### 3.859 $$\int \frac{e^{-\coth ^{-1}(a x)}}{\sqrt{c-\frac{c}{a^2 x^2}}} \, dx$$

Optimal. Leaf size=72 $\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a^2 x^2}}}-\frac{\sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{a \sqrt{c-\frac{c}{a^2 x^2}}}$

[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)
])

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Rubi [A]  time = 0.113005, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {6197, 6193, 43} $\frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a^2 x^2}}}-\frac{\sqrt{1-\frac{1}{a^2 x^2}} \log (a x+1)}{a \sqrt{c-\frac{c}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[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 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])

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 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])

Rubi steps

\begin{align*} \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}} \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]  time = 0.0301584, size = 45, normalized size = 0.62 $\frac{\sqrt{1-\frac{1}{a^2 x^2}} (a x-\log (a x+1))}{a \sqrt{c-\frac{c}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.219, size = 59, normalized size = 0.8 \begin{align*} -{\frac{ \left ( ax+1 \right ) \left ( -ax+\ln \left ( ax+1 \right ) \right ) }{{a}^{2}x}\sqrt{{\frac{ax-1}{ax+1}}}{\frac{1}{\sqrt{{\frac{c \left ({a}^{2}{x}^{2}-1 \right ) }{{a}^{2}{x}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{\sqrt{c - \frac{c}{a^{2} x^{2}}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.88431, size = 57, normalized size = 0.79 \begin{align*} \frac{\sqrt{a^{2} c}{\left (a x - \log \left (a x + 1\right )\right )}}{a^{2} c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage2