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

Optimal. Leaf size=115 \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a^2 x^2}}}+\frac{2 \sqrt{1-\frac{1}{a^2 x^2}}}{a (1-a x) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{3 \sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{a \sqrt{c-\frac{c}{a^2 x^2}}} \]

[Out]

(Sqrt[1 - 1/(a^2*x^2)]*x)/Sqrt[c - c/(a^2*x^2)] + (2*Sqrt[1 - 1/(a^2*x^2)])/(a*Sqrt[c - c/(a^2*x^2)]*(1 - a*x)
) + (3*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 = 0.123677, antiderivative size = 115, 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, 77} \[ \frac{x \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a^2 x^2}}}+\frac{2 \sqrt{1-\frac{1}{a^2 x^2}}}{a (1-a x) \sqrt{c-\frac{c}{a^2 x^2}}}+\frac{3 \sqrt{1-\frac{1}{a^2 x^2}} \log (1-a x)}{a \sqrt{c-\frac{c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[E^(3*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)] + (2*Sqrt[1 - 1/(a^2*x^2)])/(a*Sqrt[c - c/(a^2*x^2)]*(1 - a*x)
) + (3*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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{e^{3 \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^{3 \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)}{(-1+a x)^2} \, 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{2}{a (-1+a x)^2}+\frac{3}{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{2 \sqrt{1-\frac{1}{a^2 x^2}}}{a \sqrt{c-\frac{c}{a^2 x^2}} (1-a x)}+\frac{3 \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 = 0.0475806, size = 56, normalized size = 0.49 \[ \frac{\sqrt{1-\frac{1}{a^2 x^2}} \left (a x+\frac{2}{1-a x}+3 \log (1-a x)\right )}{a \sqrt{c-\frac{c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.243, size = 85, normalized size = 0.7 \begin{align*}{\frac{ \left ( ax-1 \right ) \left ({a}^{2}{x}^{2}+3\,\ln \left ( ax-1 \right ) xa-ax-3\,\ln \left ( ax-1 \right ) -2 \right ) }{ \left ( ax+1 \right ) x{a}^{2}} \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{2}}}{\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(1/((a*x-1)/(a*x+1))^(3/2)/(c-c/a^2/x^2)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{c - \frac{c}{a^{2} x^{2}}} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.54377, size = 105, normalized size = 0.91 \begin{align*} \frac{{\left (a^{2} x^{2} - a x + 3 \,{\left (a x - 1\right )} \log \left (a x - 1\right ) - 2\right )} \sqrt{a^{2} c}}{a^{3} c x - a^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^2*x^2 - a*x + 3*(a*x - 1)*log(a*x - 1) - 2)*sqrt(a^2*c)/(a^3*c*x - 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(1/((a*x-1)/(a*x+1))**(3/2)/(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*} \int \frac{1}{\sqrt{c - \frac{c}{a^{2} x^{2}}} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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