\(\int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx\) [923]

Optimal result

Integrand size = 27, antiderivative size = 108 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a \sqrt {1-\frac {1}{a^2 x^2}} x}-\frac {3 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

-(c-c/a^2/x^2)^(1/2)/a/x/(1-1/a^2/x^2)^(1/2)-3*ln(x)*(c-c/a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)+4*ln(a*x+1)*(c-c/
a^2/x^2)^(1/2)/(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6332, 6328, 90} \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=-\frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a x \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {3 \log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (a x+1)}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Int[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x),x]

[Out]

-(Sqrt[c - c/(a^2*x^2)]/(a*Sqrt[1 - 1/(a^2*x^2)]*x)) - (3*Sqrt[c - c/(a^2*x^2)]*Log[x])/Sqrt[1 - 1/(a^2*x^2)]
+ (4*Sqrt[c - c/(a^2*x^2)]*Log[1 + a*x])/Sqrt[1 - 1/(a^2*x^2)]

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -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^{-3 \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)^2}{x^2 (1+a x)} \, 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 {3 a}{x}+\frac {4 a^2}{1+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 {3 \sqrt {c-\frac {c}{a^2 x^2}} \log (x)}{\sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-\frac {c}{a^2 x^2}} \log (1+a x)}{\sqrt {1-\frac {1}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.47 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (-\frac {1}{a x}-3 \log (x)+4 \log (1+a x)\right )}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

Integrate[Sqrt[c - c/(a^2*x^2)]/(E^(3*ArcCoth[a*x])*x),x]

[Out]

(Sqrt[c - c/(a^2*x^2)]*(-(1/(a*x)) - 3*Log[x] + 4*Log[1 + a*x]))/Sqrt[1 - 1/(a^2*x^2)]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.61

method	result	size
default	\(\frac {\left (4 a \ln \left (a x +1\right ) x -3 a \ln \left (x \right ) x -1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{\left (a x -1\right )^{2}}\)	\(66\)

[In]

int((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x,x,method=_RETURNVERBOSE)

[Out]

(4*a*ln(a*x+1)*x-3*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))^(3/2)/(a*x-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.30 \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\frac {\sqrt {a^{2} c} {\left (4 \, a x \log \left (a x + 1\right ) - 3 \, a x \log \left (x\right ) - 1\right )}}{a^{2} x} \]

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x,x, algorithm="fricas")

[Out]

sqrt(a^2*c)*(4*a*x*log(a*x + 1) - 3*a*x*log(x) - 1)/(a^2*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\text {Timed out} \]

[In]

integrate((c-c/a**2/x**2)**(1/2)*((a*x-1)/(a*x+1))**(3/2)/x,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x} \,d x } \]

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x,x, algorithm="maxima")

[Out]

integrate(sqrt(c - c/(a^2*x^2))*((a*x - 1)/(a*x + 1))^(3/2)/x, x)

Giac [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int { \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x} \,d x } \]

[In]

integrate((c-c/a^2/x^2)^(1/2)*((a*x-1)/(a*x+1))^(3/2)/x,x, algorithm="giac")

[Out]

integrate(sqrt(c - c/(a^2*x^2))*((a*x - 1)/(a*x + 1))^(3/2)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{-3 \coth ^{-1}(a x)} \sqrt {c-\frac {c}{a^2 x^2}}}{x} \, dx=\int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{x} \,d x \]

[In]

int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x,x)

[Out]

int(((c - c/(a^2*x^2))^(1/2)*((a*x - 1)/(a*x + 1))^(3/2))/x, x)