Optimal. Leaf size=152 \[ -\frac{a d e \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac{d e \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}-\frac{x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (3 a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )}{3 b (2 p+5)}+\frac{e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
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Rubi [A] time = 0.136522, antiderivative size = 144, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {1652, 459, 365, 364, 12, 266, 43} \[ -\frac{a d e \left (a+b x^2\right )^{p+1}}{b^2 (p+1)}+\frac{d e \left (a+b x^2\right )^{p+2}}{b^2 (p+2)}+\frac{1}{3} x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{3 a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+\frac{e^2 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
Antiderivative was successfully verified.
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Rule 1652
Rule 459
Rule 365
Rule 364
Rule 12
Rule 266
Rule 43
Rubi steps
\begin{align*} \int x^2 (d+e x)^2 \left (a+b x^2\right )^p \, dx &=\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) \operatorname{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) \operatorname{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*}
Mathematica [A] time = 0.159155, size = 139, normalized size = 0.91 \[ \frac{1}{15} \left (a+b x^2\right )^p \left (\frac{3 e \left (e \left (p^2+3 p+2\right ) x^5 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )-\frac{5 d \left (a+b x^2\right ) \left (a-b (p+1) x^2\right )}{b^2}\right )}{(p+1) (p+2)}+5 d^2 x^3 \left (\frac{b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )\right ) \]
Antiderivative was successfully verified.
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Maple [F] time = 0.519, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( ex+d \right ) ^{2} \left ( b{x}^{2}+a \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{2}{\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e^{2} x^{4} + 2 \, d e x^{3} + d^{2} x^{2}\right )}{\left (b x^{2} + a\right )}^{p}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 39.5501, size = 430, normalized size = 2.83 \begin{align*} \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 (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \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 (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac{b x^{2} \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text{for}\: p = -2 \\- \frac{a \log{\left (- i \sqrt{a} \sqrt{\frac{1}{b}} + x \right )}}{2 b^{2}} - \frac{a \log{\left (i \sqrt{a} \sqrt{\frac{1}{b}} + x \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 ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x + d\right )}^{2}{\left (b x^{2} + a\right )}^{p} x^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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