Integrand size = 22, antiderivative size = 60 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\frac {(a+b x)^3 \left (a^2-b^2 x^2\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,5+2 p,5+p,\frac {a+b x}{2 a}\right )}{2 a b (4+p)} \]
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Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {692, 71} \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=-\frac {a^2 2^{p+3} \left (\frac {b x}{a}+1\right )^{-p-1} \left (a^2-b^2 x^2\right )^{p+1} \operatorname {Hypergeometric2F1}\left (-p-3,p+1,p+2,\frac {a-b x}{2 a}\right )}{b (p+1)} \]
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Rule 71
Rule 692
Rubi steps \begin{align*} \text {integral}& = \left (a^2 (a-b x)^{-1-p} \left (1+\frac {b x}{a}\right )^{-1-p} \left (a^2-b^2 x^2\right )^{1+p}\right ) \int (a-b x)^p \left (1+\frac {b x}{a}\right )^{3+p} \, dx \\ & = -\frac {2^{3+p} a^2 \left (1+\frac {b x}{a}\right )^{-1-p} \left (a^2-b^2 x^2\right )^{1+p} \, _2F_1\left (-3-p,1+p;2+p;\frac {a-b x}{2 a}\right )}{b (1+p)} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(155\) vs. \(2(60)=120\).
Time = 0.48 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.58 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\frac {1}{2} \left (a^2-b^2 x^2\right )^p \left (\frac {\left (-a^2+b^2 x^2\right ) \left (a^2 (7+3 p)+b^2 (1+p) x^2\right )}{b (1+p) (2+p)}+2 a^3 x \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b^2 x^2}{a^2}\right )+2 a b^2 x^3 \left (1-\frac {b^2 x^2}{a^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},\frac {b^2 x^2}{a^2}\right )\right ) \]
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\[\int \left (b x +a \right )^{3} \left (-b^{2} x^{2}+a^{2}\right )^{p}d x\]
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\[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\int { {\left (b x + a\right )}^{3} {\left (-b^{2} x^{2} + a^{2}\right )}^{p} \,d x } \]
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Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (44) = 88\).
Time = 2.07 (sec) , antiderivative size = 476, normalized size of antiderivative = 7.93 \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=a^{3} a^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + 3 a^{2} b \left (\begin {cases} \frac {x^{2} \left (a^{2}\right )^{p}}{2} & \text {for}\: b^{2} = 0 \\- \frac {\begin {cases} \frac {\left (a^{2} - b^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a^{2} - b^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 b^{2}} & \text {otherwise} \end {cases}\right ) + a a^{2 p} b^{2} x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, - p \\ \frac {5}{2} \end {matrix}\middle | {\frac {b^{2} x^{2} e^{2 i \pi }}{a^{2}}} \right )} + b^{3} \left (\begin {cases} \frac {x^{4} \left (a^{2}\right )^{p}}{4} & \text {for}\: b = 0 \\- \frac {a^{2} \log {\left (- \frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} - \frac {a^{2}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {b^{2} x^{2} \log {\left (- \frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} + \frac {b^{2} x^{2} \log {\left (\frac {a}{b} + x \right )}}{- 2 a^{2} b^{4} + 2 b^{6} x^{2}} & \text {for}\: p = -2 \\- \frac {a^{2} \log {\left (- \frac {a}{b} + x \right )}}{2 b^{4}} - \frac {a^{2} \log {\left (\frac {a}{b} + x \right )}}{2 b^{4}} - \frac {x^{2}}{2 b^{2}} & \text {for}\: p = -1 \\- \frac {a^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} - \frac {a^{2} b^{2} p x^{2} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac {b^{4} p x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} + \frac {b^{4} x^{4} \left (a^{2} - b^{2} x^{2}\right )^{p}}{2 b^{4} p^{2} + 6 b^{4} p + 4 b^{4}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\int { {\left (b x + a\right )}^{3} {\left (-b^{2} x^{2} + a^{2}\right )}^{p} \,d x } \]
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\[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\int { {\left (b x + a\right )}^{3} {\left (-b^{2} x^{2} + a^{2}\right )}^{p} \,d x } \]
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Timed out. \[ \int (a+b x)^3 \left (a^2-b^2 x^2\right )^p \, dx=\int {\left (a^2-b^2\,x^2\right )}^p\,{\left (a+b\,x\right )}^3 \,d x \]
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