Optimal. Leaf size=225 \[ \frac{25 \left (x^2+1\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+5 x+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]
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Rubi [A] time = 0.29714, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {1668, 1676, 1169, 634, 618, 204, 628} \[ \frac{25 \left (x^2+1\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (12899 \sqrt{3}-19291\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+5 x+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]
Antiderivative was successfully verified.
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Rule 1668
Rule 1676
Rule 1169
Rule 634
Rule 618
Rule 204
Rule 628
Rubi steps
\begin{align*} \int \frac{x^2 \left (4+x^2+3 x^4+5 x^6\right )}{\left (3+2 x^2+x^4\right )^2} \, dx &=\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \frac{-150-186 x^2+240 x^4}{3+2 x^2+x^4} \, dx\\ &=\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \left (240-\frac{6 \left (145+111 x^2\right )}{3+2 x^2+x^4}\right ) \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{8} \int \frac{145+111 x^2}{3+2 x^2+x^4} \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{\int \frac{145 \sqrt{2 \left (-1+\sqrt{3}\right )}-\left (145-111 \sqrt{3}\right ) x}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx}{16 \sqrt{6 \left (-1+\sqrt{3}\right )}}-\frac{\int \frac{145 \sqrt{2 \left (-1+\sqrt{3}\right )}+\left (145-111 \sqrt{3}\right ) x}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx}{16 \sqrt{6 \left (-1+\sqrt{3}\right )}}\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{96} \left (333+145 \sqrt{3}\right ) \int \frac{1}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx-\frac{1}{96} \left (333+145 \sqrt{3}\right ) \int \frac{1}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \int \frac{-\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \int \frac{\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{48} \left (333+145 \sqrt{3}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{3}\right )-x^2} \, dx,x,-\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x\right )+\frac{1}{48} \left (333+145 \sqrt{3}\right ) \operatorname{Subst}\left (\int \frac{1}{-2 \left (1+\sqrt{3}\right )-x^2} \, dx,x,\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x\right )\\ &=5 x+\frac{25 x \left (1+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{3}\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{16} \sqrt{\frac{1}{6} \left (19291+12899 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (-1+\sqrt{3}\right )}+2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )-\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{6} \left (-19291+12899 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )\\ \end{align*}
Mathematica [C] time = 0.170382, size = 121, normalized size = 0.54 \[ \frac{25 \left (x^3+x\right )}{8 \left (x^4+2 x^2+3\right )}+5 x-\frac{\left (111 \sqrt{2}-34 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1-i \sqrt{2}}}\right )}{16 \sqrt{2-2 i \sqrt{2}}}-\frac{\left (111 \sqrt{2}+34 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1+i \sqrt{2}}}\right )}{16 \sqrt{2+2 i \sqrt{2}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.018, size = 412, normalized size = 1.8 \begin{align*} 5\,x-{\frac{1}{{x}^{4}+2\,{x}^{2}+3} \left ( -{\frac{25\,{x}^{3}}{8}}-{\frac{25\,x}{8}} \right ) }-{\frac{47\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{96}}+{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{64}}-{\frac{ \left ( -94+94\,\sqrt{3} \right ) \sqrt{3}}{48\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{-34+34\,\sqrt{3}}{32\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }-{\frac{145\,\sqrt{3}}{24\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{47\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{96}}-{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{64}}-{\frac{ \left ( -94+94\,\sqrt{3} \right ) \sqrt{3}}{48\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{-34+34\,\sqrt{3}}{32\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }-{\frac{145\,\sqrt{3}}{24\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) } \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*} 5 \, x + \frac{25 \,{\left (x^{3} + x\right )}}{8 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} - \frac{1}{8} \, \int \frac{111 \, x^{2} + 145}{x^{4} + 2 \, x^{2} + 3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.75631, size = 2072, normalized size = 9.21 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0.535457, size = 51, normalized size = 0.23 \begin{align*} 5 x + \frac{25 x^{3} + 25 x}{8 x^{4} + 16 x^{2} + 24} + \operatorname{RootSum}{\left (3145728 t^{4} + 39507968 t^{2} + 166384201, \left ( t \mapsto t \log{\left (- \frac{9240576 t^{3}}{102792131} - \frac{95003488 t}{102792131} + x \right )} \right )\right )} \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 \frac{{\left (5 \, x^{6} + 3 \, x^{4} + x^{2} + 4\right )} x^{2}}{{\left (x^{4} + 2 \, x^{2} + 3\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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