3.2.95 \(\int \frac {-1+2 x+2 x^2}{(1+2 x^2) \sqrt {-x+x^4}} \, dx\)

Optimal. Leaf size=21 \[ \tan ^{-1}\left (\frac {2 \sqrt {x^4-x}}{2 x+1}\right ) \]

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Rubi [F]  time = 1.76, 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.

[In]

Int[(-1 + 2*x + 2*x^2)/((1 + 2*x^2)*Sqrt[-x + x^4]),x]

[Out]

((1 - x)*x*Sqrt[(1 + x + x^2)/(1 - (1 + Sqrt[3])*x)^2]*EllipticF[ArcCos[(1 - (1 - Sqrt[3])*x)/(1 - (1 + Sqrt[3
])*x)], (2 + Sqrt[3])/4])/(3^(1/4)*Sqrt[-(((1 - x)*x)/(1 - (1 + Sqrt[3])*x)^2)]*Sqrt[-x + x^4]) + ((1/2 + I/2)
*(I - Sqrt[2])*Sqrt[x]*Sqrt[-1 + x^3]*Defer[Subst][Defer[Int][1/(((-1)^(1/4) - 2^(1/4)*x)*Sqrt[-1 + x^6]), x],
x, Sqrt[x]])/Sqrt[-x + x^4] - ((1/2 - I/2)*(I + Sqrt[2])*Sqrt[x]*Sqrt[-1 + x^3]*Defer[Subst][Defer[Int][1/((-
(-1)^(3/4) - 2^(1/4)*x)*Sqrt[-1 + x^6]), x], x, Sqrt[x]])/Sqrt[-x + x^4] + ((1/2 + I/2)*(I - Sqrt[2])*Sqrt[x]*
Sqrt[-1 + x^3]*Defer[Subst][Defer[Int][1/(((-1)^(1/4) + 2^(1/4)*x)*Sqrt[-1 + x^6]), x], x, Sqrt[x]])/Sqrt[-x +
x^4] - ((1/2 - I/2)*(I + Sqrt[2])*Sqrt[x]*Sqrt[-1 + x^3]*Defer[Subst][Defer[Int][1/((-(-1)^(3/4) + 2^(1/4)*x)
*Sqrt[-1 + x^6]), x], x, Sqrt[x]])/Sqrt[-x + x^4]

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 {(1-x) 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 {(1-x) 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 {(1-x) 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 {(1-x) 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 {(1-x) 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 {(1-x) 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 {\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 {\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 {(1-x) 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 {(1-x) x}{\left (1-\left (1+\sqrt {3}\right ) 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}}+\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}}\\ \end {align*}

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Mathematica [C]  time = 1.28, size = 320, normalized size = 15.24 \begin {gather*} -\frac {2 \sqrt {\frac {1}{x^2}+\frac {1}{x}+1} \sqrt {\frac {x-1}{\left (1+\sqrt [3]{-1}\right ) x}} x^2 \left (\frac {\sqrt {3} \left (i \sqrt {3} x+\sqrt [3]{-1}+1\right ) F\left (\sin ^{-1}\left (\sqrt {\frac {x-(-1)^{2/3}}{\left (1+\sqrt [3]{-1}\right ) x}}\right )|\sqrt [3]{-1}\right )}{(-1)^{2/3}-x}+\frac {6 \left (\left (-2 i+\sqrt [6]{-1}+2 \sqrt {2}+(-1)^{2/3} \sqrt {2}\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 )+\frac {3}{2} \left (3 i+\sqrt {2}-\sqrt {3}-i \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 )\right )}{-3 i \sqrt {2}+2 i \sqrt {3}+\sqrt {6}}\right )}{3 \sqrt {x \left (x^3-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 + 2*x + 2*x^2)/((1 + 2*x^2)*Sqrt[-x + x^4]),x]

[Out]

(-2*Sqrt[1 + x^(-2) + x^(-1)]*Sqrt[(-1 + x)/((1 + (-1)^(1/3))*x)]*x^2*((Sqrt[3]*(1 + (-1)^(1/3) + I*Sqrt[3]*x)
*EllipticF[ArcSin[Sqrt[(-(-1)^(2/3) + x)/((1 + (-1)^(1/3))*x)]], (-1)^(1/3)])/((-1)^(2/3) - x) + (6*((-2*I + (
-1)^(1/6) + 2*Sqrt[2] + (-1)^(2/3)*Sqrt[2])*EllipticPi[(2*Sqrt[3])/(-I - 2*Sqrt[2] + Sqrt[3]), ArcSin[Sqrt[(-(
-1)^(2/3) + x)/((1 + (-1)^(1/3))*x)]], (-1)^(1/3)] + (3*(3*I + Sqrt[2] - Sqrt[3] - I*Sqrt[6])*EllipticPi[(2*Sq
rt[3])/(-I + 2*Sqrt[2] + Sqrt[3]), ArcSin[Sqrt[(-(-1)^(2/3) + x)/((1 + (-1)^(1/3))*x)]], (-1)^(1/3)])/2))/((-3
*I)*Sqrt[2] + (2*I)*Sqrt[3] + Sqrt[6])))/(3*Sqrt[x*(-1 + x^3)])

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IntegrateAlgebraic [A]  time = 1.26, size = 21, 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.

[In]

IntegrateAlgebraic[(-1 + 2*x + 2*x^2)/((1 + 2*x^2)*Sqrt[-x + x^4]),x]

[Out]

ArcTan[(2*Sqrt[-x + x^4])/(1 + 2*x)]

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fricas [A]  time = 0.54, size = 19, normalized size = 0.90 \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.

[In]

integrate((2*x^2+2*x-1)/(2*x^2+1)/(x^4-x)^(1/2),x, algorithm="fricas")

[Out]

-arctan(1/2*(2*x + 1)/sqrt(x^4 - x))

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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.

[In]

integrate((2*x^2+2*x-1)/(2*x^2+1)/(x^4-x)^(1/2),x, algorithm="giac")

[Out]

integrate((2*x^2 + 2*x - 1)/(sqrt(x^4 - x)*(2*x^2 + 1)), x)

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maple [C]  time = 0.39, size = 46, normalized size = 2.19

method	result	size
trager	\(\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x +\RootOf \left (\textit {\_Z}^{2}+1\right )+2 \sqrt {x^{4}-x}}{2 x^{2}+1}\right )\)	\(46\)
default	\(\text {Expression too large to display}\)	\(13534\)
elliptic	\(\text {Expression too large to display}\)	\(14836\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+2*x-1)/(2*x^2+1)/(x^4-x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

RootOf(_Z^2+1)*ln((2*RootOf(_Z^2+1)*x+RootOf(_Z^2+1)+2*(x^4-x)^(1/2))/(2*x^2+1))

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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.

[In]

integrate((2*x^2+2*x-1)/(2*x^2+1)/(x^4-x)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*x^2 + 2*x - 1)/(sqrt(x^4 - x)*(2*x^2 + 1)), x)

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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}{\sqrt {x^4-x}\,\left (2\,x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 2*x^2 - 1)/((x^4 - x)^(1/2)*(2*x^2 + 1)),x)

[Out]

int((2*x + 2*x^2 - 1)/((x^4 - x)^(1/2)*(2*x^2 + 1)), x)

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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.

[In]

integrate((2*x**2+2*x-1)/(2*x**2+1)/(x**4-x)**(1/2),x)

[Out]

Integral((2*x**2 + 2*x - 1)/(sqrt(x*(x - 1)*(x**2 + x + 1))*(2*x**2 + 1)), x)

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