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

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

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Rubi [A]  time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {404, 212, 206, 203} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^4]/(-1 + x^4),x]

[Out]

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

Rule 203

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

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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 404

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

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 108, normalized size = 2.04 \begin {gather*} \frac {5 x \sqrt {x^4+1} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-x^4,x^4\right )}{\left (x^4-1\right ) \left (2 x^4 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};-x^4,x^4\right )+F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-x^4,x^4\right )\right )+5 F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-x^4,x^4\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[1 + x^4]/(-1 + x^4),x]

[Out]

(5*x*Sqrt[1 + x^4]*AppellF1[1/4, -1/2, 1, 5/4, -x^4, x^4])/((-1 + x^4)*(5*AppellF1[1/4, -1/2, 1, 5/4, -x^4, x^
4] + 2*x^4*(2*AppellF1[5/4, -1/2, 2, 9/4, -x^4, x^4] + AppellF1[5/4, 1/2, 1, 9/4, -x^4, x^4])))

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IntegrateAlgebraic [A]  time = 0.23, size = 53, normalized size = 1.00 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 61, normalized size = 1.15 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (\frac {x^{4} - 2 \, \sqrt {2} \sqrt {x^{4} + 1} x + 2 \, x^{2} + 1}{x^{4} - 2 \, x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 1)/(x^4 - 1), x)

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maple [A]  time = 0.39, size = 65, normalized size = 1.23

method	result	size
elliptic	\(\frac {\left (\frac {\ln \left (-1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\ln \left (1+\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(65\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{4}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (-1+x \right ) \left (1+x \right )}\right )}{4}\)	\(72\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}-\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(362\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(x^4 + 1)/(x^4 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\sqrt {x^4+1}}{x^4-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)^(1/2)/(x^4 - 1),x)

[Out]

int((x^4 + 1)^(1/2)/(x^4 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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