3.13.39 \(\int \frac {\sqrt {1+3 x^4}}{-1+3 x^4} \, dx\)

Optimal. Leaf size=90 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {3 x^4+1}}\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3 x^4+1}}{\sqrt {3} x^2-\sqrt {2} \sqrt [4]{3} x+1}\right )}{\sqrt {2} \sqrt [4]{3}} \]

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Rubi [A]  time = 0.03, antiderivative size = 77, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {404, 212, 206, 203} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {3 x^4+1}}\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {3 x^4+1}}\right )}{2 \sqrt {2} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + 3*x^4]/(-1 + 3*x^4),x]

[Out]

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

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+3 x^4}}{-1+3 x^4} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{1-12 x^4} \, dx,x,\frac {x}{\sqrt {1+3 x^4}}\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-2 \sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt {1+3 x^4}}\right )\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+2 \sqrt {3} x^2} \, dx,x,\frac {x}{\sqrt {1+3 x^4}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {1+3 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {1+3 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{3}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 120, normalized size = 1.33 \begin {gather*} \frac {5 x \sqrt {3 x^4+1} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-3 x^4,3 x^4\right )}{\left (3 x^4-1\right ) \left (6 x^4 \left (2 F_1\left (\frac {5}{4};-\frac {1}{2},2;\frac {9}{4};-3 x^4,3 x^4\right )+F_1\left (\frac {5}{4};\frac {1}{2},1;\frac {9}{4};-3 x^4,3 x^4\right )\right )+5 F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};-3 x^4,3 x^4\right )\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[1 + 3*x^4]/(-1 + 3*x^4),x]

[Out]

(5*x*Sqrt[1 + 3*x^4]*AppellF1[1/4, -1/2, 1, 5/4, -3*x^4, 3*x^4])/((-1 + 3*x^4)*(5*AppellF1[1/4, -1/2, 1, 5/4,
-3*x^4, 3*x^4] + 6*x^4*(2*AppellF1[5/4, -1/2, 2, 9/4, -3*x^4, 3*x^4] + AppellF1[5/4, 1/2, 1, 9/4, -3*x^4, 3*x^
4])))

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IntegrateAlgebraic [A]  time = 0.43, size = 77, normalized size = 0.86 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {1+3 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{3} x}{\sqrt {1+3 x^4}}\right )}{2 \sqrt {2} \sqrt [4]{3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[1 + 3*x^4]/(-1 + 3*x^4),x]

[Out]

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

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fricas [B]  time = 0.52, size = 146, normalized size = 1.62 \begin {gather*} \frac {1}{12} \cdot 12^{\frac {3}{4}} \arctan \left (-\frac {12^{\frac {3}{4}} \sqrt {3} x^{2} - 12^{\frac {3}{4}} \sqrt {3 \, x^{4} + 1} + 2 \cdot 12^{\frac {1}{4}} \sqrt {3}}{12 \, x}\right ) - \frac {1}{48} \cdot 12^{\frac {3}{4}} \log \left (\frac {6 \cdot 12^{\frac {1}{4}} x^{3} + 12^{\frac {3}{4}} x + 2 \, \sqrt {3 \, x^{4} + 1} {\left (3 \, x^{2} + \sqrt {3}\right )}}{3 \, x^{4} - 1}\right ) + \frac {1}{48} \cdot 12^{\frac {3}{4}} \log \left (-\frac {6 \cdot 12^{\frac {1}{4}} x^{3} + 12^{\frac {3}{4}} x - 2 \, \sqrt {3 \, x^{4} + 1} {\left (3 \, x^{2} + \sqrt {3}\right )}}{3 \, x^{4} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+1)^(1/2)/(3*x^4-1),x, algorithm="fricas")

[Out]

1/12*12^(3/4)*arctan(-1/12*(12^(3/4)*sqrt(3)*x^2 - 12^(3/4)*sqrt(3*x^4 + 1) + 2*12^(1/4)*sqrt(3))/x) - 1/48*12
^(3/4)*log((6*12^(1/4)*x^3 + 12^(3/4)*x + 2*sqrt(3*x^4 + 1)*(3*x^2 + sqrt(3)))/(3*x^4 - 1)) + 1/48*12^(3/4)*lo
g(-(6*12^(1/4)*x^3 + 12^(3/4)*x - 2*sqrt(3*x^4 + 1)*(3*x^2 + sqrt(3)))/(3*x^4 - 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {3 \, x^{4} + 1}}{3 \, x^{4} - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+1)^(1/2)/(3*x^4-1),x, algorithm="giac")

[Out]

integrate(sqrt(3*x^4 + 1)/(3*x^4 - 1), x)

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maple [A]  time = 1.30, size = 86, normalized size = 0.96

method	result	size
default	\(\frac {\left (-\frac {3^{\frac {3}{4}} \ln \left (\frac {\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}}{2 x}+3^{\frac {1}{4}}}{\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}}{2 x}-3^{\frac {1}{4}}}\right )}{12}+\frac {3^{\frac {3}{4}} \arctan \left (\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}\, 3^{\frac {3}{4}}}{6 x}\right )}{6}\right ) \sqrt {2}}{2}\)	\(86\)
elliptic	\(\frac {\left (-\frac {3^{\frac {3}{4}} \ln \left (\frac {\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}}{2 x}+3^{\frac {1}{4}}}{\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}}{2 x}-3^{\frac {1}{4}}}\right )}{12}+\frac {3^{\frac {3}{4}} \arctan \left (\frac {\sqrt {3 x^{4}+1}\, \sqrt {2}\, 3^{\frac {3}{4}}}{6 x}\right )}{6}\right ) \sqrt {2}}{2}\)	\(86\)
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right ) \ln \left (\frac {x \RootOf \left (\textit {\_Z}^{4}-108\right )^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-108\right )^{2}\right )+18 \sqrt {3 x^{4}+1}}{\RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{2}+6}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{4}-108\right ) \ln \left (\frac {x \RootOf \left (\textit {\_Z}^{4}-108\right )^{3}+18 \sqrt {3 x^{4}+1}}{\RootOf \left (\textit {\_Z}^{4}-108\right )^{2} x^{2}-6}\right )}{12}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^4+1)^(1/2)/(3*x^4-1),x,method=_RETURNVERBOSE)

[Out]

1/2*(-1/12*3^(3/4)*ln((1/2*(3*x^4+1)^(1/2)*2^(1/2)/x+3^(1/4))/(1/2*(3*x^4+1)^(1/2)*2^(1/2)/x-3^(1/4)))+1/6*3^(
3/4)*arctan(1/6*(3*x^4+1)^(1/2)*2^(1/2)/x*3^(3/4)))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {3 \, x^{4} + 1}}{3 \, x^{4} - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^4+1)^(1/2)/(3*x^4-1),x, algorithm="maxima")

[Out]

integrate(sqrt(3*x^4 + 1)/(3*x^4 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {3\,x^4+1}}{3\,x^4-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^4 + 1)^(1/2)/(3*x^4 - 1),x)

[Out]

int((3*x^4 + 1)^(1/2)/(3*x^4 - 1), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {3 x^{4} + 1}}{3 x^{4} - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**4+1)**(1/2)/(3*x**4-1),x)

[Out]

Integral(sqrt(3*x**4 + 1)/(3*x**4 - 1), x)

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