Integrand size = 20, antiderivative size = 152 \[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=-\frac {a d e \left (a+b x^2\right )^{1+p}}{b^2 (1+p)}+\frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {d e \left (a+b x^2\right )^{2+p}}{b^2 (2+p)}-\frac {\left (3 a e^2-b d^2 (5+2 p)\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )}{3 b (5+2 p)} \]
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Time = 0.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1666, 470, 372, 371, 12, 272, 45} \[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=-\frac {a d e \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac {d e \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}+\frac {1}{3} x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (d^2-\frac {3 a e^2}{2 b p+5 b}\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
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Rule 12
Rule 45
Rule 272
Rule 371
Rule 372
Rule 470
Rule 1666
Rubi steps \begin{align*} \text {integral}& = \int 2 d e x^3 \left (a+b x^2\right )^p \, dx+\int x^2 \left (a+b x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx \\ & = \frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+(2 d e) \int x^3 \left (a+b x^2\right )^p \, dx-\left (-d^2+\frac {3 a e^2}{5 b+2 b p}\right ) \int x^2 \left (a+b x^2\right )^p \, dx \\ & = \frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+(d e) \text {Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )-\left (\left (-d^2+\frac {3 a e^2}{5 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^2 \left (1+\frac {b x^2}{a}\right )^p \, dx \\ & = \frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {1}{3} \left (d^2-\frac {3 a e^2}{5 b+2 b p}\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+(d e) \text {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right ) \\ & = -\frac {a d e \left (a+b x^2\right )^{1+p}}{b^2 (1+p)}+\frac {e^2 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {d e \left (a+b x^2\right )^{2+p}}{b^2 (2+p)}+\frac {1}{3} \left (d^2-\frac {3 a e^2}{5 b+2 b p}\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \\ \end{align*}
Time = 0.22 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.91 \[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {1}{15} \left (a+b x^2\right )^p \left (5 d^2 x^3 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b x^2}{a}\right )+\frac {3 e \left (-\frac {5 d \left (a+b x^2\right ) \left (a-b (1+p) x^2\right )}{b^2}+e \left (2+3 p+p^2\right ) x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right )}{(1+p) (2+p)}\right ) \]
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\[\int x^{2} \left (e x +d \right )^{2} \left (b \,x^{2}+a \right )^{p}d x\]
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\[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (126) = 252\).
Time = 8.84 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.63 \[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\frac {a^{p} d^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + \frac {a^{p} e^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + 2 d e \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \]
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\[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x^{2} \,d x } \]
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Timed out. \[ \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx=\int x^2\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
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