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

Optimal. Leaf size=74 \[ -\frac{x \left (a-\frac{1}{x}\right )}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{2 x \sqrt{1-\frac{1}{a^2 x^2}}}{c^2}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2} \]

[Out]

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

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Rubi [A]  time = 0.101428, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {6177, 823, 807, 266, 63, 208} \[ -\frac{x \left (a-\frac{1}{x}\right )}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{2 x \sqrt{1-\frac{1}{a^2 x^2}}}{c^2}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - c/(a*x))^2),x]

[Out]

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

Rule 6177

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

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{e^{-3 \coth ^{-1}(a x)}}{\left (c-\frac{c}{a x}\right )^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{c-\frac{c x}{a}}{x^2 \left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{c^3}\\ &=-\frac{\left (a-\frac{1}{x}\right ) x}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{a^2 \operatorname{Subst}\left (\int \frac{\frac{2 c}{a^2}-\frac{c x}{a^3}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{c^3}\\ &=\frac{2 \sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{\left (a-\frac{1}{x}\right ) x}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{a c^2}\\ &=\frac{2 \sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{\left (a-\frac{1}{x}\right ) x}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}+\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 a c^2}\\ &=\frac{2 \sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{\left (a-\frac{1}{x}\right ) x}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{a \operatorname{Subst}\left (\int \frac{1}{a^2-a^2 x^2} \, dx,x,\sqrt{1-\frac{1}{a^2 x^2}}\right )}{c^2}\\ &=\frac{2 \sqrt{1-\frac{1}{a^2 x^2}} x}{c^2}-\frac{\left (a-\frac{1}{x}\right ) x}{a c^2 \sqrt{1-\frac{1}{a^2 x^2}}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a c^2}\\ \end{align*}

Mathematica [A]  time = 0.0372498, size = 69, normalized size = 0.93 \[ \frac{a^2 x^2-a x \sqrt{1-\frac{1}{a^2 x^2}} \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )+a x-2}{a^2 c^2 x \sqrt{1-\frac{1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - c/(a*x))^2),x]

[Out]

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

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Maple [B]  time = 0.133, size = 250, normalized size = 3.4 \begin{align*} -{\frac{1}{2\,a{c}^{2} \left ( ax-1 \right ) } \left ( 2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}-3\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+4\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}+ \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{2}}-6\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+2\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) -3\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ) \left ({\frac{ax-1}{ax+1}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^2*a^3-3*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1
/2)*x^2*a^2+4*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x*a^2+((a*x-1)*(a*x+1))^(3/2)*(a^2)^
(1/2)-6*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x*a+2*a*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2)
)-3*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/a*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/c^2/((a*x-1)*(a*x+1))^(1/2)/(a*
x-1)

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Maxima [A]  time = 1.03599, size = 169, normalized size = 2.28 \begin{align*} -a{\left (\frac{2 \, \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{{\left (a x - 1\right )} a^{2} c^{2}}{a x + 1} - a^{2} c^{2}} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c^{2}} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c^{2}} - \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.93056, size = 163, normalized size = 2.2 \begin{align*} \frac{{\left (a x + 2\right )} \sqrt{\frac{a x - 1}{a x + 1}} - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**(3/2)/(c-c/a/x)**2,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef