\(\int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx\) [909]

Optimal result

Integrand size = 24, antiderivative size = 69 \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\frac {1}{4} \sqrt {-x^2+x^4}+\text {arctanh}\left (\frac {(-1+x) x}{\sqrt {-x^2+x^4}}\right )-\frac {1}{4} \sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt {-x^2+x^4}}{x^2}\right ) \]

[Out]

1/4*(x^4-x^2)^(1/2)+arctanh((-1+x)*x/(x^4-x^2)^(1/2))-1/4*3^(1/2)*arctanh(3^(1/2)*(x^4-x^2)^(1/2)/x^2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2059, 748, 857, 634, 212, 738} \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\frac {1}{2} \text {arctanh}\left (\frac {x^2}{\sqrt {x^4-x^2}}\right )+\frac {1}{8} \sqrt {3} \text {arctanh}\left (\frac {3-4 x^2}{2 \sqrt {3} \sqrt {x^4-x^2}}\right )+\frac {1}{4} \sqrt {x^4-x^2} \]

[In]

Int[(x*Sqrt[-x^2 + x^4])/(-3 + 2*x^2),x]

[Out]

Sqrt[-x^2 + x^4]/4 + ArcTanh[x^2/Sqrt[-x^2 + x^4]]/2 + (Sqrt[3]*ArcTanh[(3 - 4*x^2)/(2*Sqrt[3]*Sqrt[-x^2 + x^4
])])/8

Rule 212

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

Rule 634

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 738

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

Rule 748

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 2059

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {-x+x^2}}{-3+2 x} \, dx,x,x^2\right ) \\ & = \frac {1}{4} \sqrt {-x^2+x^4}-\frac {1}{8} \text {Subst}\left (\int \frac {3-4 x}{(-3+2 x) \sqrt {-x+x^2}} \, dx,x,x^2\right ) \\ & = \frac {1}{4} \sqrt {-x^2+x^4}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {-x+x^2}} \, dx,x,x^2\right )+\frac {3}{8} \text {Subst}\left (\int \frac {1}{(-3+2 x) \sqrt {-x+x^2}} \, dx,x,x^2\right ) \\ & = \frac {1}{4} \sqrt {-x^2+x^4}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {-x^2+x^4}}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{12-x^2} \, dx,x,\frac {-3+4 x^2}{\sqrt {-x^2+x^4}}\right ) \\ & = \frac {1}{4} \sqrt {-x^2+x^4}+\frac {1}{2} \text {arctanh}\left (\frac {x^2}{\sqrt {-x^2+x^4}}\right )+\frac {1}{8} \sqrt {3} \text {arctanh}\left (\frac {3-4 x^2}{2 \sqrt {3} \sqrt {-x^2+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.17 \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\frac {x \sqrt {-1+x^2} \left (x \sqrt {-1+x^2}-\sqrt {3} \text {arctanh}\left (\frac {x}{\sqrt {3} \sqrt {-1+x^2}}\right )+4 \text {arctanh}\left (\frac {\sqrt {-1+x^2}}{-1+x}\right )\right )}{4 \sqrt {x^2 \left (-1+x^2\right )}} \]

[In]

Integrate[(x*Sqrt[-x^2 + x^4])/(-3 + 2*x^2),x]

[Out]

(x*Sqrt[-1 + x^2]*(x*Sqrt[-1 + x^2] - Sqrt[3]*ArcTanh[x/(Sqrt[3]*Sqrt[-1 + x^2])] + 4*ArcTanh[Sqrt[-1 + x^2]/(
-1 + x)]))/(4*Sqrt[x^2*(-1 + x^2)])

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {\sqrt {x^{4}-x^{2}}}{4}+\frac {\ln \left (2 x^{2}-1+2 \sqrt {x^{4}-x^{2}}\right )}{4}-\frac {\sqrt {3}\, \operatorname {arctanh}\left (\frac {\left (4 x^{2}-3\right ) \sqrt {3}}{6 \sqrt {x^{4}-x^{2}}}\right )}{8}\)	\(67\)
trager	\(\frac {\sqrt {x^{4}-x^{2}}}{4}-\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x^{2}}}{x}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right ) x^{2}+6 \sqrt {x^{4}-x^{2}}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-3\right )}{2 x^{2}-3}\right )}{8}\)	\(91\)
default	\(-\frac {\sqrt {x^{4}-x^{2}}\, \left (\sqrt {6}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\sqrt {6}\, x +2\right ) \sqrt {2}}{2 \sqrt {x^{2}-1}}\right )+\sqrt {6}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (\sqrt {6}\, x -2\right ) \sqrt {2}}{2 \sqrt {x^{2}-1}}\right )-4 x \sqrt {x^{2}-1}-8 \ln \left (x +\sqrt {x^{2}-1}\right )\right )}{16 x \sqrt {x^{2}-1}}\)	\(101\)
risch	\(\frac {\sqrt {x^{2} \left (x^{2}-1\right )}}{4}+\frac {\left (\frac {\ln \left (x +\sqrt {x^{2}-1}\right )}{2}+\frac {\sqrt {6}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (1-\sqrt {6}\, \left (x +\frac {\sqrt {6}}{2}\right )\right ) \sqrt {2}}{\sqrt {4 \left (x +\frac {\sqrt {6}}{2}\right )^{2}-4 \sqrt {6}\, \left (x +\frac {\sqrt {6}}{2}\right )+2}}\right )}{16}-\frac {\sqrt {6}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\left (1+\sqrt {6}\, \left (x -\frac {\sqrt {6}}{2}\right )\right ) \sqrt {2}}{\sqrt {4 \left (x -\frac {\sqrt {6}}{2}\right )^{2}+4 \sqrt {6}\, \left (x -\frac {\sqrt {6}}{2}\right )+2}}\right )}{16}\right ) \sqrt {x^{2} \left (x^{2}-1\right )}}{x \sqrt {x^{2}-1}}\)	\(157\)

