3.7.67 \(\int \frac {x^2}{(-1+x^4) \sqrt {1+x^4}} \, dx\) [667]

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

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {504, 1225, 226, 1713, 209, 212} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((-1 + x^4)*Sqrt[1 + x^4]),x]

[Out]

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

Rule 209

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

Rule 212

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

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 504

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

Rule 1225

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

Rule 1713

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

Rubi steps

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

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Mathematica [A]
time = 0.19, size = 46, normalized size = 0.87 \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-\tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{4 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.94, size = 102, normalized size = 1.92

method	result	size
elliptic	\(\frac {\left (-\frac {\arctanh \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}-\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{4}\right ) \sqrt {2}}{2}\)	\(43\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) x +\sqrt {x^{4}+1}}{\left (1+x \right ) \left (-1+x \right )}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{8}\)	\(73\)
default	\(-\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , -i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{2 \sqrt {x^{4}+1}}\)	\(102\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-1)^(3/4)*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,I,(-I)^(1/2)/(-1)^(1/4))
+1/2*(-1)^(3/4)*(1-I*x^2)^(1/2)*(1+I*x^2)^(1/2)/(x^4+1)^(1/2)*EllipticPi((-1)^(1/4)*x,-I,(-I)^(1/2)/(-1)^(1/4)
)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 61, normalized size = 1.15 \begin {gather*} \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \frac {1}{16} \, \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^2/(x^4-1)/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

1/8*sqrt(2)*arctan(sqrt(2)*x/sqrt(x^4 + 1)) + 1/16*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \sqrt {x^{4} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/(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 {x^2}{\left (x^4-1\right )\,\sqrt {x^4+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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