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

Optimal. Leaf size=96 \[ \sqrt {\frac {1}{5} \left (2+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt {1-x^2}}\right )-\sqrt {\frac {1}{5} \left (\sqrt {5}-2\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

[Out]

-arcsin(x)-1/5*arctanh(1/2*x*(-2+2*5^(1/2))^(1/2)/(-x^2+1)^(1/2))*(-10+5*5^(1/2))^(1/2)+1/5*arctan(1/2*x*(2+2*
5^(1/2))^(1/2)/(-x^2+1)^(1/2))*(10+5*5^(1/2))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1293, 216, 1692, 377, 207, 203} \[ \sqrt {\frac {1}{5} \left (2+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt {1-x^2}}\right )-\sqrt {\frac {1}{5} \left (\sqrt {5}-2\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} x}{\sqrt {1-x^2}}\right )-\sin ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcSin[x] + Sqrt[(2 + Sqrt[5])/5]*ArcTan[(Sqrt[(1 + Sqrt[5])/2]*x)/Sqrt[1 - x^2]] - Sqrt[(-2 + Sqrt[5])/5]*Ar
cTanh[(Sqrt[(-1 + Sqrt[5])/2]*x)/Sqrt[1 - x^2]]

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])

Rule 207

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

Rule 216

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

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 1293

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Dist[(
e*f^2)/c, Int[(f*x)^(m - 2)*(d + e*x^2)^(q - 1), x], x] - Dist[f^2/c, Int[((f*x)^(m - 2)*(d + e*x^2)^(q - 1)*S
imp[a*e - (c*d - b*e)*x^2, x])/(a + b*x^2 + c*x^4), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c,
0] &&  !IntegerQ[q] && GtQ[q, 0] && GtQ[m, 1] && LeQ[m, 3]

