### 3.53 $$\int e^{2 \text{csch}^{-1}(a x)} \, dx$$

Optimal. Leaf size=47 $-\frac{2 \sqrt{\frac{1}{a^2 x^2}+1}}{a}+\frac{2 \tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{a}-\frac{2}{a^2 x}+x$

[Out]

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

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Rubi [A]  time = 0.0718435, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.75, Rules used = {6333, 6742, 266, 50, 63, 208} $-\frac{2 \sqrt{\frac{1}{a^2 x^2}+1}}{a}+\frac{2 \tanh ^{-1}\left (\sqrt{\frac{1}{a^2 x^2}+1}\right )}{a}-\frac{2}{a^2 x}+x$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*ArcCsch[a*x]),x]

[Out]

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

Rule 6333

Int[E^(ArcCsch[u_]*(n_.)), x_Symbol] :> Int[(1/u + Sqrt[1 + 1/u^2])^n, x] /; IntegerQ[n]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.0349868, size = 52, normalized size = 1.11 $-\frac{2 \sqrt{\frac{1}{a^2 x^2}+1}}{a}+\frac{2 \log \left (a x \left (\sqrt{\frac{1}{a^2 x^2}+1}+1\right )\right )}{a}-\frac{2}{a^2 x}+x$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCsch[a*x]),x]

[Out]

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

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Maple [B]  time = 0.128, size = 112, normalized size = 2.4 \begin{align*} x-2\,{\frac{1}{{a}^{2}x}}+2\,{\frac{1}{a}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( -{a}^{2} \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{3/2}+\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}{x}^{2}{a}^{2}+\ln \left ( x+\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}} \right ) x \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/a^2/x^2)^(1/2))^2,x)

[Out]

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

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Maxima [A]  time = 0.965249, size = 80, normalized size = 1.7 \begin{align*} x - \frac{2 \, \sqrt{\frac{1}{a^{2} x^{2}} + 1} - \log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right ) + \log \left (\sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right )}{a} - \frac{2}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.53242, size = 165, normalized size = 3.51 \begin{align*} \frac{a^{2} x^{2} - 2 \, a x \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x\right ) - 2 \, a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - 2 \, a x - 2}{a^{2} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.21986, size = 49, normalized size = 1.04 \begin{align*} x - \frac{2 x}{\sqrt{a^{2} x^{2} + 1}} + \frac{2 \operatorname{asinh}{\left (a x \right )}}{a} - \frac{2}{a^{2} x} - \frac{2}{a^{2} x \sqrt{a^{2} x^{2} + 1}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x - 2*x/sqrt(a**2*x**2 + 1) + 2*asinh(a*x)/a - 2/(a**2*x) - 2/(a**2*x*sqrt(a**2*x**2 + 1))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError