Optimal. Leaf size=167 \[ \frac {d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {e^3 x^3 \left (a+b x^2\right )^{1+p}}{b (5+2 p)}+\frac {3 d e^2 \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}-\frac {e \left (a e^2-b d^2 (5+2 p)\right ) x^3 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {3}{2},-p;\frac {5}{2};-\frac {b x^2}{a}\right )}{b (5+2 p)} \]
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Rubi [A]
time = 0.10, 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 = {1666, 455, 45,
470, 372, 371} \begin {gather*} \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)} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 371
Rule 372
Rule 455
Rule 470
Rule 1666
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} \text {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} \text {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*}
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Mathematica [A]
time = 0.28, size = 228, normalized size = 1.37 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (5 d \left (b^2 x^2 \left (1+\frac {b x^2}{a}\right )^p \left (d^2 (2+p)+3 e^2 (1+p) x^2\right )-3 a^2 e^2 \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+a b \left (3 e^2 p x^2 \left (1+\frac {b x^2}{a}\right )^p+d^2 (2+p) \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )\right )+10 b^2 d^2 e \left (2+3 p+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 (2+3 p+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 (1+p) (2+p)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x \left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 10.25, size = 440, normalized size = 2.63 \begin {gather*} 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 (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 ) \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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