Optimal. Leaf size=46 \[ \frac {1}{2} \log \left (x^2+2 x+3\right )-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {5 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{\sqrt {2}} \]
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Rubi [A] time = 0.13, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {6725, 203, 634, 618, 204, 628} \[ \frac {1}{2} \log \left (x^2+2 x+3\right )-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {5 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{\sqrt {2}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 204
Rule 618
Rule 628
Rule 634
Rule 6725
Rubi steps
\begin {align*} \int \frac {15-5 x+x^2+x^3}{\left (5+x^2\right ) \left (3+2 x+x^2\right )} \, dx &=\int \left (-\frac {5}{5+x^2}+\frac {6+x}{3+2 x+x^2}\right ) \, dx\\ &=-\left (5 \int \frac {1}{5+x^2} \, dx\right )+\int \frac {6+x}{3+2 x+x^2} \, dx\\ &=-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {1}{2} \int \frac {2+2 x}{3+2 x+x^2} \, dx+5 \int \frac {1}{3+2 x+x^2} \, dx\\ &=-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {1}{2} \log \left (3+2 x+x^2\right )-10 \operatorname {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,2+2 x\right )\\ &=-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {5 \tan ^{-1}\left (\frac {1+x}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {1}{2} \log \left (3+2 x+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.02, size = 46, normalized size = 1.00 \[ \frac {1}{2} \log \left (x^2+2 x+3\right )-\sqrt {5} \tan ^{-1}\left (\frac {x}{\sqrt {5}}\right )+\frac {5 \tan ^{-1}\left (\frac {x+1}{\sqrt {2}}\right )}{\sqrt {2}} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.77, size = 38, normalized size = 0.83 \[ \frac {5}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.36, size = 38, normalized size = 0.83 \[ \frac {5}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.00, size = 41, normalized size = 0.89 \[ -\sqrt {5}\, \arctan \left (\frac {\sqrt {5}\, x}{5}\right )+\frac {5 \sqrt {2}\, \arctan \left (\frac {\left (2 x +2\right ) \sqrt {2}}{4}\right )}{2}+\frac {\ln \left (x^{2}+2 x +3\right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 1.97, size = 38, normalized size = 0.83 \[ \frac {5}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (x + 1\right )}\right ) - \sqrt {5} \arctan \left (\frac {1}{5} \, \sqrt {5} x\right ) + \frac {1}{2} \, \log \left (x^{2} + 2 \, x + 3\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.00, size = 88, normalized size = 1.91 \[ \frac {\ln \left (x+1-\sqrt {2}\,1{}\mathrm {i}\right )}{2}+\frac {\ln \left (x+1+\sqrt {2}\,1{}\mathrm {i}\right )}{2}+\sqrt {5}\,\mathrm {atan}\left (\frac {2000\,\sqrt {5}}{2000\,x+1120}-\frac {224\,\sqrt {5}\,x}{2000\,x+1120}\right )-\frac {\sqrt {2}\,\ln \left (x+1-\sqrt {2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{4}+\frac {\sqrt {2}\,\ln \left (x+1+\sqrt {2}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{4} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.21, size = 51, normalized size = 1.11 \[ \frac {\log {\left (x^{2} + 2 x + 3 \right )}}{2} - \sqrt {5} \operatorname {atan}{\left (\frac {\sqrt {5} x}{5} \right )} + \frac {5 \sqrt {2} \operatorname {atan}{\left (\frac {\sqrt {2} x}{2} + \frac {\sqrt {2}}{2} \right )}}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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