3.4.36 \(\int \frac {1}{x \sqrt {-25+2 x}} \, dx\) [336]

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

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, 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 = {65, 209} \begin {gather*} \frac {2}{5} \text {ArcTan}\left (\frac {1}{5} \sqrt {2 x-25}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-25 + 2*x]),x]

[Out]

(2*ArcTan[Sqrt[-25 + 2*x]/5])/5

Rule 65

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 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 &=\text {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*}

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Mathematica [A]
time = 0.01, size = 18, normalized size = 1.00 \begin {gather*} \frac {2}{5} \tan ^{-1}\left (\frac {1}{5} \sqrt {-25+2 x}\right ) \end {gather*}

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 = 0.10, size = 13, normalized size = 0.72

method	result	size
derivativedivides	\(\frac {2 \arctan \left (\frac {\sqrt {-25+2 x}}{5}\right )}{5}\)	\(13\)
default	\(\frac {2 \arctan \left (\frac {\sqrt {-25+2 x}}{5}\right )}{5}\)	\(13\)
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) x +5 \sqrt {-25+2 x}-25 \RootOf \left (\textit {\_Z}^{2}+1\right )}{x}\right )}{5}\)	\(40\)
meijerg	\(\frac {\sqrt {-\mathrm {signum}\left (x -\frac {25}{2}\right )}\, \left (\left (-\ln \left (2\right )+\ln \left (x \right )-2 \ln \left (5\right )+i \pi \right ) \sqrt {\pi }-2 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {2 x}{25}}}{2}\right )\right )}{5 \sqrt {\pi }\, \sqrt {\mathrm {signum}\left (x -\frac {25}{2}\right )}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.16, size = 12, normalized size = 0.67 \begin {gather*} \frac {2}{5} \, \arctan \left (\frac {1}{5} \, \sqrt {2 \, x - 25}\right ) \end {gather*}

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 = 0.78, size = 12, normalized size = 0.67 \begin {gather*} \frac {2}{5} \, \arctan \left (\frac {1}{5} \, \sqrt {2 \, x - 25}\right ) \end {gather*}

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 [C] Result contains complex when optimal does not.
time = 0.43, size = 44, normalized size = 2.44 \begin {gather*} \begin {cases} \frac {2 i \operatorname {acosh}{\left (\frac {5 \sqrt {2}}{2 \sqrt {x}} \right )}}{5} & \text {for}\: \frac {1}{\left |{x}\right |} > \frac {2}{25} \\- \frac {2 \operatorname {asin}{\left (\frac {5 \sqrt {2}}{2 \sqrt {x}} \right )}}{5} & \text {otherwise} \end {cases} \end {gather*}

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

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Giac [A]
time = 1.03, size = 12, normalized size = 0.67 \begin {gather*} \frac {2}{5} \, \arctan \left (\frac {1}{5} \, \sqrt {2 \, x - 25}\right ) \end {gather*}

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))

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Mupad [B]
time = 0.15, size = 12, normalized size = 0.67 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\sqrt {2\,x-25}}{5}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(2*x - 25)^(1/2)),x)

[Out]

(2*atan((2*x - 25)^(1/2)/5))/5

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