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

Optimal. Leaf size=26 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {x^4+x^2+1}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {1698, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {x^4+x^2+1}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[3]*x)/Sqrt[1 + x^2 + x^4]]/Sqrt[3]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 1698

Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Dist[
A, Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B},
x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]

Rubi steps

\begin {align*} \int \frac {1+x^2}{\left (1-x^2\right ) \sqrt {1+x^2+x^4}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{1-3 x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right )\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {1+x^2+x^4}}\right )}{\sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 1.32, size = 522, normalized size = 20.08 \[ \frac {(-1)^{2/3} \left (-\sqrt [3]{-1} \left (\sqrt [3]{-1}-1\right )^2 \left (1+\sqrt [3]{-1}\right ) \sqrt {\sqrt [3]{-1} x^2+1} \sqrt {1-(-1)^{2/3} x^2} \operatorname {EllipticF}\left (i \sinh ^{-1}\left ((-1)^{5/6} x\right ),(-1)^{2/3}\right )-2 i \sqrt {3} \sqrt {\frac {(-1)^{2/3}-x}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}} \sqrt {\frac {x+(-1)^{2/3}}{-\sqrt [3]{-1} x+x-1}} \sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}} \left (\sqrt [3]{-1}-x\right )^2 \left (\left (1+\sqrt [3]{-1}\right ) \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}}\right ),-3\right )-2 \sqrt [3]{-1} \Pi \left (-1;\left .\sin ^{-1}\left (\sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}}\right )\right |-3\right )\right )+2 i \sqrt {3} \sqrt {\frac {(-1)^{2/3}-x}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}} \sqrt {\frac {x+(-1)^{2/3}}{-\sqrt [3]{-1} x+x-1}} \sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}} \left (\sqrt [3]{-1}-x\right )^2 \left (\left (\sqrt [3]{-1}-1\right ) \operatorname {EllipticF}\left (\sin ^{-1}\left (\sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}}\right ),-3\right )-2 \sqrt [3]{-1} \Pi \left (3;\left .\sin ^{-1}\left (\sqrt {\frac {-\sqrt [3]{-1} x+x+1}{\left (1+\sqrt [3]{-1}\right ) \left (\sqrt [3]{-1}-x\right )}}\right )\right |-3\right )\right )\right )}{\left (1-(-1)^{2/3}\right ) \sqrt {x^4+x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1)^(2/3)*(-((-1)^(1/3)*(-1 + (-1)^(1/3))^2*(1 + (-1)^(1/3))*Sqrt[1 + (-1)^(1/3)*x^2]*Sqrt[1 - (-1)^(2/3)*x^
2]*EllipticF[I*ArcSinh[(-1)^(5/6)*x], (-1)^(2/3)]) - (2*I)*Sqrt[3]*((-1)^(1/3) - x)^2*Sqrt[((-1)^(2/3) - x)/((
1 + (-1)^(1/3))*((-1)^(1/3) - x))]*Sqrt[((-1)^(2/3) + x)/(-1 + x - (-1)^(1/3)*x)]*Sqrt[(1 + x - (-1)^(1/3)*x)/
((1 + (-1)^(1/3))*((-1)^(1/3) - x))]*((1 + (-1)^(1/3))*EllipticF[ArcSin[Sqrt[(1 + x - (-1)^(1/3)*x)/((1 + (-1)
^(1/3))*((-1)^(1/3) - x))]], -3] - 2*(-1)^(1/3)*EllipticPi[-1, ArcSin[Sqrt[(1 + x - (-1)^(1/3)*x)/((1 + (-1)^(
1/3))*((-1)^(1/3) - x))]], -3]) + (2*I)*Sqrt[3]*((-1)^(1/3) - x)^2*Sqrt[((-1)^(2/3) - x)/((1 + (-1)^(1/3))*((-
1)^(1/3) - x))]*Sqrt[((-1)^(2/3) + x)/(-1 + x - (-1)^(1/3)*x)]*Sqrt[(1 + x - (-1)^(1/3)*x)/((1 + (-1)^(1/3))*(
(-1)^(1/3) - x))]*((-1 + (-1)^(1/3))*EllipticF[ArcSin[Sqrt[(1 + x - (-1)^(1/3)*x)/((1 + (-1)^(1/3))*((-1)^(1/3
) - x))]], -3] - 2*(-1)^(1/3)*EllipticPi[3, ArcSin[Sqrt[(1 + x - (-1)^(1/3)*x)/((1 + (-1)^(1/3))*((-1)^(1/3) -
x))]], -3])))/((1 - (-1)^(2/3))*Sqrt[1 + x^2 + x^4])

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IntegrateAlgebraic [A]  time = 0.21, size = 26, normalized size = 1.00 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {3} x}{\sqrt {x^4+x^2+1}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[(Sqrt[3]*x)/Sqrt[1 + x^2 + x^4]]/Sqrt[3]

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fricas [B]  time = 0.82, size = 45, normalized size = 1.73 \[ \frac {1}{6} \, \sqrt {3} \log \left (\frac {x^{4} + 2 \, \sqrt {3} \sqrt {x^{4} + x^{2} + 1} x + 4 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*sqrt(3)*log((x^4 + 2*sqrt(3)*sqrt(x^4 + x^2 + 1)*x + 4*x^2 + 1)/(x^4 - 2*x^2 + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{2} + 1}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.56, size = 31, normalized size = 1.19

method	result	size
elliptic	\(\frac {\arctanh \left (\frac {\sqrt {x^{4}+x^{2}+1}\, \sqrt {2}\, \sqrt {6}}{6 x}\right ) \sqrt {6}\, \sqrt {2}}{6}\)	\(31\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) x +\sqrt {x^{4}+x^{2}+1}}{\left (1+x \right ) \left (-1+x \right )}\right )}{3}\)	\(41\)
default	\(-\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \EllipticF \left (\frac {x \sqrt {-2+2 i \sqrt {3}}}{2}, \frac {\sqrt {-2+2 i \sqrt {3}}}{2}\right )}{\sqrt {-2+2 i \sqrt {3}}\, \sqrt {x^{4}+x^{2}+1}}+\frac {2 \sqrt {1-\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \sqrt {1-\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x^{2}}\, \EllipticPi \left (\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, x , \frac {1}{-\frac {1}{2}+\frac {i \sqrt {3}}{2}}, \frac {\sqrt {-\frac {1}{2}-\frac {i \sqrt {3}}{2}}}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}}\right )}{\sqrt {-\frac {1}{2}+\frac {i \sqrt {3}}{2}}\, \sqrt {x^{4}+x^{2}+1}}\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*arctanh(1/6*(x^4+x^2+1)^(1/2)*2^(1/2)/x*6^(1/2))*6^(1/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x^{2} + 1}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int -\frac {x^2+1}{\left (x^2-1\right )\,\sqrt {x^4+x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {x^{2}}{x^{2} \sqrt {x^{4} + x^{2} + 1} - \sqrt {x^{4} + x^{2} + 1}}\, dx - \int \frac {1}{x^{2} \sqrt {x^{4} + x^{2} + 1} - \sqrt {x^{4} + x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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