Integrand size = 18, antiderivative size = 100 \[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=-\frac {a d \left (a+b x^2\right )^{1+p}}{2 b^2 (1+p)}+\frac {d \left (a+b x^2\right )^{2+p}}{2 b^2 (2+p)}+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {778, 272, 45, 372, 371} \[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=-\frac {a d \left (a+b x^2\right )^{p+1}}{2 b^2 (p+1)}+\frac {d \left (a+b x^2\right )^{p+2}}{2 b^2 (p+2)}+\frac {1}{5} e x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \]
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Rule 45
Rule 272
Rule 371
Rule 372
Rule 778
Rubi steps \begin{align*} \text {integral}& = 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 \text {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 \text {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*}
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (-\frac {5 d \left (a+b x^2\right ) \left (a-b (1+p) x^2\right )}{b^2 (1+p) (2+p)}+2 e x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \]
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\[\int x^{3} \left (e x +d \right ) \left (b \,x^{2}+a \right )^{p}d x\]
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\[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (82) = 164\).
Time = 6.90 (sec) , antiderivative size = 364, normalized size of antiderivative = 3.64 \[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} e x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, - p \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{5} + d \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {a}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} + \frac {b x^{2} \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 a b^{2} + 2 b^{3} x^{2}} & \text {for}\: p = -2 \\- \frac {a \log {\left (x - \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} - \frac {a \log {\left (x + \sqrt {- \frac {a}{b}} \right )}}{2 b^{2}} + \frac {x^{2}}{2 b} & \text {for}\: p = -1 \\- \frac {a^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {a b p x^{2} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} p x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} + \frac {b^{2} x^{4} \left (a + b x^{2}\right )^{p}}{2 b^{2} p^{2} + 6 b^{2} p + 4 b^{2}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
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\[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{3} \,d x } \]
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Timed out. \[ \int x^3 (d+e x) \left (a+b x^2\right )^p \, dx=\int x^3\,{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]
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