\(\int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx\) [884]

Optimal result

Integrand size = 25, antiderivative size = 70 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}} x}+\frac {\sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6332, 6328, 45} \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a x \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[(E^ArcCoth[a*x]*Sqrt[c - c/(a^2*x^2)])/x,x]

[Out]

-(Sqrt[c - c/(a^2*x^2)]/(a*Sqrt[1 - 1/(a^2*x^2)]*x)) + (Sqrt[c - c/(a^2*x^2)]*Log[x])/Sqrt[1 - 1/(a^2*x^2)]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6328

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u/x^(
2*p))*(-1 + a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !
IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]

Rule 6332

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_), x_Symbol] :> Dist[c^IntPart[p]*((c + d/x^2
)^FracPart[p]/(1 - 1/(a^2*x^2))^FracPart[p]), Int[u*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a
, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !IntegerQ[n/2] &&  !(IntegerQ[p] || GtQ[c, 0])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.59 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{a x}+\log (x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Integrate[(E^ArcCoth[a*x]*Sqrt[c - c/(a^2*x^2)])/x,x]

[Out]

(Sqrt[c - c/(a^2*x^2)]*(-(1/(a*x)) + Log[x]))/Sqrt[1 - 1/(a^2*x^2)]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.71

method	result	size
default	\(\frac {\left (a \ln \left (x \right ) x -1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.30 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\sqrt {a^{2} c} {\left (a x \log \left (x\right ) - 1\right )}}{a^{2} x} \]

[In]

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

[Out]

sqrt(a^2*c)*(a*x*log(x) - 1)/(a^2*x)

Sympy [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)*(c-c/a**2/x**2)**(1/2)/x,x)

[Out]

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

Maxima [F]

\[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}}}{x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(c - c/(a^2*x^2))/(x*sqrt((a*x - 1)/(a*x + 1))), x)

Giac [F]

\[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}}}{x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

integrate(sqrt(c - c/(a^2*x^2))/(x*sqrt((a*x - 1)/(a*x + 1))), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}}{x\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

[In]

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

[Out]

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