left (d^2-\frac {a e^2}{3 c+2 c p}\right ) x \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {c x^2}{a}\right ) \\ \end{align*}
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00 \[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\frac {\left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \left (3 c d^2 (1+p) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )+e \left (3 d \left (c x^2 \left (1+\frac {c x^2}{a}\right )^p+a \left (-1+\left (1+\frac {c x^2}{a}\right )^p\right )\right )+c e (1+p) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {c x^2}{a}\right )\right )\right )}{3 c (1+p)} \]
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\[\int \left (e x +d \right )^{2} \left (c \,x^{2}+a \right )^{p}d x\]
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\[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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Time = 5.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73 \[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=a^{p} d^{2} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{p} e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )}}{3} + 2 d e \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\begin {cases} \frac {\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + c x^{2} \right )} & \text {otherwise} \end {cases}}{2 c} & \text {otherwise} \end {cases}\right ) \]
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\[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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\[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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Timed out. \[ \int (d+e x)^2 \left (a+c x^2\right )^p \, dx=\int {\left (c\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
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