[In]

int(x*(x^4-x^2)^(1/2)/(2*x^2-3),x,method=_RETURNVERBOSE)

[Out]

1/4*(x^4-x^2)^(1/2)+1/4*ln(2*x^2-1+2*(x^4-x^2)^(1/2))-1/8*3^(1/2)*arctanh(1/6*(4*x^2-3)*3^(1/2)/(x^4-x^2)^(1/2
))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.36 \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\frac {1}{8} \, \sqrt {3} \log \left (\frac {8 \, x^{2} - \sqrt {3} {\left (4 \, x^{2} - 3\right )} - 2 \, \sqrt {x^{4} - x^{2}} {\left (2 \, \sqrt {3} - 3\right )} - 6}{2 \, x^{2} - 3}\right ) + \frac {1}{4} \, \sqrt {x^{4} - x^{2}} - \frac {1}{2} \, \log \left (-\frac {x^{2} - \sqrt {x^{4} - x^{2}}}{x}\right ) \]

[In]

integrate(x*(x^4-x^2)^(1/2)/(2*x^2-3),x, algorithm="fricas")

[Out]

1/8*sqrt(3)*log((8*x^2 - sqrt(3)*(4*x^2 - 3) - 2*sqrt(x^4 - x^2)*(2*sqrt(3) - 3) - 6)/(2*x^2 - 3)) + 1/4*sqrt(
x^4 - x^2) - 1/2*log(-(x^2 - sqrt(x^4 - x^2))/x)

Sympy [F]

\[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\int \frac {x \sqrt {x^{2} \left (x - 1\right ) \left (x + 1\right )}}{2 x^{2} - 3}\, dx \]

[In]

integrate(x*(x**4-x**2)**(1/2)/(2*x**2-3),x)

[Out]

Integral(x*sqrt(x**2*(x - 1)*(x + 1))/(2*x**2 - 3), x)

Maxima [F]

\[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\int { \frac {\sqrt {x^{4} - x^{2}} x}{2 \, x^{2} - 3} \,d x } \]

[In]

integrate(x*(x^4-x^2)^(1/2)/(2*x^2-3),x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 - x^2)*x/(2*x^2 - 3), x)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.72 \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=-\frac {1}{24} \, \sqrt {3} {\left (-2 i \, \sqrt {3} \pi - 3 \, \log \left (-\frac {\sqrt {3} + 3}{\sqrt {3} - 3}\right )\right )} \mathrm {sgn}\left (x\right ) + \frac {1}{4} \, \sqrt {x^{2} - 1} x \mathrm {sgn}\left (x\right ) - \frac {1}{8} \, \sqrt {3} \log \left (\frac {{\left | 2 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 2 \, \sqrt {3} - 4 \right |}}{{\left | 2 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 2 \, \sqrt {3} - 4 \right |}}\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{4} \, \log \left ({\left (x - \sqrt {x^{2} - 1}\right )}^{2}\right ) \mathrm {sgn}\left (x\right ) \]

[In]

integrate(x*(x^4-x^2)^(1/2)/(2*x^2-3),x, algorithm="giac")

[Out]

-1/24*sqrt(3)*(-2*I*sqrt(3)*pi - 3*log(-(sqrt(3) + 3)/(sqrt(3) - 3)))*sgn(x) + 1/4*sqrt(x^2 - 1)*x*sgn(x) - 1/
8*sqrt(3)*log(abs(2*(x - sqrt(x^2 - 1))^2 - 2*sqrt(3) - 4)/abs(2*(x - sqrt(x^2 - 1))^2 + 2*sqrt(3) - 4))*sgn(x
) - 1/4*log((x - sqrt(x^2 - 1))^2)*sgn(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x \sqrt {-x^2+x^4}}{-3+2 x^2} \, dx=\int \frac {x\,\sqrt {x^4-x^2}}{2\,x^2-3} \,d x \]

[In]

int((x*(x^4 - x^2)^(1/2))/(2*x^2 - 3),x)

[Out]

int((x*(x^4 - x^2)^(1/2))/(2*x^2 - 3), x)