Optimal. Leaf size=167 \[ \frac{d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}-\frac{e x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (a e^2-b d^2 (2 p+5)\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )}{b (2 p+5)}+\frac{e^3 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
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Rubi [A] time = 0.149759, antiderivative size = 159, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {1652, 444, 43, 459, 365, 364} \[ \frac{d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac{3 d e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+e x^3 \left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (d^2-\frac{a e^2}{2 b p+5 b}\right ) \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+\frac{e^3 x^3 \left (a+b x^2\right )^{p+1}}{b (2 p+5)} \]
Antiderivative was successfully verified.
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Rule 1652
Rule 444
Rule 43
Rule 459
Rule 365
Rule 364
Rubi steps
\begin{align*} \int x (d+e x)^3 \left (a+b x^2\right )^p \, dx &=\int x \left (a+b x^2\right )^p \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^2 \left (a+b x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx\\ &=\frac{e^3 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac{1}{2} \operatorname{Subst}\left (\int (a+b x)^p \left (d^3+3 d e^2 x\right ) \, dx,x,x^2\right )+\left (3 e \left (d^2-\frac{a e^2}{5 b+2 b p}\right )\right ) \int x^2 \left (a+b x^2\right )^p \, dx\\ &=\frac{e^3 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{\left (b d^3-3 a d e^2\right ) (a+b x)^p}{b}+\frac{3 d e^2 (a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right )+\left (3 e \left (d^2-\frac{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{d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac{e^3 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac{3 d e^2 \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+e \left (d^2-\frac{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 = 0.244, size = 228, normalized size = 1.37 \[ \frac{\left (a+b x^2\right )^p \left (\frac{b x^2}{a}+1\right )^{-p} \left (5 d \left (-3 a^2 e^2 \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+b^2 x^2 \left (\frac{b x^2}{a}+1\right )^p \left (d^2 (p+2)+3 e^2 (p+1) x^2\right )+a b \left (d^2 (p+2) \left (\left (\frac{b x^2}{a}+1\right )^p-1\right )+3 e^2 p x^2 \left (\frac{b x^2}{a}+1\right )^p\right )\right )+10 b^2 d^2 e \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac{3}{2},-p;\frac{5}{2};-\frac{b x^2}{a}\right )+2 b^2 e^3 \left (p^2+3 p+2\right ) x^5 \, _2F_1\left (\frac{5}{2},-p;\frac{7}{2};-\frac{b x^2}{a}\right )\right )}{10 b^2 (p+1) (p+2)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.536, size = 0, normalized size = 0. \begin{align*} \int x \left ( ex+d \right ) ^{3} \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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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^{3} x^{4} + 3 \, d e^{2} x^{3} + 3 \, d^{2} e x^{2} + d^{3} x\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 [A] time = 33.897, size = 471, normalized size = 2.82 \begin{align*} a^{p} d^{2} e 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 )} + \frac{a^{p} e^{3} 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} + d^{3} \left (\begin{cases} \frac{a^{p} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{2} \right )} & \text{otherwise} \end{cases}}{2 b} & \text{otherwise} \end{cases}\right ) + 3 d e^{2} \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 )}^{3}{\left (b x^{2} + a\right )}^{p} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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