\(\int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx\) [885]

Optimal result

Integrand size = 25, antiderivative size = 46 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}{2 a \sqrt {1-\frac {1}{a^2 x^2}} x^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6332, 6328, 37} \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=-\frac {(a x+1)^2 \sqrt {c-\frac {c}{a^2 x^2}}}{2 a x^2 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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^2} \, dx}{\sqrt {1-\frac {1}{a^2 x^2}}} \\ & = \frac {\sqrt {c-\frac {c}{a^2 x^2}} \int \frac {1+a x}{x^3} \, dx}{a \sqrt {1-\frac {1}{a^2 x^2}}} \\ & = -\frac {\sqrt {c-\frac {c}{a^2 x^2}} (1+a x)^2}{2 a \sqrt {1-\frac {1}{a^2 x^2}} x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{2 a x^2}-\frac {1}{x}\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.15

method	result	size
gosper	\(-\frac {\left (2 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{2 x \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(53\)
default	\(-\frac {\left (2 a x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}}{2 x \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(53\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.46 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=-\frac {\sqrt {a^{2} c} {\left (2 \, a x + 1\right )}}{2 \, a^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.37 \[ \int \frac {e^{\coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x^2} \, dx=\frac {\left (x\,\sqrt {c-\frac {c}{a^2\,x^2}}+\frac {\sqrt {c-\frac {c}{a^2\,x^2}}}{2\,a}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{\frac {x}{a}-x^2} \]

[In]

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

[Out]

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