Rule 1692

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandInteg
rand[Px*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, q}, x] && PolyQ[Px, x^2] && NeQ[b
^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {x^2 \sqrt {1-x^2}}{-1+x^2+x^4} \, dx &=-\int \frac {1}{\sqrt {1-x^2}} \, dx-\int \frac {1-2 x^2}{\sqrt {1-x^2} \left (-1+x^2+x^4\right )} \, dx\\ &=-\sin ^{-1}(x)-\int \left (\frac {-2+\frac {4}{\sqrt {5}}}{\sqrt {1-x^2} \left (1-\sqrt {5}+2 x^2\right )}+\frac {-2-\frac {4}{\sqrt {5}}}{\sqrt {1-x^2} \left (1+\sqrt {5}+2 x^2\right )}\right ) \, dx\\ &=-\sin ^{-1}(x)+\frac {1}{5} \left (2 \left (5-2 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {1-x^2} \left (1-\sqrt {5}+2 x^2\right )} \, dx+\frac {1}{5} \left (2 \left (5+2 \sqrt {5}\right )\right ) \int \frac {1}{\sqrt {1-x^2} \left (1+\sqrt {5}+2 x^2\right )} \, dx\\ &=-\sin ^{-1}(x)+\frac {1}{5} \left (2 \left (5-2 \sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {5}-\left (-3+\sqrt {5}\right ) x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right )+\frac {1}{5} \left (2 \left (5+2 \sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {5}-\left (-3-\sqrt {5}\right ) x^2} \, dx,x,\frac {x}{\sqrt {1-x^2}}\right )\\ &=-\sin ^{-1}(x)+\sqrt {\frac {1}{5} \left (2+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} x}{\sqrt {1-x^2}}\right )-\sqrt {\frac {1}{5} \left (-2+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} x}{\sqrt {1-x^2}}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.50, size = 743, normalized size = 7.74 \[ \frac {i \sqrt {5 \left (\sqrt {5}-2\right )} \log \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sqrt {1-x^2}-i \sqrt {2 \left (1+\sqrt {5}\right )} x+2\right )+2 i \sqrt {\sqrt {5}-2} \log \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sqrt {1-x^2}-i \sqrt {2 \left (1+\sqrt {5}\right )} x+2\right )-i \sqrt {5 \left (\sqrt {5}-2\right )} \log \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sqrt {1-x^2}+i \sqrt {2 \left (1+\sqrt {5}\right )} x+2\right )-2 i \sqrt {\sqrt {5}-2} \log \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sqrt {1-x^2}+i \sqrt {2 \left (1+\sqrt {5}\right )} x+2\right )-\sqrt {5 \left (2+\sqrt {5}\right )} \log \left (\sqrt {2} \sqrt {\left (\sqrt {5}-3\right ) \left (x^2-1\right )}-\sqrt {2 \left (\sqrt {5}-1\right )} x+2\right )+2 \sqrt {2+\sqrt {5}} \log \left (\sqrt {2} \sqrt {\left (\sqrt {5}-3\right ) \left (x^2-1\right )}-\sqrt {2 \left (\sqrt {5}-1\right )} x+2\right )+\sqrt {5 \left (2+\sqrt {5}\right )} \log \left (\sqrt {2} \sqrt {\left (\sqrt {5}-3\right ) \left (x^2-1\right )}+\sqrt {2 \left (\sqrt {5}-1\right )} x+2\right )-2 \sqrt {2+\sqrt {5}} \log \left (\sqrt {2} \sqrt {\left (\sqrt {5}-3\right ) \left (x^2-1\right )}+\sqrt {2 \left (\sqrt {5}-1\right )} x+2\right )+\left (\sqrt {5}-2\right ) \sqrt {2+\sqrt {5}} \log \left (x-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )}\right )-\sqrt {5 \left (2+\sqrt {5}\right )} \log \left (x+\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )}\right )+2 \sqrt {2+\sqrt {5}} \log \left (x+\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )}\right )-i \sqrt {5 \left (\sqrt {5}-2\right )} \log \left (x-i \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}\right )-2 i \sqrt {\sqrt {5}-2} \log \left (x-i \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}\right )+i \sqrt {5 \left (\sqrt {5}-2\right )} \log \left (x+i \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}\right )+2 i \sqrt {\sqrt {5}-2} \log \left (x+i \sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )}\right )-2 \sqrt {5} \sin ^{-1}(x)}{2 \sqrt {5}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[5]*ArcSin[x] + (-2 + Sqrt[5])*Sqrt[2 + Sqrt[5]]*Log[-Sqrt[(-1 + Sqrt[5])/2] + x] + 2*Sqrt[2 + Sqrt[5]
]*Log[Sqrt[(-1 + Sqrt[5])/2] + x] - Sqrt[5*(2 + Sqrt[5])]*Log[Sqrt[(-1 + Sqrt[5])/2] + x] - (2*I)*Sqrt[-2 + Sq
rt[5]]*Log[(-I)*Sqrt[(1 + Sqrt[5])/2] + x] - I*Sqrt[5*(-2 + Sqrt[5])]*Log[(-I)*Sqrt[(1 + Sqrt[5])/2] + x] + (2
*I)*Sqrt[-2 + Sqrt[5]]*Log[I*Sqrt[(1 + Sqrt[5])/2] + x] + I*Sqrt[5*(-2 + Sqrt[5])]*Log[I*Sqrt[(1 + Sqrt[5])/2]
+ x] + (2*I)*Sqrt[-2 + Sqrt[5]]*Log[2 - I*Sqrt[2*(1 + Sqrt[5])]*x + Sqrt[2*(3 + Sqrt[5])]*Sqrt[1 - x^2]] + I*
Sqrt[5*(-2 + Sqrt[5])]*Log[2 - I*Sqrt[2*(1 + Sqrt[5])]*x + Sqrt[2*(3 + Sqrt[5])]*Sqrt[1 - x^2]] - (2*I)*Sqrt[-
2 + Sqrt[5]]*Log[2 + I*Sqrt[2*(1 + Sqrt[5])]*x + Sqrt[2*(3 + Sqrt[5])]*Sqrt[1 - x^2]] - I*Sqrt[5*(-2 + Sqrt[5]
)]*Log[2 + I*Sqrt[2*(1 + Sqrt[5])]*x + Sqrt[2*(3 + Sqrt[5])]*Sqrt[1 - x^2]] + 2*Sqrt[2 + Sqrt[5]]*Log[2 - Sqrt
[2*(-1 + Sqrt[5])]*x + Sqrt[2]*Sqrt[(-3 + Sqrt[5])*(-1 + x^2)]] - Sqrt[5*(2 + Sqrt[5])]*Log[2 - Sqrt[2*(-1 + S
qrt[5])]*x + Sqrt[2]*Sqrt[(-3 + Sqrt[5])*(-1 + x^2)]] - 2*Sqrt[2 + Sqrt[5]]*Log[2 + Sqrt[2*(-1 + Sqrt[5])]*x +
Sqrt[2]*Sqrt[(-3 + Sqrt[5])*(-1 + x^2)]] + Sqrt[5*(2 + Sqrt[5])]*Log[2 + Sqrt[2*(-1 + Sqrt[5])]*x + Sqrt[2]*S
qrt[(-3 + Sqrt[5])*(-1 + x^2)]])/(2*Sqrt[5])

