Integrand size = 20, antiderivative size = 83 \[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e (1+p)}+d x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right ) \]
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Time = 0.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {655, 252, 251} \[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=d x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right )-\frac {\left (d^2-e^2 x^2\right )^{p+1}}{2 e (p+1)} \]
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Rule 251
Rule 252
Rule 655
Rubi steps \begin{align*} \text {integral}& = -\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e (1+p)}+d \int \left (d^2-e^2 x^2\right )^p \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e (1+p)}+\left (d \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p}\right ) \int \left (1-\frac {e^2 x^2}{d^2}\right )^p \, dx \\ & = -\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e (1+p)}+d x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \, _2F_1\left (\frac {1}{2},-p;\frac {3}{2};\frac {e^2 x^2}{d^2}\right ) \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=-\frac {\left (d^2-e^2 x^2\right )^{1+p}}{2 e (1+p)}+d x \left (d^2-e^2 x^2\right )^p \left (1-\frac {e^2 x^2}{d^2}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {e^2 x^2}{d^2}\right ) \]
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\[\int \left (e x +d \right ) \left (-e^{2} x^{2}+d^{2}\right )^{p}d x\]
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\[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]
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Time = 1.13 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=d d^{2 p} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {e^{2} x^{2} e^{2 i \pi }}{d^{2}}} \right )} + e \left (\begin {cases} \frac {x^{2} \left (d^{2}\right )^{p}}{2} & \text {for}\: e^{2} = 0 \\- \frac {\begin {cases} \frac {\left (d^{2} - e^{2} x^{2}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (d^{2} - e^{2} x^{2} \right )} & \text {otherwise} \end {cases}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) \]
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\[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]
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\[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\int { {\left (e x + d\right )} {\left (-e^{2} x^{2} + d^{2}\right )}^{p} \,d x } \]
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Time = 13.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int (d+e x) \left (d^2-e^2 x^2\right )^p \, dx=\frac {d\,x\,{\left (d^2-e^2\,x^2\right )}^p\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},-p;\ \frac {3}{2};\ \frac {e^2\,x^2}{d^2}\right )}{{\left (1-\frac {e^2\,x^2}{d^2}\right )}^p}-\frac {{\left (d^2-e^2\,x^2\right )}^{p+1}}{2\,e\,\left (p+1\right )} \]
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