Optimal. Leaf size=72 \[ \frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{6} \log \left (2 x^2-x+2\right )+x-\frac {1}{3} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.09, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {2075, 634, 618, 204, 628} \[ \frac {1}{3} \log \left (x^2+x+1\right )+\frac {1}{6} \log \left (2 x^2-x+2\right )+x-\frac {1}{3} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{3 \sqrt {3}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 204
Rule 618
Rule 628
Rule 634
Rule 2075
Rubi steps
\begin {align*} \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+3 x^2+x^3+2 x^4} \, dx &=\int \left (1+\frac {2 (-2+x)}{3 \left (1+x+x^2\right )}+\frac {2 (1+x)}{3 \left (2-x+2 x^2\right )}\right ) \, dx\\ &=x+\frac {2}{3} \int \frac {-2+x}{1+x+x^2} \, dx+\frac {2}{3} \int \frac {1+x}{2-x+2 x^2} \, dx\\ &=x+\frac {1}{6} \int \frac {-1+4 x}{2-x+2 x^2} \, dx+\frac {1}{3} \int \frac {1+2 x}{1+x+x^2} \, dx+\frac {5}{6} \int \frac {1}{2-x+2 x^2} \, dx-\frac {5}{3} \int \frac {1}{1+x+x^2} \, dx\\ &=x+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{6} \log \left (2-x+2 x^2\right )-\frac {5}{3} \operatorname {Subst}\left (\int \frac {1}{-15-x^2} \, dx,x,-1+4 x\right )+\frac {10}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=x-\frac {1}{3} \sqrt {\frac {5}{3}} \tan ^{-1}\left (\frac {1-4 x}{\sqrt {15}}\right )-\frac {10 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\frac {1}{3} \log \left (1+x+x^2\right )+\frac {1}{6} \log \left (2-x+2 x^2\right )\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.02, size = 69, normalized size = 0.96 \[ \frac {1}{18} \left (3 \left (2 \log \left (x^2+x+1\right )+\log \left (2 x^2-x+2\right )+6 x\right )-20 \sqrt {3} \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )+2 \sqrt {15} \tan ^{-1}\left (\frac {4 x-1}{\sqrt {15}}\right )\right ) \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 1.04, size = 62, normalized size = 0.86 \[ \frac {1}{9} \, \sqrt {5} \sqrt {3} \arctan \left (\frac {1}{15} \, \sqrt {5} \sqrt {3} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x + \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
giac [A] time = 0.31, size = 56, normalized size = 0.78 \[ \frac {1}{9} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x + \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maple [A] time = 0.00, size = 57, normalized size = 0.79 \[ x +\frac {\sqrt {15}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {15}}{15}\right )}{9}-\frac {10 \sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{9}+\frac {\ln \left (x^{2}+x +1\right )}{3}+\frac {\ln \left (2 x^{2}-x +2\right )}{6} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 1.59, size = 56, normalized size = 0.78 \[ \frac {1}{9} \, \sqrt {15} \arctan \left (\frac {1}{15} \, \sqrt {15} {\left (4 \, x - 1\right )}\right ) - \frac {10}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + x + \frac {1}{6} \, \log \left (2 \, x^{2} - x + 2\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 2.29, size = 80, normalized size = 1.11 \[ x+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,5{}\mathrm {i}}{9}\right )-\ln \left (x-\frac {1}{4}-\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {15}\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{4}+\frac {\sqrt {15}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{6}+\frac {\sqrt {15}\,1{}\mathrm {i}}{18}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 0.23, size = 75, normalized size = 1.04 \[ x + \frac {\log {\left (x^{2} - \frac {x}{2} + 1 \right )}}{6} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} + \frac {\sqrt {15} \operatorname {atan}{\left (\frac {4 \sqrt {15} x}{15} - \frac {\sqrt {15}}{15} \right )}}{9} - \frac {10 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{9} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________