________________________________________________________________________________________

fricas [B]  time = 1.09, size = 290, normalized size = 3.02 \[ \frac {2}{5} \, \sqrt {5} \sqrt {\sqrt {5} + 2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {-x^{2} + 1} {\left (\sqrt {5} - 3\right )} + \sqrt {5} - 3\right )} \sqrt {\sqrt {5} + 2} \sqrt {\frac {x^{4} - 4 \, x^{2} - \sqrt {5} {\left (x^{4} - 2 \, x^{2}\right )} - 2 \, {\left (\sqrt {5} x^{2} - x^{2} + 2\right )} \sqrt {-x^{2} + 1} + 4}{x^{4}}} + 2 \, \sqrt {-x^{2} + 1} \sqrt {\sqrt {5} + 2} {\left (\sqrt {5} - 3\right )}}{4 \, x}\right ) + \frac {1}{10} \, \sqrt {5} \sqrt {\sqrt {5} - 2} \log \left (-\frac {2 \, x^{2} + {\left (\sqrt {-x^{2} + 1} {\left (\sqrt {5} x + x\right )} - \sqrt {5} x - x\right )} \sqrt {\sqrt {5} - 2} + 2 \, \sqrt {-x^{2} + 1} - 2}{x^{2}}\right ) - \frac {1}{10} \, \sqrt {5} \sqrt {\sqrt {5} - 2} \log \left (-\frac {2 \, x^{2} - {\left (\sqrt {-x^{2} + 1} {\left (\sqrt {5} x + x\right )} - \sqrt {5} x - x\right )} \sqrt {\sqrt {5} - 2} + 2 \, \sqrt {-x^{2} + 1} - 2}{x^{2}}\right ) + 2 \, \arctan \left (\frac {\sqrt {-x^{2} + 1} - 1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*sqrt(5)*sqrt(sqrt(5) + 2)*arctan(1/4*(sqrt(2)*(sqrt(-x^2 + 1)*(sqrt(5) - 3) + sqrt(5) - 3)*sqrt(sqrt(5) +
2)*sqrt((x^4 - 4*x^2 - sqrt(5)*(x^4 - 2*x^2) - 2*(sqrt(5)*x^2 - x^2 + 2)*sqrt(-x^2 + 1) + 4)/x^4) + 2*sqrt(-x^
2 + 1)*sqrt(sqrt(5) + 2)*(sqrt(5) - 3))/x) + 1/10*sqrt(5)*sqrt(sqrt(5) - 2)*log(-(2*x^2 + (sqrt(-x^2 + 1)*(sqr
t(5)*x + x) - sqrt(5)*x - x)*sqrt(sqrt(5) - 2) + 2*sqrt(-x^2 + 1) - 2)/x^2) - 1/10*sqrt(5)*sqrt(sqrt(5) - 2)*l
og(-(2*x^2 - (sqrt(-x^2 + 1)*(sqrt(5)*x + x) - sqrt(5)*x - x)*sqrt(sqrt(5) - 2) + 2*sqrt(-x^2 + 1) - 2)/x^2) +
2*arctan((sqrt(-x^2 + 1) - 1)/x)

