Integrand size = 18, antiderivative size = 125 \[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {a^2 e \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac {a e \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {e \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {1}{5} d x^5 \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \]
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Time = 0.06 (sec) , antiderivative size = 125, 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, 372, 371, 272, 45} \[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {a^2 e \left (a+b x^2\right )^{p+1}}{2 b^3 (p+1)}-\frac {a e \left (a+b x^2\right )^{p+2}}{b^3 (p+2)}+\frac {e \left (a+b x^2\right )^{p+3}}{2 b^3 (p+3)}+\frac {1}{5} d x^5 \left (a+b x^2\right )^p \left (\frac {b x^2}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right ) \]
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Rule 45
Rule 272
Rule 371
Rule 372
Rule 778
Rubi steps \begin{align*} \text {integral}& = d \int x^4 \left (a+b x^2\right )^p \, dx+e \int x^5 \left (a+b x^2\right )^p \, dx \\ & = \frac {1}{2} e \text {Subst}\left (\int x^2 (a+b x)^p \, dx,x,x^2\right )+\left (d \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} d 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} e \text {Subst}\left (\int \left (\frac {a^2 (a+b x)^p}{b^2}-\frac {2 a (a+b x)^{1+p}}{b^2}+\frac {(a+b x)^{2+p}}{b^2}\right ) \, dx,x,x^2\right ) \\ & = \frac {a^2 e \left (a+b x^2\right )^{1+p}}{2 b^3 (1+p)}-\frac {a e \left (a+b x^2\right )^{2+p}}{b^3 (2+p)}+\frac {e \left (a+b x^2\right )^{3+p}}{2 b^3 (3+p)}+\frac {1}{5} d 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 = 112, normalized size of antiderivative = 0.90 \[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {1}{10} \left (a+b x^2\right )^p \left (\frac {5 e \left (a+b x^2\right ) \left (2 a^2-2 a b (1+p) x^2+b^2 \left (2+3 p+p^2\right ) x^4\right )}{b^3 (1+p) (2+p) (3+p)}+2 d x^5 \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b x^2}{a}\right )\right ) \]
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\[\int x^{4} \left (e x +d \right ) \left (b \,x^{2}+a \right )^{p}d x\]
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\[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 921 vs. \(2 (104) = 208\).
Time = 8.51 (sec) , antiderivative size = 950, normalized size of antiderivative = 7.60 \[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\frac {a^{p} d 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} + e \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 ) \]
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\[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \]
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\[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (b x^{2} + a\right )}^{p} x^{4} \,d x } \]
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Timed out. \[ \int x^4 (d+e x) \left (a+b x^2\right )^p \, dx=\int x^4\,{\left (b\,x^2+a\right )}^p\,\left (d+e\,x\right ) \,d x \]
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