3.336 \(\int \frac{1}{x \sqrt{-25+2 x}} \, dx\)

Optimal. Leaf size=18 \[ \frac{2}{5} \tan ^{-1}\left (\frac{1}{5} \sqrt{2 x-25}\right ) \]

[Out]

(2*ArcTan[Sqrt[-25 + 2*x]/5])/5

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Rubi [A]  time = 0.002827, antiderivative size = 18, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {63, 203} \[ \frac{2}{5} \tan ^{-1}\left (\frac{1}{5} \sqrt{2 x-25}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-25 + 2*x]),x]

[Out]

(2*ArcTan[Sqrt[-25 + 2*x]/5])/5

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rubi steps

\begin{align*} \int \frac{1}{x \sqrt{-25+2 x}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\frac{25}{2}+\frac{x^2}{2}} \, dx,x,\sqrt{-25+2 x}\right )\\ &=\frac{2}{5} \tan ^{-1}\left (\frac{1}{5} \sqrt{-25+2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0024075, size = 18, normalized size = 1. \[ \frac{2}{5} \tan ^{-1}\left (\frac{1}{5} \sqrt{2 x-25}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Sqrt[-25 + 2*x]),x]

[Out]

(2*ArcTan[Sqrt[-25 + 2*x]/5])/5

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Maple [A]  time = 0.006, size = 13, normalized size = 0.7 \begin{align*}{\frac{2}{5}\arctan \left ({\frac{1}{5}\sqrt{-25+2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(-25+2*x)^(1/2),x)

[Out]

2/5*arctan(1/5*(-25+2*x)^(1/2))

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Maxima [A]  time = 1.40573, size = 16, normalized size = 0.89 \begin{align*} \frac{2}{5} \, \arctan \left (\frac{1}{5} \, \sqrt{2 \, x - 25}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-25+2*x)^(1/2),x, algorithm="maxima")

[Out]

2/5*arctan(1/5*sqrt(2*x - 25))

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Fricas [A]  time = 1.85513, size = 43, normalized size = 2.39 \begin{align*} \frac{2}{5} \, \arctan \left (\frac{1}{5} \, \sqrt{2 \, x - 25}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-25+2*x)^(1/2),x, algorithm="fricas")

[Out]

2/5*arctan(1/5*sqrt(2*x - 25))

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Sympy [A]  time = 0.979219, size = 44, normalized size = 2.44 \begin{align*} \begin{cases} \frac{2 i \operatorname{acosh}{\left (\frac{5 \sqrt{2}}{2 \sqrt{x}} \right )}}{5} & \text{for}\: \frac{25}{2 \left |{x}\right |} > 1 \\- \frac{2 \operatorname{asin}{\left (\frac{5 \sqrt{2}}{2 \sqrt{x}} \right )}}{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-25+2*x)**(1/2),x)

[Out]

Piecewise((2*I*acosh(5*sqrt(2)/(2*sqrt(x)))/5, 25/(2*Abs(x)) > 1), (-2*asin(5*sqrt(2)/(2*sqrt(x)))/5, True))

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Giac [A]  time = 1.0588, size = 16, normalized size = 0.89 \begin{align*} \frac{2}{5} \, \arctan \left (\frac{1}{5} \, \sqrt{2 \, x - 25}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(-25+2*x)^(1/2),x, algorithm="giac")

[Out]

2/5*arctan(1/5*sqrt(2*x - 25))