Integrand size = 17, antiderivative size = 178 \[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\frac {e (d+e x)^2 \left (a+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-c d^2 (5+2 p)\right )-2 c d e (1+p) (3+p) x\right ) \left (a+c x^2\right )^{1+p}}{2 c^2 (2+p) \left (3+5 p+2 p^2\right )}-\frac {d \left (3 a e^2-c d^2 (3+2 p)\right ) x \left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )}{c (3+2 p)} \]
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Time = 0.10 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.95, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {757, 794, 252, 251} \[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=-\frac {e \left (a+c x^2\right )^{p+1} \left ((2 p+3) \left (a e^2-c d^2 (2 p+5)\right )-2 c d e (p+1) (p+3) x\right )}{2 c^2 (p+2) \left (2 p^2+5 p+3\right )}+d x \left (a+c x^2\right )^p \left (\frac {c x^2}{a}+1\right )^{-p} \left (d^2-\frac {3 a e^2}{2 c p+3 c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )+\frac {e (d+e x)^2 \left (a+c x^2\right )^{p+1}}{2 c (p+2)} \]
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Rule 251
Rule 252
Rule 757
Rule 794
Rubi steps \begin{align*} \text {integral}& = \frac {e (d+e x)^2 \left (a+c x^2\right )^{1+p}}{2 c (2+p)}+\frac {\int (d+e x) \left (-2 \left (a e^2-c d^2 (2+p)\right )+2 c d e (3+p) x\right ) \left (a+c x^2\right )^p \, dx}{2 c (2+p)} \\ & = \frac {e (d+e x)^2 \left (a+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-c d^2 (5+2 p)\right )-2 c d e (1+p) (3+p) x\right ) \left (a+c x^2\right )^{1+p}}{2 c^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac {3 a e^2}{3 c+2 c p}\right )\right ) \int \left (a+c x^2\right )^p \, dx \\ & = \frac {e (d+e x)^2 \left (a+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-c d^2 (5+2 p)\right )-2 c d e (1+p) (3+p) x\right ) \left (a+c x^2\right )^{1+p}}{2 c^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac {3 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 {e (d+e x)^2 \left (a+c x^2\right )^{1+p}}{2 c (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-c d^2 (5+2 p)\right )-2 c d e (1+p) (3+p) x\right ) \left (a+c x^2\right )^{1+p}}{2 c^2 (2+p) \left (3+5 p+2 p^2\right )}+d \left (d^2-\frac {3 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 = 0.47 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.25 \[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\frac {\left (a+c x^2\right )^p \left (1+\frac {c x^2}{a}\right )^{-p} \left (2 c^2 d^3 \left (2+3 p+p^2\right ) x \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {c x^2}{a}\right )+e \left (c^2 x^2 \left (1+\frac {c x^2}{a}\right )^p \left (3 d^2 (2+p)+e^2 (1+p) x^2\right )-a^2 e^2 \left (-1+\left (1+\frac {c x^2}{a}\right )^p\right )+a c \left (e^2 p x^2 \left (1+\frac {c x^2}{a}\right )^p+3 d^2 (2+p) \left (-1+\left (1+\frac {c x^2}{a}\right )^p\right )\right )+2 c^2 d e \left (2+3 p+p^2\right ) x^3 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {c x^2}{a}\right )\right )\right )}{2 c^2 (1+p) (2+p)} \]
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\[\int \left (e x +d \right )^{3} \left (c \,x^{2}+a \right )^{p}d x\]
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\[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (151) = 302\).
Time = 6.49 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.46 \[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=a^{p} d^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )} + a^{p} d e^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {c x^{2} e^{i \pi }}{a}} \right )} + 3 d^{2} e \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: c = 0 \\\frac {\begin {cases} \frac {\left (a + c x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + c x^{2} \right )} & \text {otherwise} \end {cases}}{2 c} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: c = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{c}} \right )}}{2 a c^{2} + 2 c^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{c}} \right )}}{2 a c^{2} + 2 c^{3} x^{2}} + \frac {a}{2 a c^{2} + 2 c^{3} x^{2}} + \frac {c x^{2} \log {\left (x - \sqrt {- \frac {a}{c}} \right )}}{2 a c^{2} + 2 c^{3} x^{2}} + \frac {c x^{2} \log {\left (x + \sqrt {- \frac {a}{c}} \right )}}{2 a c^{2} + 2 c^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{c}} \right )}}{2 c^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{c}} \right )}}{2 c^{2}} + \frac {x^{2}}{2 c} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + c x^{2}\right )^{p}}{2 c^{2} p^{2} + 6 c^{2} p + 4 c^{2}} + \frac {a c p x^{2} \left (a + c x^{2}\right )^{p}}{2 c^{2} p^{2} + 6 c^{2} p + 4 c^{2}} + \frac {c^{2} p x^{4} \left (a + c x^{2}\right )^{p}}{2 c^{2} p^{2} + 6 c^{2} p + 4 c^{2}} + \frac {c^{2} x^{4} \left (a + c x^{2}\right )^{p}}{2 c^{2} p^{2} + 6 c^{2} p + 4 c^{2}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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\[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{3} {\left (c x^{2} + a\right )}^{p} \,d x } \]
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Timed out. \[ \int (d+e x)^3 \left (a+c x^2\right )^p \, dx=\int {\left (c\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]
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