3.1018 \(\int \frac{x^2}{\sqrt{-1+x^4} (1+x^4)} \, dx\)

Optimal. Leaf size=49 \[ -\frac{1}{4} \tan ^{-1}\left (\frac{x^2+1}{x \sqrt{x^4-1}}\right )-\frac{1}{4} \tanh ^{-1}\left (\frac{1-x^2}{x \sqrt{x^4-1}}\right ) \]

[Out]

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

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Rubi [C]  time = 0.119633, antiderivative size = 47, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {490, 1211, 222, 1699, 206, 203} \[ \left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) x}{\sqrt{x^4-1}}\right )-\left (\frac{1}{8}+\frac{i}{8}\right ) \tan ^{-1}\left (\frac{(1+i) x}{\sqrt{x^4-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[-1 + x^4]*(1 + x^4)),x]

[Out]

(-1/8 - I/8)*ArcTan[((1 + I)*x)/Sqrt[-1 + x^4]] + (1/8 + I/8)*ArcTanh[((1 + I)*x)/Sqrt[-1 + x^4]]

Rule 490

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), In
t[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 1211

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 222

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-(a*b), 2]}, Simp[(Sqrt[-a + q*x^2]*Sqrt[(a + q*x^2
)/q]*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2])/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]), x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 1699

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]

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 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])

Rubi steps

\begin{align*} \int \frac{x^2}{\sqrt{-1+x^4} \left (1+x^4\right )} \, dx &=-\left (\frac{1}{2} \int \frac{1}{\left (i-x^2\right ) \sqrt{-1+x^4}} \, dx\right )+\frac{1}{2} \int \frac{1}{\left (i+x^2\right ) \sqrt{-1+x^4}} \, dx\\ &=-\left (\frac{1}{4} i \int \frac{i-x^2}{\left (i+x^2\right ) \sqrt{-1+x^4}} \, dx\right )+\frac{1}{4} i \int \frac{i+x^2}{\left (i-x^2\right ) \sqrt{-1+x^4}} \, dx\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{i-2 x^2} \, dx,x,\frac{x}{\sqrt{-1+x^4}}\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{i+2 x^2} \, dx,x,\frac{x}{\sqrt{-1+x^4}}\right )\\ &=\left (-\frac{1}{8}-\frac{i}{8}\right ) \tan ^{-1}\left (\frac{(1+i) x}{\sqrt{-1+x^4}}\right )+\left (\frac{1}{8}+\frac{i}{8}\right ) \tanh ^{-1}\left (\frac{(1+i) x}{\sqrt{-1+x^4}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0195402, size = 46, normalized size = 0.94 \[ \frac{x^3 \sqrt{1-x^4} F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};x^4,-x^4\right )}{3 \sqrt{x^4-1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^3*Sqrt[1 - x^4]*AppellF1[3/4, 1/2, 1, 7/4, x^4, -x^4])/(3*Sqrt[-1 + x^4])

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Maple [B]  time = 0.016, size = 88, normalized size = 1.8 \begin{align*}{\frac{1}{8}\arctan \left ( 1+{\frac{1}{x}\sqrt{{x}^{4}-1}} \right ) }-{\frac{1}{8}\arctan \left ( -{\frac{1}{x}\sqrt{{x}^{4}-1}}+1 \right ) }+{\frac{1}{16}\ln \left ({ \left ({\frac{{x}^{4}-1}{2\,{x}^{2}}}+{\frac{1}{x}\sqrt{{x}^{4}-1}}+1 \right ) \left ({\frac{{x}^{4}-1}{2\,{x}^{2}}}-{\frac{1}{x}\sqrt{{x}^{4}-1}}+1 \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*arctan(1+(x^4-1)^(1/2)/x)-1/8*arctan(-(x^4-1)^(1/2)/x+1)+1/16*ln((1/2*(x^4-1)/x^2+(x^4-1)^(1/2)/x+1)/(1/2*
(x^4-1)/x^2-(x^4-1)^(1/2)/x+1))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (x^{4} + 1\right )} \sqrt{x^{4} - 1}}\,{d x} \end{align*}

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/((x^4 + 1)*sqrt(x^4 - 1)), x)

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Fricas [A]  time = 2.20115, size = 132, normalized size = 2.69 \begin{align*} \frac{1}{4} \, \arctan \left (\frac{\sqrt{x^{4} - 1} x}{x^{2} + 1}\right ) + \frac{1}{8} \, \log \left (\frac{x^{4} + 2 \, x^{2} + 2 \, \sqrt{x^{4} - 1} x - 1}{x^{4} + 1}\right ) \end{align*}

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]

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

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

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\left (x^{4} + 1\right )} \sqrt{x^{4} - 1}}\,{d x} \end{align*}

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/((x^4 + 1)*sqrt(x^4 - 1)), x)