Optimal. Leaf size=466 \[ \frac{b e x^{-2 p-5} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{1}{2} (-2 p-5),-p-1,\frac{1}{2} (-2 p-3),-\frac{e x^2}{d}\right )}{c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}-\frac{b e^2 x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \text{Hypergeometric2F1}\left (\frac{1}{2} (-2 p-3),-p-1,\frac{1}{2} (-2 p-1),-\frac{e x^2}{d}\right )}{c d^2 (p+1) (p+2) (p+3) (2 p+3)}-\frac{e^2 x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (p+1) (p+2) (p+3)}+\frac{e x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (p+2) (p+3)}-\frac{x^{-2 (p+3)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+3)}-\frac{b x^{-2 p-5} \left (c^4 d^2 \left (p^2+3 p+2\right )+2 c^2 d e (p+1)+2 e^2\right ) \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-5);1,-p-1;\frac{1}{2} (-2 p-3);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)} \]
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Rubi [A] time = 1.42578, antiderivative size = 466, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {271, 264, 4976, 12, 6725, 365, 364, 511, 510} \[ -\frac{e^2 x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (p+1) (p+2) (p+3)}+\frac{e x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (p+2) (p+3)}-\frac{x^{-2 (p+3)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+3)}-\frac{b x^{-2 p-5} \left (c^4 d^2 \left (p^2+3 p+2\right )+2 c^2 d e (p+1)+2 e^2\right ) \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} F_1\left (\frac{1}{2} (-2 p-5);1,-p-1;\frac{1}{2} (-2 p-3);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}+\frac{b e x^{-2 p-5} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac{1}{2} (-2 p-5),-p-1;\frac{1}{2} (-2 p-3);-\frac{e x^2}{d}\right )}{c^3 d^2 (p+1) (p+2) (p+3) (2 p+5)}-\frac{b e^2 x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac{e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac{1}{2} (-2 p-3),-p-1;\frac{1}{2} (-2 p-1);-\frac{e x^2}{d}\right )}{c d^2 (p+1) (p+2) (p+3) (2 p+3)} \]
Antiderivative was successfully verified.
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Rule 271
Rule 264
Rule 4976
Rule 12
Rule 6725
Rule 365
Rule 364
Rule 511
Rule 510
Rubi steps
\begin{align*} \int x^{-7-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-(b c) \int \frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (-d^2 \left (2+3 p+p^2\right )+2 d e (1+p) x^2-2 e^2 x^4\right )}{2 d^3 (1+p) (2+p) (3+p) \left (1+c^2 x^2\right )} \, dx\\ &=-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-\frac{(b c) \int \frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (-d^2 \left (2+3 p+p^2\right )+2 d e (1+p) x^2-2 e^2 x^4\right )}{1+c^2 x^2} \, dx}{2 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}-\frac{(b c) \int \left (\frac{2 e \left (e+c^2 d (1+p)\right ) x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^4}-\frac{2 e^2 x^{2-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^2}+\frac{\left (-2 c^4 d^2-2 c^2 d e-2 e^2-3 c^4 d^2 p-2 c^2 d e p-c^4 d^2 p^2\right ) x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac{\left (b e^2\right ) \int x^{2-2 (3+p)} \left (d+e x^2\right )^{1+p} \, dx}{c d^3 (1+p) (2+p) (3+p)}-\frac{\left (b e \left (e+c^2 d (1+p)\right )\right ) \int x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \, dx}{c^3 d^3 (1+p) (2+p) (3+p)}+\frac{\left (b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right )\right ) \int \frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c^3 d^3 (1+p) (2+p) (3+p)}\\ &=-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac{\left (b e^2 \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p}\right ) \int x^{2-2 (3+p)} \left (1+\frac{e x^2}{d}\right )^{1+p} \, dx}{c d^2 (1+p) (2+p) (3+p)}-\frac{\left (b e \left (e+c^2 d (1+p)\right ) \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p}\right ) \int x^{-2 (3+p)} \left (1+\frac{e x^2}{d}\right )^{1+p} \, dx}{c^3 d^2 (1+p) (2+p) (3+p)}+\frac{\left (b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right ) \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p}\right ) \int \frac{x^{-2 (3+p)} \left (1+\frac{e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c^3 d^2 (1+p) (2+p) (3+p)}\\ &=-\frac{b \left (2 e^2+2 c^2 d e (1+p)+c^4 d^2 \left (2+3 p+p^2\right )\right ) x^{-5-2 p} \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p} F_1\left (\frac{1}{2} (-5-2 p);1,-1-p;\frac{1}{2} (-3-2 p);-c^2 x^2,-\frac{e x^2}{d}\right )}{2 c^3 d^2 (1+p) (2+p) (3+p) (5+2 p)}-\frac{e^2 x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^3 (1+p) (2+p) (3+p)}+\frac{e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{d^2 (2+p) (3+p)}-\frac{x^{-2 (3+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (3+p)}+\frac{b e \left (e+c^2 d (1+p)\right ) x^{-5-2 p} \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p} \, _2F_1\left (\frac{1}{2} (-5-2 p),-1-p;\frac{1}{2} (-3-2 p);-\frac{e x^2}{d}\right )}{c^3 d^2 (1+p) (2+p) (3+p) (5+2 p)}-\frac{b e^2 x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac{e x^2}{d}\right )^{-p} \, _2F_1\left (\frac{1}{2} (-3-2 p),-1-p;\frac{1}{2} (-1-2 p);-\frac{e x^2}{d}\right )}{c d^2 (1+p) (2+p) (3+p) (3+2 p)}\\ \end{align*}
Mathematica [F] time = 4.4568, size = 0, normalized size = 0. \[ \int x^{-7-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.874, size = 0, normalized size = 0. \begin{align*} \int{x}^{-7-2\,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^{7}}\,{d x} - \frac{{\left (2 \, e^{3} x^{6} - 2 \, d e^{2} p x^{4} +{\left (p^{2} + p\right )} d^{2} e x^{2} +{\left (p^{2} + 3 \, p + 2\right )} d^{3}\right )} a e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \left (x\right )\right )}}{2 \,{\left (p^{3} + 6 \, p^{2} + 11 \, p + 6\right )} d^{3} x^{6}} \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 - 7}, 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 - 7}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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