Optimal. Leaf size=232 \[ \frac{5 x^3}{3}-\frac{25 \left (x^2+3\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{2} \left (26499 \sqrt{3}-14395\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{2} \left (26499 \sqrt{3}-14395\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )-17 x-\frac{1}{16} \sqrt{\frac{1}{2} \left (14395+26499 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )+\frac{1}{16} \sqrt{\frac{1}{2} \left (14395+26499 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]
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Rubi [A] time = 0.292392, antiderivative size = 232, 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} \[ \frac{5 x^3}{3}-\frac{25 \left (x^2+3\right ) x}{8 \left (x^4+2 x^2+3\right )}-\frac{1}{32} \sqrt{\frac{1}{2} \left (26499 \sqrt{3}-14395\right )} \log \left (x^2-\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )+\frac{1}{32} \sqrt{\frac{1}{2} \left (26499 \sqrt{3}-14395\right )} \log \left (x^2+\sqrt{2 \left (\sqrt{3}-1\right )} x+\sqrt{3}\right )-17 x-\frac{1}{16} \sqrt{\frac{1}{2} \left (14395+26499 \sqrt{3}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (\sqrt{3}-1\right )}-2 x}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right )+\frac{1}{16} \sqrt{\frac{1}{2} \left (14395+26499 \sqrt{3}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (\sqrt{3}-1\right )}}{\sqrt{2 \left (1+\sqrt{3}\right )}}\right ) \]
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^4 \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 (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \frac{450-150 x^2-336 x^4+240 x^6}{3+2 x^2+x^4} \, dx\\ &=-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{48} \int \left (-816+240 x^2+\frac{6 \left (483+127 x^2\right )}{3+2 x^2+x^4}\right ) \, dx\\ &=-17 x+\frac{5 x^3}{3}-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{8} \int \frac{483+127 x^2}{3+2 x^2+x^4} \, dx\\ &=-17 x+\frac{5 x^3}{3}-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{\int \frac{483 \sqrt{2 \left (-1+\sqrt{3}\right )}-\left (483-127 \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{483 \sqrt{2 \left (-1+\sqrt{3}\right )}+\left (483-127 \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 )}}\\ &=-17 x+\frac{5 x^3}{3}-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}+\frac{1}{32} \left (127+161 \sqrt{3}\right ) \int \frac{1}{\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2} \, dx+\frac{1}{32} \left (127+161 \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}{2} \left (-14395+26499 \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}{2} \left (-14395+26499 \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\\ &=-17 x+\frac{5 x^3}{3}-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{32} \sqrt{\frac{1}{2} \left (-14395+26499 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{2} \left (-14395+26499 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{16} \left (-127-161 \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}{16} \left (-127-161 \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 )\\ &=-17 x+\frac{5 x^3}{3}-\frac{25 x \left (3+x^2\right )}{8 \left (3+2 x^2+x^4\right )}-\frac{1}{16} \sqrt{\frac{1}{2} \left (14395+26499 \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}{2} \left (14395+26499 \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}{2} \left (-14395+26499 \sqrt{3}\right )} \log \left (\sqrt{3}-\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )+\frac{1}{32} \sqrt{\frac{1}{2} \left (-14395+26499 \sqrt{3}\right )} \log \left (\sqrt{3}+\sqrt{2 \left (-1+\sqrt{3}\right )} x+x^2\right )\\ \end{align*}
Mathematica [C] time = 0.161821, size = 129, normalized size = 0.56 \[ \frac{5 x^3}{3}-\frac{25 \left (x^2+3\right ) x}{8 \left (x^4+2 x^2+3\right )}-17 x+\frac{\left (127 \sqrt{2}-356 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1-i \sqrt{2}}}\right )}{16 \sqrt{2-2 i \sqrt{2}}}+\frac{\left (127 \sqrt{2}+356 i\right ) \tan ^{-1}\left (\frac{x}{\sqrt{1+i \sqrt{2}}}\right )}{16 \sqrt{2+2 i \sqrt{2}}} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.022, size = 416, normalized size = 1.8 \begin{align*}{\frac{5\,{x}^{3}}{3}}-17\,x+{\frac{1}{{x}^{4}+2\,{x}^{2}+3} \left ( -{\frac{25\,{x}^{3}}{8}}-{\frac{75\,x}{8}} \right ) }-{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{64}}-{\frac{89\,\ln \left ({x}^{2}+\sqrt{3}-x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{32}}-{\frac{ \left ( -34+34\,\sqrt{3} \right ) \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{-178+178\,\sqrt{3}}{16\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{161\,\sqrt{3}}{8\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x-\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{17\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}\sqrt{3}}{64}}+{\frac{89\,\ln \left ({x}^{2}+\sqrt{3}+x\sqrt{-2+2\,\sqrt{3}} \right ) \sqrt{-2+2\,\sqrt{3}}}{32}}-{\frac{ \left ( -34+34\,\sqrt{3} \right ) \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{-178+178\,\sqrt{3}}{16\,\sqrt{2+2\,\sqrt{3}}}\arctan \left ({\frac{2\,x+\sqrt{-2+2\,\sqrt{3}}}{\sqrt{2+2\,\sqrt{3}}}} \right ) }+{\frac{161\,\sqrt{3}}{8\,\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*} \frac{5}{3} \, x^{3} - 17 \, x - \frac{25 \,{\left (x^{3} + 3 \, x\right )}}{8 \,{\left (x^{4} + 2 \, x^{2} + 3\right )}} + \frac{1}{8} \, \int \frac{127 \, x^{2} + 483}{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 = 1.77465, size = 2102, normalized size = 9.06 \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 = 0.532538, size = 58, normalized size = 0.25 \begin{align*} \frac{5 x^{3}}{3} - 17 x - \frac{25 x^{3} + 75 x}{8 x^{4} + 16 x^{2} + 24} + \operatorname{RootSum}{\left (1048576 t^{4} + 29480960 t^{2} + 2106591003, \left ( t \mapsto t \log{\left (\frac{557056 t^{3}}{816619683} + \frac{166600064 t}{816619683} + 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^{4}}{{\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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