Integrand size = 29, antiderivative size = 19 \[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\arctan \left (\frac {2 \sqrt {x+x^4}}{-1+2 x}\right ) \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {-1-2 x+2 x^2}{\sqrt {x} \left (1+2 x^2\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \left (\frac {1}{\sqrt {x} \sqrt {1+x^3}}-\frac {2 (1+x)}{\sqrt {x} \left (1+2 x^2\right ) \sqrt {1+x^3}}\right ) \, dx}{\sqrt {x+x^4}} \\ & = \frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}-\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1+x}{\sqrt {x} \left (1+2 x^2\right ) \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}} \\ & = -\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \int \left (\frac {i-\frac {1}{\sqrt {2}}}{2 \sqrt {x} \left (i-\sqrt {2} x\right ) \sqrt {1+x^3}}+\frac {i+\frac {1}{\sqrt {2}}}{2 \sqrt {x} \left (i+\sqrt {2} x\right ) \sqrt {1+x^3}}\right ) \, dx}{\sqrt {x+x^4}}+\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (\left (2 i-\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \left (i-\sqrt {2} x\right ) \sqrt {1+x^3}} \, dx}{2 \sqrt {x+x^4}}-\frac {\left (\left (2 i+\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \left (i+\sqrt {2} x\right ) \sqrt {1+x^3}} \, dx}{2 \sqrt {x+x^4}} \\ & = \frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (\left (2 i-\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (i-\sqrt {2} x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}-\frac {\left (\left (2 i+\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (i+\sqrt {2} x^2\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}-\frac {\left (\left (2 i-\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \left (-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}}-\frac {\left (\left (2 i+\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \left (-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = \frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}+\frac {\left ((-1)^{3/4} \left (2 i-\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}}+\frac {\left ((-1)^{3/4} \left (2 i-\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}}+\frac {\left (\sqrt [4]{-1} \left (2 i+\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}}+\frac {\left (\sqrt [4]{-1} \left (2 i+\sqrt {2}\right ) \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\left (-(-1)^{3/4}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{2 \sqrt {x+x^4}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(41\) vs. \(2(19)=38\).
Time = 21.95 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\frac {2 \sqrt {1+\frac {1}{x^3}} x^2 \arctan \left (\frac {\sqrt {1+\frac {1}{x^3}}}{1+\frac {1}{x}}\right )}{\sqrt {x+x^4}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 3.66 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.42
method | result | size |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {x^{4}+x}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{2 x^{2}+1}\right )\) | \(46\) |
default | \(\text {Expression too large to display}\) | \(13532\) |
elliptic | \(\text {Expression too large to display}\) | \(14832\) |
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Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=-\arctan \left (\frac {2 \, x - 1}{2 \, \sqrt {x^{4} + x}}\right ) \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\int \frac {2 x^{2} - 2 x - 1}{\sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (2 x^{2} + 1\right )}\, dx \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\int { \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (2 \, x^{2} + 1\right )}} \,d x } \]
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\[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\int { \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (2 \, x^{2} + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx=\int -\frac {-2\,x^2+2\,x+1}{\left (2\,x^2+1\right )\,\sqrt {x^4+x}} \,d x \]
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