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]

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

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Rubi [A]  time = 0.19, antiderivative size = 188, normalized size of antiderivative = 1.00, 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 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 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 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]

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 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 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]

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*}

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Mathematica [C]  time = 0.03, 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)

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fricas [A]  time = 0.52, size = 247, normalized size = 1.31 \[ \frac {1}{16} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + 2 \, x^{2} + 2 \, \sqrt {2}\right ) - \frac {1}{16} \cdot 2^{\frac {1}{4}} \sqrt {2 \, \sqrt {2} + 4} {\left (\sqrt {2} - 2\right )} \log \left (-2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + 2 \, x^{2} + 2 \, \sqrt {2}\right ) - \frac {1}{4} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + 2 \, x^{2} + 2 \, \sqrt {2}} \sqrt {2 \, \sqrt {2} + 4} - \sqrt {2} - 1\right ) - \frac {1}{4} \cdot 2^{\frac {3}{4}} \sqrt {2 \, \sqrt {2} + 4} \arctan \left (-\frac {1}{2} \cdot 2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + \frac {1}{2} \cdot 2^{\frac {1}{4}} \sqrt {-2^{\frac {3}{4}} x \sqrt {2 \, \sqrt {2} + 4} + 2 \, x^{2} + 2 \, \sqrt {2}} \sqrt {2 \, \sqrt {2} + 4} + \sqrt {2} + 1\right ) \]

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)

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giac [A]  time = 1.49, size = 147, normalized size = 0.78 \[ \frac {1}{4} \, \sqrt {2 \, \sqrt {2} + 2} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \, x + 2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {2} + 2} \arctan \left (\frac {2^{\frac {3}{4}} {\left (2 \, x - 2^{\frac {1}{4}} \sqrt {\sqrt {2} + 2}\right )}}{2 \, \sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {2 \, \sqrt {2} - 2} \log \left (x^{2} + 2^{\frac {1}{4}} x \sqrt {\sqrt {2} + 2} + \sqrt {2}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} - 2} \log \left (x^{2} - 2^{\frac {1}{4}} x \sqrt {\sqrt {2} + 2} + \sqrt {2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.10, size = 308, normalized size = 1.64 \[ \frac {\sqrt {2}\, \left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 x -\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 x -\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2}\, \left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 x +\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}-\frac {\left (2+2 \sqrt {2}\right ) \arctan \left (\frac {2 x +\sqrt {2+2 \sqrt {2}}}{\sqrt {-2+2 \sqrt {2}}}\right )}{4 \sqrt {-2+2 \sqrt {2}}}+\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x^{2}-\sqrt {2+2 \sqrt {2}}\, x +\sqrt {2}\right )}{8}-\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x^{2}-\sqrt {2+2 \sqrt {2}}\, x +\sqrt {2}\right )}{8}-\frac {\sqrt {2+2 \sqrt {2}}\, \sqrt {2}\, \ln \left (x^{2}+\sqrt {2+2 \sqrt {2}}\, x +\sqrt {2}\right )}{8}+\frac {\sqrt {2+2 \sqrt {2}}\, \ln \left (x^{2}+\sqrt {2+2 \sqrt {2}}\, x +\sqrt {2}\right )}{8} \]

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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{{\left (x^{2} - 1\right )}^{2} + 1}\,{d x} \]

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)

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mupad [B]  time = 2.30, size = 101, normalized size = 0.54 \[ \mathrm {atanh}\left (32\,x\,{\left (\sqrt {-\frac {\sqrt {2}}{32}-\frac {1}{32}}+\sqrt {\frac {\sqrt {2}}{32}-\frac {1}{32}}\right )}^3\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{32}-\frac {1}{32}}+2\,\sqrt {\frac {\sqrt {2}}{32}-\frac {1}{32}}\right )+\mathrm {atanh}\left (32\,x\,{\left (\sqrt {-\frac {\sqrt {2}}{32}-\frac {1}{32}}-\sqrt {\frac {\sqrt {2}}{32}-\frac {1}{32}}\right )}^3\right )\,\left (2\,\sqrt {-\frac {\sqrt {2}}{32}-\frac {1}{32}}-2\,\sqrt {\frac {\sqrt {2}}{32}-\frac {1}{32}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/((x^2 - 1)^2 + 1),x)

[Out]

atanh(32*x*((- 2^(1/2)/32 - 1/32)^(1/2) + (2^(1/2)/32 - 1/32)^(1/2))^3)*(2*(- 2^(1/2)/32 - 1/32)^(1/2) + 2*(2^
(1/2)/32 - 1/32)^(1/2)) + atanh(32*x*((- 2^(1/2)/32 - 1/32)^(1/2) - (2^(1/2)/32 - 1/32)^(1/2))^3)*(2*(- 2^(1/2
)/32 - 1/32)^(1/2) - 2*(2^(1/2)/32 - 1/32)^(1/2))

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sympy [A]  time = 0.52, size = 24, normalized size = 0.13 \[ \operatorname {RootSum} {\left (128 t^{4} + 16 t^{2} + 1, \left (t \mapsto t \log {\left (64 t^{3} + 4 t + x \right )} \right )\right )} \]

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)))

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