________________________________________________________________________________________

giac [B]  time = 0.72, size = 209, normalized size = 2.18 \[ -\frac {1}{2} \, \pi \mathrm {sgn}\relax (x) - \frac {1}{5} \, \sqrt {5 \, \sqrt {5} + 10} \arctan \left (-\frac {\frac {x}{\sqrt {-x^{2} + 1} - 1} - \frac {\sqrt {-x^{2} + 1} - 1}{x}}{\sqrt {2 \, \sqrt {5} + 2}}\right ) - \frac {1}{10} \, \sqrt {5 \, \sqrt {5} - 10} \log \left ({\left | \sqrt {2 \, \sqrt {5} - 2} - \frac {x}{\sqrt {-x^{2} + 1} - 1} + \frac {\sqrt {-x^{2} + 1} - 1}{x} \right |}\right ) + \frac {1}{10} \, \sqrt {5 \, \sqrt {5} - 10} \log \left ({\left | -\sqrt {2 \, \sqrt {5} - 2} - \frac {x}{\sqrt {-x^{2} + 1} - 1} + \frac {\sqrt {-x^{2} + 1} - 1}{x} \right |}\right ) - \arctan \left (-\frac {x {\left (\frac {{\left (\sqrt {-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 1\right )}}{2 \, {\left (\sqrt {-x^{2} + 1} - 1\right )}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*pi*sgn(x) - 1/5*sqrt(5*sqrt(5) + 10)*arctan(-(x/(sqrt(-x^2 + 1) - 1) - (sqrt(-x^2 + 1) - 1)/x)/sqrt(2*sqr
t(5) + 2)) - 1/10*sqrt(5*sqrt(5) - 10)*log(abs(sqrt(2*sqrt(5) - 2) - x/(sqrt(-x^2 + 1) - 1) + (sqrt(-x^2 + 1)
- 1)/x)) + 1/10*sqrt(5*sqrt(5) - 10)*log(abs(-sqrt(2*sqrt(5) - 2) - x/(sqrt(-x^2 + 1) - 1) + (sqrt(-x^2 + 1) -
1)/x)) - arctan(-1/2*x*((sqrt(-x^2 + 1) - 1)^2/x^2 - 1)/(sqrt(-x^2 + 1) - 1))

