Optimal. Leaf size=176 \[ -\frac {e \left (a e^2-3 b d^2 (2+p)\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p) (2+p)}+\frac {3 d e^2 x \left (a+b x^2\right )^{1+p}}{b (3+2 p)}+\frac {e^3 x^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac {d \left (3 a e^2-b d^2 (3+2 p)\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )}{b (3+2 p)} \]
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Rubi [A]
time = 0.10, antiderivative size = 169, normalized size of antiderivative = 0.96, number of steps
used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {757, 794, 252,
251} \begin {gather*} -\frac {e \left (a+b x^2\right )^{p+1} \left ((2 p+3) \left (a e^2-b d^2 (2 p+5)\right )-2 b d e (p+1) (p+3) x\right )}{2 b^2 (p+2) \left (2 p^2+5 p+3\right )}+d x \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+3 b}\right ) \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+\frac {e (d+e x)^2 \left (a+b x^2\right )^{p+1}}{2 b (p+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 251
Rule 252
Rule 757
Rule 794
Rubi steps
\begin {align*} \int (d+e x)^3 \left (a+b x^2\right )^p \, dx &=\frac {e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}+\frac {\int (d+e x) \left (-2 \left (a e^2-b d^2 (2+p)\right )+2 b d e (3+p) x\right ) \left (a+b x^2\right )^p \, dx}{2 b (2+p)}\\ &=\frac {e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac {3 a e^2}{3 b+2 b p}\right )\right ) \int \left (a+b x^2\right )^p \, dx\\ &=\frac {e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+\left (d \left (d^2-\frac {3 a e^2}{3 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=\frac {e (d+e x)^2 \left (a+b x^2\right )^{1+p}}{2 b (2+p)}-\frac {e \left ((3+2 p) \left (a e^2-b d^2 (5+2 p)\right )-2 b d e (1+p) (3+p) x\right ) \left (a+b x^2\right )^{1+p}}{2 b^2 (2+p) \left (3+5 p+2 p^2\right )}+d \left (d^2-\frac {3 a e^2}{3 b+2 b p}\right ) x \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )\\ \end {align*}
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Mathematica [A]
time = 0.28, size = 223, normalized size = 1.27 \begin {gather*} \frac {\left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (2 b^2 d^3 \left (2+3 p+p^2\right ) x \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};-\frac {b x^2}{a}\right )+e \left (b^2 x^2 \left (1+\frac {b x^2}{a}\right )^p \left (3 d^2 (2+p)+e^2 (1+p) x^2\right )-a^2 e^2 \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )+a b \left (e^2 p x^2 \left (1+\frac {b x^2}{a}\right )^p+3 d^2 (2+p) \left (-1+\left (1+\frac {b x^2}{a}\right )^p\right )\right )+2 b^2 d 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 )\right )\right )}{2 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 \[\int \left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}\, dx\]
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 [B] Leaf count of result is larger than twice the leaf count of optimal. 337 vs.
\(2 (148) = 296\).
time = 6.71, size = 437, normalized size = 2.48 \begin {gather*} a^{p} d^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + a^{p} d e^{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 d^{2} e \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^{3} \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 {\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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