Integrand size = 27, antiderivative size = 14 \[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=2 \arctan \left (x \sqrt {1+x^3}\right ) \]
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\[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{\sqrt {1+x^3} \left (1+x^2+x^5\right )}+\frac {5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx+5 \int \frac {x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx \\ \end{align*}
Time = 1.69 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=2 \arctan \left (x \sqrt {1+x^3}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 2.34 (sec) , antiderivative size = 60, normalized size of antiderivative = 4.29
method | result | size |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{5}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}+2 x \sqrt {x^{3}+1}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{5}+x^{2}+1}\right )\) | \(60\) |
default | \(-\sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{2}+1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}+1\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{4}-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}\right ) \left (3-i \sqrt {3}\right ) \sqrt {\frac {1+x}{3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x -i \sqrt {3}}{-3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x +i \sqrt {3}}{-3+i \sqrt {3}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{4}}{2}+\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{3}}{2}-\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{4} \sqrt {3}}{2}-\frac {i \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {3}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\right )\) | \(197\) |
elliptic | \(-\sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (\textit {\_Z}^{5}+\textit {\_Z}^{2}+1\right )}{\sum }\frac {\underline {\hspace {1.25 ex}}\alpha \left (\underline {\hspace {1.25 ex}}\alpha ^{3}+1\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{4}-\underline {\hspace {1.25 ex}}\alpha ^{3}+\underline {\hspace {1.25 ex}}\alpha ^{2}\right ) \left (3-i \sqrt {3}\right ) \sqrt {\frac {1+x}{3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x -i \sqrt {3}}{-3-i \sqrt {3}}}\, \sqrt {\frac {-1+2 x +i \sqrt {3}}{-3+i \sqrt {3}}}\, \operatorname {EllipticPi}\left (\sqrt {\frac {1+x}{\frac {3}{2}-\frac {i \sqrt {3}}{2}}}, -\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{4}}{2}+\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{3}}{2}-\frac {3 \underline {\hspace {1.25 ex}}\alpha ^{2}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{4} \sqrt {3}}{2}-\frac {i \underline {\hspace {1.25 ex}}\alpha ^{3} \sqrt {3}}{2}+\frac {i \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}}{2}, \sqrt {\frac {-\frac {3}{2}+\frac {i \sqrt {3}}{2}}{-\frac {3}{2}-\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {x^{3}+1}}\right )\) | \(197\) |
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Leaf count of result is larger than twice the leaf count of optimal. 25 vs. \(2 (12) = 24\).
Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\arctan \left (\frac {{\left (x^{5} + x^{2} - 1\right )} \sqrt {x^{3} + 1}}{2 \, {\left (x^{4} + x\right )}}\right ) \]
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\[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\int \frac {5 x^{3} + 2}{\sqrt {\left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (x^{5} + x^{2} + 1\right )}\, dx \]
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\[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\int { \frac {5 \, x^{3} + 2}{{\left (x^{5} + x^{2} + 1\right )} \sqrt {x^{3} + 1}} \,d x } \]
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\[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\int { \frac {5 \, x^{3} + 2}{{\left (x^{5} + x^{2} + 1\right )} \sqrt {x^{3} + 1}} \,d x } \]
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Time = 1.01 (sec) , antiderivative size = 163, normalized size of antiderivative = 11.64 \[ \int \frac {2+5 x^3}{\sqrt {1+x^3} \left (1+x^2+x^5\right )} \, dx=\sum _{k=1}^5\left (-\frac {\sqrt {6}\,\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {-\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x+1\right )}\,\Pi \left (\frac {3+\sqrt {3}\,1{}\mathrm {i}}{2\,\left (\mathrm {root}\left (z^5+z^2+1,z,k\right )+1\right )};\mathrm {asin}\left (\frac {\sqrt {6}\,\sqrt {-\left (-3+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (x+1\right )}}{6}\right )\middle |\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {3-3\,x+\sqrt {3}\,x\,1{}\mathrm {i}+\sqrt {3}\,1{}\mathrm {i}}\,\sqrt {3-3\,x-\sqrt {3}\,x\,1{}\mathrm {i}-\sqrt {3}\,1{}\mathrm {i}}}{18\,\sqrt {x^3+1}\,\left (\mathrm {root}\left (z^5+z^2+1,z,k\right )+1\right )\,\mathrm {root}\left (z^5+z^2+1,z,k\right )}\right ) \]
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