Optimal. Leaf size=285 \[ \frac {e x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+2)}-\frac {b x^{-2 p-3} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} F_1\left (\frac {1}{2} (-2 p-3);1,-p-1;\frac {1}{2} (-2 p-1);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (p+1) (p+2) (2 p+3)}+\frac {b e x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-3),-p-1;\frac {1}{2} (-2 p-1);-\frac {e x^2}{d}\right )}{2 c d \left (2 p^3+9 p^2+13 p+6\right )} \]
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Rubi [A] time = 0.37, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {271, 264, 4976, 12, 584, 365, 364, 511, 510} \[ \frac {e x^{-2 (p+1)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (p+1) (p+2)}-\frac {x^{-2 (p+2)} \left (d+e x^2\right )^{p+1} \left (a+b \tan ^{-1}(c x)\right )}{2 d (p+2)}-\frac {b x^{-2 p-3} \left (c^2 d (p+1)+e\right ) \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} F_1\left (\frac {1}{2} (-2 p-3);1,-p-1;\frac {1}{2} (-2 p-1);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (p+1) (p+2) (2 p+3)}+\frac {b e x^{-2 p-3} \left (d+e x^2\right )^p \left (\frac {e x^2}{d}+1\right )^{-p} \, _2F_1\left (\frac {1}{2} (-2 p-3),-p-1;\frac {1}{2} (-2 p-1);-\frac {e x^2}{d}\right )}{2 c d \left (2 p^3+9 p^2+13 p+6\right )} \]
Antiderivative was successfully verified.
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Rule 12
Rule 264
Rule 271
Rule 364
Rule 365
Rule 510
Rule 511
Rule 584
Rule 4976
Rubi steps
\begin {align*} \int x^{-5-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{2 d^2 (1+p) (2+p) \left (1+c^2 x^2\right )} \, dx\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b c) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (-d-d p+e x^2\right )}{1+c^2 x^2} \, dx}{2 d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b c) \int \left (\frac {e x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2}+\frac {\left (-e+c^2 (-d-d p)\right ) x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{c^2 \left (1+c^2 x^2\right )}\right ) \, dx}{2 d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {(b e) \int x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \, dx}{2 c d^2 (1+p) (2+p)}+\frac {\left (b \left (e+c^2 d (1+p)\right )\right ) \int \frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d^2 (1+p) (2+p)}\\ &=\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}-\frac {\left (b e \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p}\right ) \int x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p} \, dx}{2 c d (1+p) (2+p)}+\frac {\left (b \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 \frac {x^{-2 (2+p)} \left (1+\frac {e x^2}{d}\right )^{1+p}}{1+c^2 x^2} \, dx}{2 c d (1+p) (2+p)}\\ &=-\frac {b \left (e+c^2 d (1+p)\right ) x^{-3-2 p} \left (d+e x^2\right )^p \left (1+\frac {e x^2}{d}\right )^{-p} F_1\left (\frac {1}{2} (-3-2 p);1,-1-p;\frac {1}{2} (-1-2 p);-c^2 x^2,-\frac {e x^2}{d}\right )}{2 c d (1+p) (2+p) (3+2 p)}+\frac {e x^{-2 (1+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d^2 (1+p) (2+p)}-\frac {x^{-2 (2+p)} \left (d+e x^2\right )^{1+p} \left (a+b \tan ^{-1}(c x)\right )}{2 d (2+p)}+\frac {b e 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 )}{2 c d \left (6+13 p+9 p^2+2 p^3\right )}\\ \end {align*}
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Mathematica [F] time = 4.63, size = 0, normalized size = 0.00 \[ \int x^{-5-2 p} \left (d+e x^2\right )^p \left (a+b \tan ^{-1}(c x)\right ) \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 5}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \arctan \left (c x\right ) + a\right )} {\left (e x^{2} + d\right )}^{p} x^{-2 \, p - 5}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.55, size = 0, normalized size = 0.00 \[ \int x^{-5-2 p} \left (e \,x^{2}+d \right )^{p} \left (a +b \arctan \left (c x \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {\arctan \left (c x\right ) e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \relax (x)\right )}}{x^{5}}\,{d x} + \frac {{\left (e^{2} x^{4} - d e p x^{2} - d^{2} {\left (p + 1\right )}\right )} a e^{\left (p \log \left (e x^{2} + d\right ) - 2 \, p \log \relax (x)\right )}}{2 \, {\left (p^{2} + 3 \, p + 2\right )} d^{2} x^{4}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^p}{x^{2\,p+5}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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