### 3.55 $$\int \frac{e^{-3 \coth ^{-1}(a x)}}{x^2} \, dx$$

Optimal. Leaf size=53 $\frac{2 \left (a-\frac{1}{x}\right )^2}{a \sqrt{1-\frac{1}{a^2 x^2}}}+3 a \sqrt{1-\frac{1}{a^2 x^2}}+3 a \csc ^{-1}(a x)$

[Out]

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

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Rubi [A]  time = 0.0775462, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.417, Rules used = {6169, 853, 669, 641, 216} $\frac{2 \left (a-\frac{1}{x}\right )^2}{a \sqrt{1-\frac{1}{a^2 x^2}}}+3 a \sqrt{1-\frac{1}{a^2 x^2}}+3 a \csc ^{-1}(a x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 6169

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/
a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]

Rule 853

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

Rule 669

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

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0673987, size = 41, normalized size = 0.77 $\frac{a \sqrt{1-\frac{1}{a^2 x^2}} (5 a x+1)}{a x+1}+3 a \sin ^{-1}\left (\frac{1}{a x}\right )$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Sqrt[1 - 1/(a^2*x^2)]*(1 + 5*a*x))/(1 + a*x) + 3*a*ArcSin[1/(a*x)]

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Maple [B]  time = 0.131, size = 592, normalized size = 11.2 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^4*a^4+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3+(a^2*x^2-1)^(3/2)*(a^2)^(
1/2)*x^2*a^2-5*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3+ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3
*a^4-3*a^3*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))-ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1
/2))*x^3*a^4+2*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x*a+2*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^2*a^2+2*(a^2)^(
1/2)*(a^2*x^2-1)^(3/2)*x*a-7*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^2*a^2+2*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/
(a^2)^(1/2))*x^2*a^3-6*a^2*x^2*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(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+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x*a+(a^2*x^2-1)^(3/2)*(a^2)^(1/2)-3*(a^2)^(
1/2)*(a^2*x^2-1)^(1/2)*x*a+ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-3*a*x*(a^2)^(1/2)*arcta
n(1/(a^2*x^2-1)^(1/2))-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)/x/(a*x-1)/((a*x-1)*(a*x+1))^(1/2)

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Maxima [A]  time = 1.5446, size = 97, normalized size = 1.83 \begin{align*} 2 \, a{\left (2 \, \sqrt{\frac{a x - 1}{a x + 1}} + \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{\frac{a x - 1}{a x + 1} + 1} - 3 \, \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(2*sqrt((a*x - 1)/(a*x + 1)) + sqrt((a*x - 1)/(a*x + 1))/((a*x - 1)/(a*x + 1) + 1) - 3*arctan(sqrt((a*x -
1)/(a*x + 1))))

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Fricas [A]  time = 1.51105, size = 116, normalized size = 2.19 \begin{align*} -\frac{6 \, a x \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) -{\left (5 \, a x + 1\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(6*a*x*arctan(sqrt((a*x - 1)/(a*x + 1))) - (5*a*x + 1)*sqrt((a*x - 1)/(a*x + 1)))/x

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(((a*x - 1)/(a*x + 1))**(3/2)/x**2, x)

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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)/x^2,x, algorithm="giac")

[Out]

undef