### 3.671 $$\int \frac{e^{\coth ^{-1}(a x)} \sqrt{c-a^2 c x^2}}{x} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.138636, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.12, Rules used = {6192, 6193, 43} $\frac{\sqrt{c-a^2 c x^2}}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\log (x) \sqrt{c-a^2 c x^2}}{a x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6192

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c + d*x^2)^p/(x^(2*p)*(
1 - 1/(a^2*x^2))^p), Int[u*x^(2*p)*(1 - 1/(a^2*x^2))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x]
&& EqQ[a^2*c + d, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

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

Mathematica [A]  time = 0.0203672, size = 42, normalized size = 0.61 $\frac{\sqrt{c-a^2 c x^2} (a x+\log (x))}{a x \sqrt{1-\frac{1}{a^2 x^2}}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.138, size = 44, normalized size = 0.6 \begin{align*}{\frac{ax+\ln \left ( x \right ) }{ax+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.72308, size = 42, normalized size = 0.61 \begin{align*} \frac{\sqrt{-a^{2} c}{\left (a x + \log \left (x\right )\right )}}{a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-a^2*c)*(a*x + log(x))/a

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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