Integrand size = 33, antiderivative size = 230 \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=\frac {1}{14} \left (7-5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x-\frac {\left (19 i+7 \sqrt {7}\right ) \arctan \left (\frac {1-i \sqrt {7}+8 x}{\sqrt {2 \left (35+i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35+i \sqrt {7}\right )}}+\frac {\left (19 i-7 \sqrt {7}\right ) \arctan \left (\frac {1+i \sqrt {7}+8 x}{\sqrt {2 \left (35-i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35-i \sqrt {7}\right )}}+\frac {1}{28} \left (7+5 i \sqrt {7}\right ) \log \left (4+\left (1-i \sqrt {7}\right ) x+4 x^2\right )+\frac {1}{28} \left (7-5 i \sqrt {7}\right ) \log \left (4+\left (1+i \sqrt {7}\right ) x+4 x^2\right ) \]
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Time = 0.26 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2112, 787, 648, 632, 210, 642} \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=-\frac {\left (7 \sqrt {7}+19 i\right ) \arctan \left (\frac {8 x-i \sqrt {7}+1}{\sqrt {2 \left (35+i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35+i \sqrt {7}\right )}}+\frac {\left (-7 \sqrt {7}+19 i\right ) \arctan \left (\frac {8 x+i \sqrt {7}+1}{\sqrt {2 \left (35-i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35-i \sqrt {7}\right )}}+\frac {1}{28} \left (7+5 i \sqrt {7}\right ) \log \left (4 x^2+\left (1-i \sqrt {7}\right ) x+4\right )+\frac {1}{28} \left (7-5 i \sqrt {7}\right ) \log \left (4 x^2+\left (1+i \sqrt {7}\right ) x+4\right )+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7-5 i \sqrt {7}\right ) x \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 787
Rule 2112
Rubi steps \begin{align*} \text {integral}& = \frac {i \int \frac {x \left (9-5 i \sqrt {7}+\left (10-2 i \sqrt {7}\right ) x\right )}{4+\left (1-i \sqrt {7}\right ) x+4 x^2} \, dx}{\sqrt {7}}-\frac {i \int \frac {x \left (9+5 i \sqrt {7}+\left (10+2 i \sqrt {7}\right ) x\right )}{4+\left (1+i \sqrt {7}\right ) x+4 x^2} \, dx}{\sqrt {7}} \\ & = \frac {1}{14} \left (7-5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x+\frac {i \int \frac {-4 \left (10-2 i \sqrt {7}\right )+\left (-\left (\left (1-i \sqrt {7}\right ) \left (10-2 i \sqrt {7}\right )\right )+4 \left (9-5 i \sqrt {7}\right )\right ) x}{4+\left (1-i \sqrt {7}\right ) x+4 x^2} \, dx}{4 \sqrt {7}}-\frac {i \int \frac {-4 \left (10+2 i \sqrt {7}\right )+\left (-\left (\left (1+i \sqrt {7}\right ) \left (10+2 i \sqrt {7}\right )\right )+4 \left (9+5 i \sqrt {7}\right )\right ) x}{4+\left (1+i \sqrt {7}\right ) x+4 x^2} \, dx}{4 \sqrt {7}} \\ & = \frac {1}{14} \left (7-5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x-\frac {1}{28} \left (-7+5 i \sqrt {7}\right ) \int \frac {1+i \sqrt {7}+8 x}{4+\left (1+i \sqrt {7}\right ) x+4 x^2} \, dx+\frac {1}{28} \left (7+5 i \sqrt {7}\right ) \int \frac {1-i \sqrt {7}+8 x}{4+\left (1-i \sqrt {7}\right ) x+4 x^2} \, dx+\frac {1}{14} \left (-49+19 i \sqrt {7}\right ) \int \frac {1}{4+\left (1+i \sqrt {7}\right ) x+4 x^2} \, dx-\frac {1}{14} \left (49+19 i \sqrt {7}\right ) \int \frac {1}{4+\left (1-i \sqrt {7}\right ) x+4 x^2} \, dx \\ & = \frac {1}{14} \left (7-5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x+\frac {1}{28} \left (7+5 i \sqrt {7}\right ) \log \left (4+\left (1-i \sqrt {7}\right ) x+4 x^2\right )+\frac {1}{28} \left (7-5 i \sqrt {7}\right ) \log \left (4+\left (1+i \sqrt {7}\right ) x+4 x^2\right )+\frac {1}{7} \left (49-19 i \sqrt {7}\right ) \text {Subst}\left (\int \frac {1}{-2 \left (35-i \sqrt {7}\right )-x^2} \, dx,x,1+i \sqrt {7}+8 x\right )+\frac {1}{7} \left (49+19 i \sqrt {7}\right ) \text {Subst}\left (\int \frac {1}{-2 \left (35+i \sqrt {7}\right )-x^2} \, dx,x,1-i \sqrt {7}+8 x\right ) \\ & = \frac {1}{14} \left (7-5 i \sqrt {7}\right ) x+\frac {1}{14} \left (7+5 i \sqrt {7}\right ) x-\frac {\left (19 i+7 \sqrt {7}\right ) \tan ^{-1}\left (\frac {1-i \sqrt {7}+8 x}{\sqrt {2 \left (35+i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35+i \sqrt {7}\right )}}+\frac {\left (19 i-7 \sqrt {7}\right ) \tan ^{-1}\left (\frac {1+i \sqrt {7}+8 x}{\sqrt {2 \left (35-i \sqrt {7}\right )}}\right )}{\sqrt {14 \left (35-i \sqrt {7}\right )}}+\frac {1}{28} \left (7+5 i \sqrt {7}\right ) \log \left (4+\left (1-i \sqrt {7}\right ) x+4 x^2\right )+\frac {1}{28} \left (7-5 i \sqrt {7}\right ) \log \left (4+\left (1+i \sqrt {7}\right ) x+4 x^2\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.41 \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=x+2 \text {RootSum}\left [2+\text {$\#$1}+5 \text {$\#$1}^2+\text {$\#$1}^3+2 \text {$\#$1}^4\&,\frac {-\log (x-\text {$\#$1})+2 \log (x-\text {$\#$1}) \text {$\#$1}-2 \log (x-\text {$\#$1}) \text {$\#$1}^2+\log (x-\text {$\#$1}) \text {$\#$1}^3}{1+10 \text {$\#$1}+3 \text {$\#$1}^2+8 \text {$\#$1}^3}\&\right ] \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.27
method | result | size |
default | \(x +2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{4}+\textit {\_Z}^{3}+5 \textit {\_Z}^{2}+\textit {\_Z} +2\right )}{\sum }\frac {\left (\textit {\_R}^{3}-2 \textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R}^{3}+3 \textit {\_R}^{2}+10 \textit {\_R} +1}\right )\) | \(62\) |
risch | \(x +2 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{4}+\textit {\_Z}^{3}+5 \textit {\_Z}^{2}+\textit {\_Z} +2\right )}{\sum }\frac {\left (\textit {\_R}^{3}-2 \textit {\_R}^{2}+2 \textit {\_R} -1\right ) \ln \left (x -\textit {\_R} \right )}{8 \textit {\_R}^{3}+3 \textit {\_R}^{2}+10 \textit {\_R} +1}\right )\) | \(62\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1190 vs. \(2 (151) = 302\).
Time = 1.00 (sec) , antiderivative size = 1190, normalized size of antiderivative = 5.17 \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=\text {Too large to display} \]
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Time = 0.52 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.21 \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=x + \operatorname {RootSum} {\left (343 t^{4} - 343 t^{3} + 294 t^{2} - 336 t + 128, \left ( t \mapsto t \log {\left (\frac {3773 t^{3}}{304} - \frac {1029 t^{2}}{304} + \frac {1001 t}{152} + x - \frac {121}{19} \right )} \right )\right )} \]
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\[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=\int { \frac {{\left (2 \, x^{3} + 3 \, x^{2} + x + 5\right )} x}{2 \, x^{4} + x^{3} + 5 \, x^{2} + x + 2} \,d x } \]
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\[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=\int { \frac {{\left (2 \, x^{3} + 3 \, x^{2} + x + 5\right )} x}{2 \, x^{4} + x^{3} + 5 \, x^{2} + x + 2} \,d x } \]
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Time = 9.48 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.80 \[ \int \frac {x \left (5+x+3 x^2+2 x^3\right )}{2+x+5 x^2+x^3+2 x^4} \, dx=x+\left (\sum _{k=1}^4\ln \left (\frac {115\,\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )}{8}+15\,x-\frac {\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )\,x\,137}{8}+\frac {{\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )}^2\,x\,133}{8}-\frac {{\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )}^3\,x\,147}{4}-\frac {189\,{\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )}^2}{16}+\frac {49\,{\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )}^3}{16}-4\right )\,\mathrm {root}\left (z^4-z^3+\frac {6\,z^2}{7}-\frac {48\,z}{49}+\frac {128}{343},z,k\right )\right ) \]
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