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

Optimal. Leaf size=53 \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {398} \begin {gather*} \frac {\tan ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((2 - x^2)*(-1 + x^2)^(1/4)),x]

[Out]

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

Rule 398

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b^2/a), 4]}, Simp[(b*Ar
cTan[(q*x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(2*Sqrt[2]*a*d*q), x] + Simp[(b*ArcTanh[(q*x)/(Sqrt[2]*(a + b*x^2)^(1
/4))])/(2*Sqrt[2]*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]

Rubi steps

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

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Mathematica [C]  time = 0.14, size = 115, normalized size = 2.17 \begin {gather*} -\frac {6 x F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )}{\left (x^2-2\right ) \sqrt [4]{x^2-1} \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};x^2,\frac {x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};x^2,\frac {x^2}{2}\right )\right )+6 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};x^2,\frac {x^2}{2}\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-6*x*AppellF1[1/2, 1/4, 1, 3/2, x^2, x^2/2])/((-2 + x^2)*(-1 + x^2)^(1/4)*(6*AppellF1[1/2, 1/4, 1, 3/2, x^2,
x^2/2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, x^2, x^2/2] + AppellF1[3/2, 5/4, 1, 5/2, x^2, x^2/2])))

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IntegrateAlgebraic [A]  time = 0.10, size = 55, normalized size = 1.04 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {x}{\sqrt {2} \sqrt [4]{x^2-1}}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{x^2-1}}{x}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 8.37, size = 91, normalized size = 1.72 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (-\frac {x^{4} + 2 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{2} - 1} x^{2} + 4 \, \sqrt {2} {\left (x^{2} - 1\right )}^{\frac {3}{4}} x + 4 \, x^{2} - 4}{x^{4} - 4 \, x^{2} + 4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(sqrt(2)*(x^2 - 1)^(1/4)/x) + 1/8*sqrt(2)*log(-(x^4 + 2*sqrt(2)*(x^2 - 1)^(1/4)*x^3 + 4*sqr
t(x^2 - 1)*x^2 + 4*sqrt(2)*(x^2 - 1)^(3/4)*x + 4*x^2 - 4)/(x^4 - 4*x^2 + 4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.22, size = 121, normalized size = 2.28 \begin {gather*} -\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {-\sqrt {x^{2}-1}\, x -x +\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}-2\right )+\left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}-2\right )}{x^{2}-2}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {-\sqrt {x^{2}-1}\, x +x +\left (x^{2}-1\right )^{\frac {3}{4}} \RootOf \left (\textit {\_Z}^{2}+2\right )-\left (x^{2}-1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{2}+2\right )}{x^{2}-2}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*RootOf(_Z^2-2)*ln(-(RootOf(_Z^2-2)*(x^2-1)^(3/4)-(x^2-1)^(1/2)*x+RootOf(_Z^2-2)*(x^2-1)^(1/4)-x)/(x^2-2))
-1/4*RootOf(_Z^2+2)*ln(-(RootOf(_Z^2+2)*(x^2-1)^(3/4)-(x^2-1)^(1/2)*x-RootOf(_Z^2+2)*(x^2-1)^(1/4)+x)/(x^2-2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {1}{{\left (x^2-1\right )}^{1/4}\,\left (x^2-2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/((x^2 - 1)^(1/4)*(x^2 - 2)),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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