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 = 0.15, 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 43
Rule 364
Rule 365
Rule 444
Rule 459
Rule 1652
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*}
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Mathematica [A] time = 0.23, 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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fricas [F] time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{4} + 3 \, d e^{2} x^{3} + 3 \, d^{2} e x^{2} + d^{3} x\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]
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 \[ \int {\left (e x + d\right )}^{3} {\left (b x^{2} + a\right )}^{p} x\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (e x +d \right )^{3} x \left (b \,x^{2}+a \right )^{p}\, dx \]
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 \[ \frac {{\left (b x^{2} + a\right )}^{p + 1} d^{3}}{2 \, b {\left (p + 1\right )}} + \int {\left (e^{3} x^{4} + 3 \, d e^{2} x^{3} + 3 \, d^{2} e x^{2}\right )} {\left (b x^{2} + a\right )}^{p}\,{d x} \]
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 \[ \int x\,{\left (b\,x^2+a\right )}^p\,{\left (d+e\,x\right )}^3 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 22.80, size = 471, normalized size = 2.82 \[ a^{p} d^{2} e x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )} + \frac {a^{p} e^{3} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + d^{3} \left (\begin {cases} \frac {a^{p} x^{2}}{2} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{2} \right )} & \text {otherwise} \end {cases}}{2 b} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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