\(\int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx\) [2437]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 18, antiderivative size = 36 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=-\frac {\arctan \left (\frac {4-5 x}{2 \sqrt {2} \sqrt {-2+5 x-3 x^2}}\right )}{\sqrt {2}} \]

[Out]

-1/2*arctan(1/4*(4-5*x)*2^(1/2)/(-3*x^2+5*x-2)^(1/2))*2^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {738, 210} \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=-\frac {\arctan \left (\frac {4-5 x}{2 \sqrt {2} \sqrt {-3 x^2+5 x-2}}\right )}{\sqrt {2}} \]

[In]

Int[1/(x*Sqrt[-2 + 5*x - 3*x^2]),x]

[Out]

-(ArcTan[(4 - 5*x)/(2*Sqrt[2]*Sqrt[-2 + 5*x - 3*x^2])]/Sqrt[2])

Rule 210

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

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]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{-8-x^2} \, dx,x,\frac {-4+5 x}{\sqrt {-2+5 x-3 x^2}}\right )\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {4-5 x}{2 \sqrt {2} \sqrt {-2+5 x-3 x^2}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {-4+10 x-6 x^2}}{-2+3 x}\right ) \]

[In]

Integrate[1/(x*Sqrt[-2 + 5*x - 3*x^2]),x]

[Out]

-(Sqrt[2]*ArcTan[Sqrt[-4 + 10*x - 6*x^2]/(-2 + 3*x)])

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.81

method result size
default \(\frac {\sqrt {2}\, \arctan \left (\frac {\left (-4+5 x \right ) \sqrt {2}}{4 \sqrt {-3 x^{2}+5 x -2}}\right )}{2}\) \(29\)
trager \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right )+4 \sqrt {-3 x^{2}+5 x -2}}{x}\right )}{2}\) \(46\)

[In]

int(1/x/(-3*x^2+5*x-2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*2^(1/2)*arctan(1/4*(-4+5*x)*2^(1/2)/(-3*x^2+5*x-2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-3 \, x^{2} + 5 \, x - 2} {\left (5 \, x - 4\right )}}{4 \, {\left (3 \, x^{2} - 5 \, x + 2\right )}}\right ) \]

[In]

integrate(1/x/(-3*x^2+5*x-2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*arctan(1/4*sqrt(2)*sqrt(-3*x^2 + 5*x - 2)*(5*x - 4)/(3*x^2 - 5*x + 2))

Sympy [F]

\[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=\int \frac {1}{x \sqrt {- \left (x - 1\right ) \left (3 x - 2\right )}}\, dx \]

[In]

integrate(1/x/(-3*x**2+5*x-2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=\frac {1}{2} \, \sqrt {2} \arcsin \left (\frac {5 \, x}{{\left | x \right |}} - \frac {4}{{\left | x \right |}}\right ) \]

[In]

integrate(1/x/(-3*x^2+5*x-2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(2)*arcsin(5*x/abs(x) - 4/abs(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=-\frac {1}{3} \, \sqrt {6} \sqrt {3} \arctan \left (\frac {1}{12} \, \sqrt {6} {\left (\frac {5 \, {\left (2 \, \sqrt {3} \sqrt {-3 \, x^{2} + 5 \, x - 2} - 1\right )}}{6 \, x - 5} - 1\right )}\right ) \]

[In]

integrate(1/x/(-3*x^2+5*x-2)^(1/2),x, algorithm="giac")

[Out]

-1/3*sqrt(6)*sqrt(3)*arctan(1/12*sqrt(6)*(5*(2*sqrt(3)*sqrt(-3*x^2 + 5*x - 2) - 1)/(6*x - 5) - 1))

Mupad [B] (verification not implemented)

Time = 10.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \sqrt {-2+5 x-3 x^2}} \, dx=\frac {\sqrt {2}\,\ln \left (\frac {5\,x-4+\sqrt {2}\,\sqrt {-3\,x^2+5\,x-2}\,2{}\mathrm {i}}{x}\right )\,1{}\mathrm {i}}{2} \]

[In]

int(1/(x*(5*x - 3*x^2 - 2)^(1/2)),x)

[Out]

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