Optimal. Leaf size=207 \[ -\frac {a d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {e^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {d \left (b d^2-6 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {3 d e^2 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}-\frac {e \left (5 a e^2-3 b d^2 (7+2 p)\right ) x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )}{5 b (7+2 p)} \]
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Rubi [A]
time = 0.13, antiderivative size = 201, normalized size of antiderivative = 0.97, number of steps
used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1666, 457, 78,
470, 372, 371} \begin {gather*} -\frac {a d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}+\frac {d \left (b d^2-6 a e^2\right ) \left (a+b x^2\right )^{p+2}}{2 b^3 (p+2)}+\frac {3 d e^2 \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \left (3 d^2-\frac {5 a e^2}{2 b p+7 b}\right ) \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )+\frac {e^3 x^5 \left (a+b x^2\right )^{p+1}}{b (2 p+7)} \end {gather*}
Antiderivative was successfully verified.
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Rule 78
Rule 371
Rule 372
Rule 457
Rule 470
Rule 1666
Rubi steps
\begin {align*} \int x^3 (d+e x)^3 \left (a+b x^2\right )^p \, dx &=\int x^3 \left (a+b x^2\right )^p \left (d^3+3 d e^2 x^2\right ) \, dx+\int x^4 \left (a+b x^2\right )^p \left (3 d^2 e+e^3 x^2\right ) \, dx\\ &=\frac {e^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {1}{2} \text {Subst}\left (\int x (a+b x)^p \left (d^3+3 d e^2 x\right ) \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac {5 a e^2}{7 b+2 b p}\right )\right ) \int x^4 \left (a+b x^2\right )^p \, dx\\ &=\frac {e^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {1}{2} \text {Subst}\left (\int \left (\frac {a d \left (-b d^2+3 a e^2\right ) (a+b x)^p}{b^2}+\frac {\left (b d^3-6 a d e^2\right ) (a+b x)^{1+p}}{b^2}+\frac {3 d e^2 (a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^2\right )+\left (e \left (3 d^2-\frac {5 a e^2}{7 b+2 b p}\right ) \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p}\right ) \int x^4 \left (1+\frac {b x^2}{a}\right )^p \, dx\\ &=-\frac {a d \left (b d^2-3 a e^2\right ) \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}+\frac {e^3 x^5 \left (a+b x^2\right )^{1+p}}{b (7+2 p)}+\frac {d \left (b d^2-6 a e^2\right ) \left (a+b x^2\right )^{2+p}}{2 b^3 (2+p)}+\frac {3 d e^2 \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {1}{5} e \left (3 d^2-\frac {5 a e^2}{7 b+2 b p}\right ) x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )\\ \end {align*}
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Mathematica [A]
time = 0.29, size = 196, normalized size = 0.95 \begin {gather*} \frac {1}{70} \left (a+b x^2\right )^p \left (\frac {35 d^3 \left (a+b x^2\right ) \left (-a+b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+\frac {105 d e^2 \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+42 d^2 e x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )+10 e^3 x^7 \left (1+\frac {b x^2}{a}\right )^{-p} \, _2F_1\left (\frac {7}{2},-p;\frac {9}{2};-\frac {b x^2}{a}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{3} \left (e x +d \right )^{3} \left (b \,x^{2}+a \right )^{p}\, dx\]
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 = 19.07, size = 1329, normalized size = 6.42 \begin {gather*} \frac {3 a^{p} d^{2} e 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} + \frac {a^{p} e^{3} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, - p \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{7} + d^{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 ) + 3 d e^{2} \left (\begin {cases} \frac {a^{p} x^{6}}{6} & \text {for}\: b = 0 \\\frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {3 a^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {4 a b x^{2}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} + \frac {2 b^{2} x^{4} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{4 a^{2} b^{3} + 8 a b^{4} x^{2} + 4 b^{5} x^{4}} & \text {for}\: p = -3 \\- \frac {2 a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -2 \\\frac {a^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} + \frac {a^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{3}} - \frac {a x^{2}}{2 b^{2}} + \frac {x^{4}}{4 b} & \text {for}\: p = -1 \\\frac {2 a^{3} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {a b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {b^{3} p^{2} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {3 b^{3} p x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} + \frac {2 b^{3} x^{6} \left (a + b x^{2}\right )^{p}}{2 b^{3} p^{3} + 12 b^{3} p^{2} + 22 b^{3} p + 12 b^{3}} & \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 = -0.00 \begin {gather*} \int x^3\,{\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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