________________________________________________________________________________________

maple [B]  time = 0.09, size = 160, normalized size = 1.67 \[ -\frac {\sqrt {5}\, \arctanh \left (\frac {\sqrt {-x^{2}+1}-1}{\sqrt {2+\sqrt {5}}\, x}\right )}{5 \sqrt {2+\sqrt {5}}}+\frac {\sqrt {\sqrt {5}-2}\, \sqrt {5}\, \arctanh \left (\frac {\sqrt {-x^{2}+1}-1}{\sqrt {\sqrt {5}-2}\, x}\right )}{5}+2 \arctan \left (\frac {\sqrt {-x^{2}+1}-1}{x}\right )-\frac {\sqrt {2+\sqrt {5}}\, \sqrt {5}\, \arctan \left (\frac {\sqrt {-x^{2}+1}-1}{\sqrt {2+\sqrt {5}}\, x}\right )}{5}-\frac {\sqrt {5}\, \arctan \left (\frac {\sqrt {-x^{2}+1}-1}{\sqrt {\sqrt {5}-2}\, x}\right )}{5 \sqrt {\sqrt {5}-2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(2+5^(1/2))^(1/2)*5^(1/2)*arctan(((-x^2+1)^(1/2)-1)/x/(2+5^(1/2))^(1/2))+1/5*(5^(1/2)-2)^(1/2)*5^(1/2)*ar
ctanh(((-x^2+1)^(1/2)-1)/x/(5^(1/2)-2)^(1/2))-1/5*5^(1/2)/(2+5^(1/2))^(1/2)*arctanh(((-x^2+1)^(1/2)-1)/x/(2+5^
(1/2))^(1/2))-1/5*5^(1/2)/(5^(1/2)-2)^(1/2)*arctan(((-x^2+1)^(1/2)-1)/x/(5^(1/2)-2)^(1/2))+2*arctan(((-x^2+1)^
(1/2)-1)/x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{2} + 1} x^{2}}{x^{4} + x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.50, size = 383, normalized size = 3.99 \[ -\mathrm {asin}\relax (x)-\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}-1\right )\,1{}\mathrm {i}}{\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}\right )\,\left (\sqrt {5}-2\right )}{\left (2\,\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}+4\,{\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}}+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}-1\right )\,1{}\mathrm {i}}{\sqrt {\frac {\sqrt {5}}{2}+\frac {3}{2}}}-\sqrt {1-x^2}\,1{}\mathrm {i}}{x-\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}}\right )\,\left (\sqrt {5}+2\right )}{\left (2\,\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}+4\,{\left (-\frac {\sqrt {5}}{2}-\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {\sqrt {5}}{2}+\frac {3}{2}}}+\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}+1\right )\,1{}\mathrm {i}}{\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}}\right )\,\left (\sqrt {5}-2\right )}{\left (2\,\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}}+4\,{\left (\frac {\sqrt {5}}{2}-\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {3}{2}-\frac {\sqrt {5}}{2}}}-\frac {\ln \left (\frac {\frac {\left (x\,\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}+1\right )\,1{}\mathrm {i}}{\sqrt {\frac {\sqrt {5}}{2}+\frac {3}{2}}}+\sqrt {1-x^2}\,1{}\mathrm {i}}{x+\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}}\right )\,\left (\sqrt {5}+2\right )}{\left (2\,\sqrt {-\frac {\sqrt {5}}{2}-\frac {1}{2}}+4\,{\left (-\frac {\sqrt {5}}{2}-\frac {1}{2}\right )}^{3/2}\right )\,\sqrt {\frac {\sqrt {5}}{2}+\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log((((x*(- 5^(1/2)/2 - 1/2)^(1/2) - 1)*1i)/(5^(1/2)/2 + 3/2)^(1/2) - (1 - x^2)^(1/2)*1i)/(x - (- 5^(1/2)/2 -
1/2)^(1/2)))*(5^(1/2) + 2))/((2*(- 5^(1/2)/2 - 1/2)^(1/2) + 4*(- 5^(1/2)/2 - 1/2)^(3/2))*(5^(1/2)/2 + 3/2)^(1
/2)) - (log((((x*(5^(1/2)/2 - 1/2)^(1/2) - 1)*1i)/(3/2 - 5^(1/2)/2)^(1/2) - (1 - x^2)^(1/2)*1i)/(x - (5^(1/2)/
2 - 1/2)^(1/2)))*(5^(1/2) - 2))/((2*(5^(1/2)/2 - 1/2)^(1/2) + 4*(5^(1/2)/2 - 1/2)^(3/2))*(3/2 - 5^(1/2)/2)^(1/
2)) - asin(x) + (log((((x*(5^(1/2)/2 - 1/2)^(1/2) + 1)*1i)/(3/2 - 5^(1/2)/2)^(1/2) + (1 - x^2)^(1/2)*1i)/(x +
(5^(1/2)/2 - 1/2)^(1/2)))*(5^(1/2) - 2))/((2*(5^(1/2)/2 - 1/2)^(1/2) + 4*(5^(1/2)/2 - 1/2)^(3/2))*(3/2 - 5^(1/
2)/2)^(1/2)) - (log((((x*(- 5^(1/2)/2 - 1/2)^(1/2) + 1)*1i)/(5^(1/2)/2 + 3/2)^(1/2) + (1 - x^2)^(1/2)*1i)/(x +
(- 5^(1/2)/2 - 1/2)^(1/2)))*(5^(1/2) + 2))/((2*(- 5^(1/2)/2 - 1/2)^(1/2) + 4*(- 5^(1/2)/2 - 1/2)^(3/2))*(5^(1
/2)/2 + 3/2)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \sqrt {- \left (x - 1\right ) \left (x + 1\right )}}{x^{4} + x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________