Optimal. Leaf size=129 \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+1)}-\frac{b c x^{-2 p-1} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-1);1,-p-1;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (2 p^2+3 p+1\right )} \]
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Rubi [A] time = 0.166492, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {264, 4976, 12, 511, 510} \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+1)}-\frac{b c x^{-2 p-1} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-1);1,-p-1;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (2 p^2+3 p+1\right )} \]
Antiderivative was successfully verified.
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Rule 264
Rule 4976
Rule 12
Rule 511
Rule 510
Rubi steps
\begin{align*} \int x^{-3-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}-(b c) \int -\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{2 d (1+p) \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}+\frac{(b c) \int \frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 d (1+p)}\\ &=-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}+\frac{\left (b c \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p}\right ) \int \frac{x^{-2 (1+p)} \left (1+\frac{e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 (1+p)}\\ &=-\frac{b c x^{-1-2 p} \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p} F_1\left (\frac{1}{2} (-1-2 p);1,-1-p;\frac{1}{2} (1-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 \left (1+3 p+2 p^2\right )}-\frac{x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (1+p)}\\ \end{align*}
Mathematica [A] time = 0.434221, size = 166, normalized size = 1.29 \[ -\frac{x^{-2 (p+1)} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \left (b e x \text{Hypergeometric2F1}\left (-p-\frac{1}{2},-p,\frac{1}{2}-p,-\frac{e x^2}{d}\right )+c (2 p+1) \left (d+e x^2\right ) \left (\frac{e x^2}{d}+1\right )^p \left (a+b \tan ^{-1}(c x)\right )+b x \left (c^2 d-e\right ) F_1\left (-p-\frac{1}{2};-p,1;\frac{1}{2}-p;-\frac{e x^2}{d},-c^2 x^2\right )\right )}{2 c d (p+1) (2 p+1)} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.886, size = 0, normalized size = 0. \begin{align*} \int{x}^{-3-2\,p} \left ( e{x}^{2}+d \right ) ^{p} \left ( a+b\arctan \left ( cx \right ) \right ) \, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} b \int \frac{\arctan \left (c x\right ) e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \left (x\right )\right )}}{x^{3}}\,{d x} - \frac{{\left (e x^{2} + d\right )} a e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \left (x\right )\right )}}{2 \, d{\left (p + 1\right )} x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \arctan \left (c x\right ) + a\right )}{\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \arctan \left (c x\right ) + a\right )}{\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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