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

Optimal. Leaf size=69 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

[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)

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Rubi [A]  time = 0.26, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6197, 6193, 43} \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}}}{a x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\log (x) \sqrt {c-\frac {c}{a^2 x^2}}}{\sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c/(a^2*x^2)]/(E^ArcCoth[a*x]*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 43

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 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), 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 6197

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*} \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}} \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*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 0.59 \[ \frac {\sqrt {c-\frac {c}{a^2 x^2}} \left (a \log (x)+\frac {1}{x}\right )}{a \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 21, normalized size = 0.30 \[ \frac {\sqrt {a^{2} c} {\left (a x \log \relax (x) + 1\right )}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a^2*c)*(a*x*log(x) + 1)/(a^2*x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [sign(x),sign(a*
x+1)]sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 50, normalized size = 0.72 \[ \frac {\left (a \ln \relax (x ) x +1\right ) \sqrt {\frac {c \left (a^{2} x^{2}-1\right )}{a^{2} x^{2}}}\, \sqrt {\frac {a x -1}{a x +1}}}{a x -1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c - \frac {c}{a^{2} x^{2}}} \sqrt {\frac {a x - 1}{a x + 1}}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {c-\frac {c}{a^2\,x^2}}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {\frac {a x - 1}{a x + 1}} \sqrt {- c \left (-1 + \frac {1}{a x}\right ) \left (1 + \frac {1}{a x}\right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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