\(\int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx\) [859]

Optimal result

Integrand size = 24, antiderivative size = 72 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{a \sqrt {c-\frac {c}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6332, 6328, 45} \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (a x+1)}{a \sqrt {c-\frac {c}{a^2 x^2}}} \]

[In]

Int[1/(E^ArcCoth[a*x]*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/Sqrt[c - c/(a^2*x^2)] - (Sqrt[1 - 1/(a^2*x^2)]*Log[1 + a*x])/(a*Sqrt[c - c/(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 {1-\frac {1}{a^2 x^2}} \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {1-\frac {1}{a^2 x^2}}} \, dx}{\sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\left (a \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \frac {x}{1+a x} \, dx}{\sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\left (a \sqrt {1-\frac {1}{a^2 x^2}}\right ) \int \left (\frac {1}{a}-\frac {1}{a (1+a x)}\right ) \, dx}{\sqrt {c-\frac {c}{a^2 x^2}}} \\ & = \frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{\sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-\frac {1}{a^2 x^2}} \log (1+a x)}{a \sqrt {c-\frac {c}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.60 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (x-\frac {\log (1+a x)}{a}\right )}{\sqrt {c-\frac {c}{a^2 x^2}}} \]

[In]

Integrate[1/(E^ArcCoth[a*x]*Sqrt[c - c/(a^2*x^2)]),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (-a x +\ln \left (a x +1\right )\right )}{\sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, x \,a^{2}}\)	\(59\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.36 \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\frac {\sqrt {a^{2} c} {\left (a x - \log \left (a x + 1\right )\right )}}{a^{2} c} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{-\coth ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 x^2}}} \, dx=\text {Exception raised: NotImplementedError} \]

[In]

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

[Out]

Exception raised: NotImplementedError >> unable to parse Giac output: Bad Argument Type

Mupad [F(-1)]

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

[In]

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

[Out]

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