3.491 \(\int \frac{x^2}{1+(-1+x^2)^2} \, dx\)

Optimal. Leaf size=188 \[ \frac{\log \left (x^2-\sqrt{2 \left (1+\sqrt{2}\right )} x+\sqrt{2}\right )}{4 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (x^2+\sqrt{2 \left (1+\sqrt{2}\right )} x+\sqrt{2}\right )}{4 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 x}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.188943, antiderivative size = 188, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {1989, 1127, 1161, 618, 204, 1164, 628} \[ \frac{\log \left (x^2-\sqrt{2 \left (1+\sqrt{2}\right )} x+\sqrt{2}\right )}{4 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{\log \left (x^2+\sqrt{2 \left (1+\sqrt{2}\right )} x+\sqrt{2}\right )}{4 \sqrt{2 \left (1+\sqrt{2}\right )}}-\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{\sqrt{2 \left (1+\sqrt{2}\right )}-2 x}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right )+\frac{1}{2} \sqrt{\frac{1}{2} \left (1+\sqrt{2}\right )} \tan ^{-1}\left (\frac{2 x+\sqrt{2 \left (1+\sqrt{2}\right )}}{\sqrt{2 \left (\sqrt{2}-1\right )}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + (-1 + x^2)^2),x]

[Out]

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

Rule 1989

Int[(u_)^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[(d*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{d, m, p}, x] &&
TrinomialQ[u, x] &&  !TrinomialMatchQ[u, x]

Rule 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1164

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

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

Mathematica [C]  time = 0.0307766, size = 39, normalized size = 0.21 \[ -\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-1-i}}\right )}{(-1-i)^{3/2}}-\frac{\tan ^{-1}\left (\frac{x}{\sqrt{-1+i}}\right )}{(-1+i)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + (-1 + x^2)^2),x]

[Out]

-(ArcTan[x/Sqrt[-1 - I]]/(-1 - I)^(3/2)) - ArcTan[x/Sqrt[-1 + I]]/(-1 + I)^(3/2)

________________________________________________________________________________________

Maple [B]  time = 0.077, size = 308, normalized size = 1.6 \begin{align*} -{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{2+2\,\sqrt{2}} \right ) }{8}}+{\frac{\sqrt{2} \left ( 2+2\,\sqrt{2} \right ) }{4\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}\ln \left ({x}^{2}+\sqrt{2}+x\sqrt{2+2\,\sqrt{2}} \right ) }{8}}-{\frac{2+2\,\sqrt{2}}{4\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x+\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }+{\frac{\sqrt{2+2\,\sqrt{2}}\sqrt{2}\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{2+2\,\sqrt{2}} \right ) }{8}}+{\frac{\sqrt{2} \left ( 2+2\,\sqrt{2} \right ) }{4\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) }-{\frac{\sqrt{2+2\,\sqrt{2}}\ln \left ({x}^{2}+\sqrt{2}-x\sqrt{2+2\,\sqrt{2}} \right ) }{8}}-{\frac{2+2\,\sqrt{2}}{4\,\sqrt{-2+2\,\sqrt{2}}}\arctan \left ({\frac{2\,x-\sqrt{2+2\,\sqrt{2}}}{\sqrt{-2+2\,\sqrt{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/((x^2 - 1)^2 + 1), x)

________________________________________________________________________________________

Fricas [A]  time = 1.08493, size = 771, normalized size = 4.1 \begin{align*} \frac{1}{16} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 2\right )} \log \left (2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}\right ) - \frac{1}{16} \cdot 2^{\frac{1}{4}} \sqrt{2 \, \sqrt{2} + 4}{\left (\sqrt{2} - 2\right )} \log \left (-2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}\right ) - \frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}} \sqrt{2 \, \sqrt{2} + 4} - \sqrt{2} - 1\right ) - \frac{1}{4} \cdot 2^{\frac{3}{4}} \sqrt{2 \, \sqrt{2} + 4} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + \frac{1}{2} \cdot 2^{\frac{1}{4}} \sqrt{-2^{\frac{3}{4}} x \sqrt{2 \, \sqrt{2} + 4} + 2 \, x^{2} + 2 \, \sqrt{2}} \sqrt{2 \, \sqrt{2} + 4} + \sqrt{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*2^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(2^(3/4)*x*sqrt(2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2)) - 1/16*2
^(1/4)*sqrt(2*sqrt(2) + 4)*(sqrt(2) - 2)*log(-2^(3/4)*x*sqrt(2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2)) - 1/4*2^(3/4)
*sqrt(2*sqrt(2) + 4)*arctan(-1/2*2^(3/4)*x*sqrt(2*sqrt(2) + 4) + 1/2*2^(1/4)*sqrt(2^(3/4)*x*sqrt(2*sqrt(2) + 4
) + 2*x^2 + 2*sqrt(2))*sqrt(2*sqrt(2) + 4) - sqrt(2) - 1) - 1/4*2^(3/4)*sqrt(2*sqrt(2) + 4)*arctan(-1/2*2^(3/4
)*x*sqrt(2*sqrt(2) + 4) + 1/2*2^(1/4)*sqrt(-2^(3/4)*x*sqrt(2*sqrt(2) + 4) + 2*x^2 + 2*sqrt(2))*sqrt(2*sqrt(2)
+ 4) + sqrt(2) + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.457658, size = 24, normalized size = 0.13 \begin{align*} \operatorname{RootSum}{\left (128 t^{4} + 16 t^{2} + 1, \left ( t \mapsto t \log{\left (64 t^{3} + 4 t + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(128*_t**4 + 16*_t**2 + 1, Lambda(_t, _t*log(64*_t**3 + 4*_t + x)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2/((x^2 - 1)^2 + 1), x)