### 3.2437 $$\int \frac{1}{x \sqrt{-2+5 x-3 x^2}} \, dx$$

Optimal. Leaf size=36 $-\frac{\tan ^{-1}\left (\frac{4-5 x}{2 \sqrt{2} \sqrt{-3 x^2+5 x-2}}\right )}{\sqrt{2}}$

[Out]

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

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Rubi [A]  time = 0.0108753, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {724, 204} $-\frac{\tan ^{-1}\left (\frac{4-5 x}{2 \sqrt{2} \sqrt{-3 x^2+5 x-2}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[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 724

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 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])

Rubi steps

\begin{align*} \int \frac{1}{x \sqrt{-2+5 x-3 x^2}} \, dx &=-\left (2 \operatorname{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]  time = 0.0097152, size = 30, normalized size = 0.83 $\frac{\tan ^{-1}\left (\frac{5 x-4}{2 \sqrt{-6 x^2+10 x-4}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

ArcTan[(-4 + 5*x)/(2*Sqrt[-4 + 10*x - 6*x^2])]/Sqrt[2]

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Maple [A]  time = 0.042, size = 29, normalized size = 0.8 \begin{align*}{\frac{\sqrt{2}}{2}\arctan \left ({\frac{ \left ( 5\,x-4 \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{-3\,{x}^{2}+5\,x-2}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.49993, size = 27, normalized size = 0.75 \begin{align*} \frac{1}{2} \, \sqrt{2} \arcsin \left (\frac{5 \, x}{{\left | x \right |}} - \frac{4}{{\left | x \right |}}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Fricas [A]  time = 2.03531, size = 115, normalized size = 3.19 \begin{align*} -\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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{- \left (x - 1\right ) \left (3 x - 2\right )}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13379, size = 59, normalized size = 1.64 \begin{align*} -\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 ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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))