Optimal. Leaf size=113 \[ \frac {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {2}{3} d e x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \]
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Rubi [A] time = 0.10, antiderivative size = 113, normalized size of antiderivative = 1.00, 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, 12, 365, 364} \[ \frac {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {e^2 \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {2}{3} d e x^3 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right ) \]
Antiderivative was successfully verified.
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Rule 12
Rule 43
Rule 364
Rule 365
Rule 444
Rule 1652
Rubi steps
\begin {align*} \int x (d+e x)^2 \left (a+b x^2\right )^p \, dx &=\int 2 d e x^2 \left (a+b x^2\right )^p \, dx+\int x \left (a+b x^2\right )^p \left (d^2+e^2 x^2\right ) \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int (a+b x)^p \left (d^2+e^2 x\right ) \, dx,x,x^2\right )+(2 d e) \int x^2 \left (a+b x^2\right )^p \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {\left (b d^2-a e^2\right ) (a+b x)^p}{b}+\frac {e^2 (a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right )+\left (2 d e \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 {\left (b d^2-a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e^2 \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {2}{3} d e 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*}
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Mathematica [A] time = 0.18, size = 184, normalized size = 1.63 \[ \frac {\left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (-3 a^2 e^2 \left (\left (\frac {b x^2}{a}+1\right )^p-1\right )+3 b^2 x^2 \left (\frac {b x^2}{a}+1\right )^p \left (d^2 (p+2)+e^2 (p+1) x^2\right )+4 b^2 d e \left (p^2+3 p+2\right ) x^3 \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )+3 a b \left (d^2 (p+2) \left (\left (\frac {b x^2}{a}+1\right )^p-1\right )+e^2 p x^2 \left (\frac {b x^2}{a}+1\right )^p\right )\right )}{6 b^2 (p+1) (p+2)} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{2} x^{3} + 2 \, d e x^{2} + d^{2} x\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e x + d\right )}^{2} {\left (b x^{2} + a\right )}^{p} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.04, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{2} x \left (b \,x^{2}+a \right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (b x^{2} + a\right )}^{p + 1} d^{2}}{2 \, b {\left (p + 1\right )}} + \int {\left (e^{2} x^{3} + 2 \, d e x^{2}\right )} {\left (b x^{2} + a\right )}^{p}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 12.91, size = 439, normalized size = 3.88 \[ \frac {2 a^{p} d e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3} + d^{2} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) + e^{2} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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