\(\int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx\) [2425]

Optimal result

Integrand size = 27, antiderivative size = 77 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=\frac {(a-b x) \log (x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}}-\frac {(a-b x) \log (a-b x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}} \]

[Out]

(-b*x+a)*ln(x)/a/(-(b*x-a)^2)^(1/2)-(-b*x+a)*ln(-b*x+a)/a/(-(b*x-a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {660, 36, 29, 31} \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=\frac {\log (x) (a-b x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}}-\frac {(a-b x) \log (a-b x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}} \]

[In]

Int[1/(x*Sqrt[-a^2 + 2*a*b*x - b^2*x^2]),x]

[Out]

((a - b*x)*Log[x])/(a*Sqrt[-a^2 + 2*a*b*x - b^2*x^2]) - ((a - b*x)*Log[a - b*x])/(a*Sqrt[-a^2 + 2*a*b*x - b^2*
x^2])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (a b-b^2 x\right ) \int \frac {1}{x \left (a b-b^2 x\right )} \, dx}{\sqrt {-a^2+2 a b x-b^2 x^2}} \\ & = \frac {\left (a b-b^2 x\right ) \int \frac {1}{x} \, dx}{a b \sqrt {-a^2+2 a b x-b^2 x^2}}+\frac {\left (b \left (a b-b^2 x\right )\right ) \int \frac {1}{a b-b^2 x} \, dx}{a \sqrt {-a^2+2 a b x-b^2 x^2}} \\ & = \frac {(a-b x) \log (x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}}-\frac {(a-b x) \log (a-b x)}{a \sqrt {-a^2+2 a b x-b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {-b^2} x}{a}-\frac {\sqrt {-a^2+2 a b x-b^2 x^2}}{a}\right )}{a} \]

[In]

Integrate[1/(x*Sqrt[-a^2 + 2*a*b*x - b^2*x^2]),x]

[Out]

(-2*ArcTan[(Sqrt[-b^2]*x)/a - Sqrt[-a^2 + 2*a*b*x - b^2*x^2]/a])/a

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.45

method	result	size
default	\(\frac {\left (-b x +a \right ) \left (\ln \left (x \right )-\ln \left (-b x +a \right )\right )}{\sqrt {-\left (-b x +a \right )^{2}}\, a}\)	\(35\)
risch	\(-\frac {\left (-b x +a \right ) \ln \left (b x -a \right )}{\sqrt {-\left (-b x +a \right )^{2}}\, a}+\frac {\left (-b x +a \right ) \ln \left (-x \right )}{\sqrt {-\left (-b x +a \right )^{2}}\, a}\)	\(59\)

[In]

int(1/x/(-(b*x-a)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(-b*x+a)*(ln(x)-ln(-b*x+a))/(-(-b*x+a)^2)^(1/2)/a

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=-\sqrt {-\frac {1}{a^{2}}} \log \left (\frac {i \, a^{2} \sqrt {-\frac {1}{a^{2}}} + 2 \, b x - a}{2 \, b}\right ) + \sqrt {-\frac {1}{a^{2}}} \log \left (\frac {-i \, a^{2} \sqrt {-\frac {1}{a^{2}}} + 2 \, b x - a}{2 \, b}\right ) \]

[In]

integrate(1/x/(-(b*x-a)^2)^(1/2),x, algorithm="fricas")

[Out]

-sqrt(-1/a^2)*log(1/2*(I*a^2*sqrt(-1/a^2) + 2*b*x - a)/b) + sqrt(-1/a^2)*log(1/2*(-I*a^2*sqrt(-1/a^2) + 2*b*x
- a)/b)

Sympy [F]

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

[In]

integrate(1/x/(-(b*x-a)**2)**(1/2),x)

[Out]

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=-\frac {i \, \left (-1\right )^{-2 \, a b x + 2 \, a^{2}} \log \left (-\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a} \]

[In]

integrate(1/x/(-(b*x-a)^2)^(1/2),x, algorithm="maxima")

[Out]

-I*(-1)^(-2*a*b*x + 2*a^2)*log(-2*a*b*x/abs(x) + 2*a^2/abs(x))/a

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=\frac {i \, \log \left ({\left | b x - a \right |}\right )}{a \mathrm {sgn}\left (-b x + a\right )} - \frac {i \, \log \left ({\left | x \right |}\right )}{a \mathrm {sgn}\left (-b x + a\right )} \]

[In]

integrate(1/x/(-(b*x-a)^2)^(1/2),x, algorithm="giac")

[Out]

I*log(abs(b*x - a))/(a*sgn(-b*x + a)) - I*log(abs(x))/(a*sgn(-b*x + a))

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x \sqrt {-a^2+2 a b x-b^2 x^2}} \, dx=-\frac {\ln \left (a\,b-\frac {a^2}{x}+\frac {\sqrt {-a^2}\,\sqrt {-a^2+2\,a\,b\,x-b^2\,x^2}}{x}\right )}{\sqrt {-a^2}} \]

[In]

int(1/(x*(-(a - b*x)^2)^(1/2)),x)

[Out]

-log(a*b - a^2/x + ((-a^2)^(1/2)*(2*a*b*x - b^2*x^2 - a^2)^(1/2))/x)/(-a^2)^(1/2)