Optimal. Leaf size=19 \[ \tan ^{-1}\left (\frac {2 \sqrt {x^4+x}}{2 x-1}\right ) \]
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Rubi [F] time = 1.90, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {-1-2 x+2 x^2}{\left (1+2 x^2\right ) \sqrt {x+x^4}} \, dx &=\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 ) \operatorname {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}} F\left (\cos ^{-1}\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}} F\left (\cos ^{-1}\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 ) \operatorname {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 ) \operatorname {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}} F\left (\cos ^{-1}\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 ) \operatorname {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 ) \operatorname {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}} F\left (\cos ^{-1}\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}
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Mathematica [C] time = 1.49, size = 307, normalized size = 16.16 \begin {gather*} -\frac {2 \sqrt {\frac {1}{x^2}-\frac {1}{x}+1} \sqrt {\frac {\frac {1}{x}+1}{1+\sqrt [3]{-1}}} x^2 \left (\frac {i \sqrt {3} \left (\sqrt {3} x+(-1)^{5/6}+i\right ) F\left (\sin ^{-1}\left (\sqrt {\frac {x+(-1)^{2/3}}{\left (1+\sqrt [3]{-1}\right ) x}}\right )|\sqrt [3]{-1}\right )}{x+(-1)^{2/3}}-\frac {3 i \left (\sqrt {2}-i\right ) \Pi \left (\frac {2 \sqrt {3}}{-i-2 \sqrt {2}+\sqrt {3}};\sin ^{-1}\left (\sqrt {\frac {x+(-1)^{2/3}}{\left (1+\sqrt [3]{-1}\right ) x}}\right )|\sqrt [3]{-1}\right )}{(-1)^{5/6}+\sqrt {2}}+\frac {3 \left (5+i \sqrt {2}+i \sqrt {3}+\sqrt {6}\right ) \Pi \left (\frac {2 \sqrt {3}}{-i+2 \sqrt {2}+\sqrt {3}};\sin ^{-1}\left (\sqrt {\frac {x+(-1)^{2/3}}{\left (1+\sqrt [3]{-1}\right ) x}}\right )|\sqrt [3]{-1}\right )}{5 i+2 \sqrt {2}+\sqrt {3}+2 i \sqrt {6}}\right )}{3 \sqrt {x^4+x}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 1.23, size = 19, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {2 \sqrt {x+x^4}}{-1+2 x}\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.50, size = 17, normalized size = 0.89 \begin {gather*} -\arctan \left (\frac {2 \, x - 1}{2 \, \sqrt {x^{4} + x}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (2 \, x^{2} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.38, size = 46, normalized size = 2.42
method | result | size |
trager | \(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x +2 \sqrt {x^{4}+x}-\RootOf \left (\textit {\_Z}^{2}+1\right )}{2 x^{2}+1}\right )\) | \(46\) |
default | \(\text {Expression too large to display}\) | \(13532\) |
elliptic | \(\text {Expression too large to display}\) | \(14832\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{2} - 2 \, x - 1}{\sqrt {x^{4} + x} {\left (2 \, x^{2} + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {-2\,x^2+2\,x+1}{\left (2\,x^2+1\right )\,\sqrt {x^4+x}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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