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

Optimal. Leaf size=43 \[ -\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^2+2}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{3} x}{\sqrt{x^2+2}}\right )}{2 \sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0197251, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {1175, 377, 206, 203} \[ -\frac{1}{2} \tan ^{-1}\left (\frac{x}{\sqrt{x^2+2}}\right )-\frac{\tanh ^{-1}\left (\frac{\sqrt{3} x}{\sqrt{x^2+2}}\right )}{2 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 + x^2]*(-1 + x^4)),x]

[Out]

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

Rule 1175

Int[((d_) + (e_.)*(x_)^2)^(q_)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2*r), In
t[(d + e*x^2)^q/(r - c*x^2), x], x] - Dist[c/(2*r), Int[(d + e*x^2)^q/(r + c*x^2), x], x]] /; FreeQ[{a, c, d,
e, q}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

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]

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

Mathematica [C]  time = 0.0454008, size = 96, normalized size = 2.23 \[ \frac{1}{12} \left (-3 \tan ^{-1}\left (\frac{-x+2 i}{\sqrt{x^2+2}}\right )+3 \tan ^{-1}\left (\frac{x+2 i}{\sqrt{x^2+2}}\right )+\sqrt{3} \tanh ^{-1}\left (\frac{2-x}{\sqrt{3} \sqrt{x^2+2}}\right )-\sqrt{3} \tanh ^{-1}\left (\frac{x+2}{\sqrt{3} \sqrt{x^2+2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTan[(2*I - x)/Sqrt[2 + x^2]] + 3*ArcTan[(2*I + x)/Sqrt[2 + x^2]] + Sqrt[3]*ArcTanh[(2 - x)/(Sqrt[3]*Sqr
t[2 + x^2])] - Sqrt[3]*ArcTanh[(2 + x)/(Sqrt[3]*Sqrt[2 + x^2])])/12

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Maple [B]  time = 0.018, size = 70, normalized size = 1.6 \begin{align*} -{\frac{1}{2}\arctan \left ({x{\frac{1}{\sqrt{{x}^{2}+2}}}} \right ) }+{\frac{\sqrt{3}}{12}{\it Artanh} \left ({\frac{ \left ( 4-2\,x \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{ \left ( 1+x \right ) ^{2}+1-2\,x}}}} \right ) }-{\frac{\sqrt{3}}{12}{\it Artanh} \left ({\frac{ \left ( 4+2\,x \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{ \left ( -1+x \right ) ^{2}+1+2\,x}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(x/(x^2+2)^(1/2))+1/12*3^(1/2)*arctanh(1/6*(4-2*x)*3^(1/2)/((1+x)^2+1-2*x)^(1/2))-1/12*3^(1/2)*arct
anh(1/6*(4+2*x)*3^(1/2)/((-1+x)^2+1+2*x)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.02215, size = 189, normalized size = 4.4 \begin{align*} \frac{1}{12} \, \sqrt{3} \log \left (\frac{4 \, x^{2} - \sqrt{3}{\left (2 \, x^{2} + 1\right )} - \sqrt{x^{2} + 2}{\left (2 \, \sqrt{3} x - 3 \, x\right )} + 2}{x^{2} - 1}\right ) - \frac{1}{2} \, \arctan \left (-x^{2} + \sqrt{x^{2} + 2} x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.11023, size = 100, normalized size = 2.33 \begin{align*} -\frac{1}{12} \, \sqrt{3} \log \left (\frac{{\left | 2 \,{\left (x - \sqrt{x^{2} + 2}\right )}^{2} - 4 \, \sqrt{3} - 8 \right |}}{{\left | 2 \,{\left (x - \sqrt{x^{2} + 2}\right )}^{2} + 4 \, \sqrt{3} - 8 \right |}}\right ) + \frac{1}{2} \, \arctan \left (\frac{1}{2} \,{\left (x - \sqrt{x^{2} + 2}\right )}^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*log(abs(2*(x - sqrt(x^2 + 2))^2 - 4*sqrt(3) - 8)/abs(2*(x - sqrt(x^2 + 2))^2 + 4*sqrt(3) - 8)) +
1/2*arctan(1/2*(x - sqrt(x^2 + 2))^2)