Optimal. Leaf size=100 \[ -\frac {a d \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {d \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \]
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Rubi [A] time = 0.06, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {764, 266, 43, 365, 364} \[ -\frac {a d \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {d \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right ) \]
Antiderivative was successfully verified.
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Rule 43
Rule 266
Rule 364
Rule 365
Rule 764
Rubi steps
\begin {align*} \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx &=d \int x^3 \left (a+b x^2\right )^p \, dx+e \int x^4 \left (a+b x^2\right )^p \, dx\\ &=\frac {1}{2} d \operatorname {Subst}\left (\int x (a+b x)^p \, dx,x,x^2\right )+\left (e \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 {1}{5} e 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 )+\frac {1}{2} d \operatorname {Subst}\left (\int \left (-\frac {a (a+b x)^p}{b}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,x^2\right )\\ &=-\frac {a d \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {d \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {1}{5} e 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 = 0.07, size = 87, normalized size = 0.87 \[ \frac {1}{10} \left (a+b x^2\right )^p \left (2 e x^5 \left (\frac {b x^2}{a}+1\right )^{-p} \, _2F_1\left (\frac {5}{2},-p;\frac {7}{2};-\frac {b x^2}{a}\right )-\frac {5 d \left (a+b x^2\right ) \left (a-b (p+1) x^2\right )}{b^2 (p+1) (p+2)}\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e x^{4} + d x^{3}\right )} {\left (b x^{2} + a\right )}^{p}, x\right ) \]
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 \[ \int {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{3}\,{d x} \]
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 \[ \int \left (e x +d \right ) x^{3} \left (b \,x^{2}+a \right )^{p}\, dx \]
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 \[ e \int {\left (b x^{2} + a\right )}^{p} x^{4}\,{d x} + \frac {{\left (b^{2} {\left (p + 1\right )} x^{4} + a b p x^{2} - a^{2}\right )} {\left (b x^{2} + a\right )}^{p} d}{2 \, {\left (p^{2} + 3 \, p + 2\right )} b^{2}} \]
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 \[ \int x^3\,{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [C] time = 16.78, size = 394, normalized size = 3.94 \[ \frac {a^{p} e x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + d \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} - \frac {a \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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