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

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

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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 = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {377, 212, 206, 203} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(Sqrt[2]*x)/(-1 + x^4)^(1/4)]/(2*Sqrt[2]) + ArcTanh[(Sqrt[2]*x)/(-1 + x^4)^(1/4)]/(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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 44, normalized size = 0.83 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4-1}}\right )+\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4-1}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.22, size = 53, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{-1+x^4}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 7.66, size = 144, normalized size = 2.72 \begin {gather*} -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {2 \, {\left (2 \, \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {1}{4}} x^{3} + \sqrt {2} {\left (x^{4} - 1\right )}^{\frac {3}{4}} x\right )}}{3 \, x^{4} + 1}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (-\frac {73 \, x^{8} - 58 \, x^{4} + 4 \, \sqrt {2} {\left (13 \, x^{5} - x\right )} {\left (x^{4} - 1\right )}^{\frac {3}{4}} + 8 \, \sqrt {2} {\left (7 \, x^{7} - 3 \, x^{3}\right )} {\left (x^{4} - 1\right )}^{\frac {1}{4}} + 16 \, {\left (5 \, x^{6} - x^{2}\right )} \sqrt {x^{4} - 1} + 1}{9 \, x^{8} + 6 \, x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*sqrt(2)*arctan(2*(2*sqrt(2)*(x^4 - 1)^(1/4)*x^3 + sqrt(2)*(x^4 - 1)^(3/4)*x)/(3*x^4 + 1)) + 1/16*sqrt(2)*
log(-(73*x^8 - 58*x^4 + 4*sqrt(2)*(13*x^5 - x)*(x^4 - 1)^(3/4) + 8*sqrt(2)*(7*x^7 - 3*x^3)*(x^4 - 1)^(1/4) + 1
6*(5*x^6 - x^2)*sqrt(x^4 - 1) + 1)/(9*x^8 + 6*x^4 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.10, size = 160, normalized size = 3.02

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\frac {4 \RootOf \left (\textit {\_Z}^{2}-2\right ) \sqrt {x^{4}-1}\, x^{2}+5 \RootOf \left (\textit {\_Z}^{2}-2\right ) x^{4}+4 \left (x^{4}-1\right )^{\frac {3}{4}} x +8 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}-2\right )}{3 x^{4}+1}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-4 \RootOf \left (\textit {\_Z}^{2}+2\right ) \sqrt {x^{4}-1}\, x^{2}+5 \RootOf \left (\textit {\_Z}^{2}+2\right ) x^{4}+4 \left (x^{4}-1\right )^{\frac {3}{4}} x -8 x^{3} \left (x^{4}-1\right )^{\frac {1}{4}}-\RootOf \left (\textit {\_Z}^{2}+2\right )}{3 x^{4}+1}\right )}{8}\)	\(160\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*RootOf(_Z^2-2)*ln((4*RootOf(_Z^2-2)*(x^4-1)^(1/2)*x^2+5*RootOf(_Z^2-2)*x^4+4*(x^4-1)^(3/4)*x+8*x^3*(x^4-1)
^(1/4)-RootOf(_Z^2-2))/(3*x^4+1))+1/8*RootOf(_Z^2+2)*ln((-4*RootOf(_Z^2+2)*(x^4-1)^(1/2)*x^2+5*RootOf(_Z^2+2)*
x^4+4*(x^4-1)^(3/4)*x-8*x^3*(x^4-1)^(1/4)-RootOf(_Z^2+2))/(3*x^4+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{{\left (x^4-1\right )}^{1/4}\,\left (3\,x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (3 x^{4} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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