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

Optimal. Leaf size=43 \[ \frac{2 x \sqrt{\frac{1}{a^2 x^2}+1}}{a}+\frac{2 \log (x)}{a^2}-\frac{2 \text{csch}^{-1}(a x)}{a^2}+\frac{x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.15791, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6338, 6742, 242, 277, 215} \[ \frac{2 x \sqrt{\frac{1}{a^2 x^2}+1}}{a}+\frac{2 \log (x)}{a^2}-\frac{2 \text{csch}^{-1}(a x)}{a^2}+\frac{x^2}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6338

Int[E^(ArcCsch[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(1/u + Sqrt[1 + 1/u^2])^n, x] /; FreeQ[m, x] && Int
egerQ[n]

Rule 6742

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

Rule 242

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 215

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

Rubi steps

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

Mathematica [A]  time = 0.0472147, size = 44, normalized size = 1.02 \[ \frac{a x \left (4 \sqrt{\frac{1}{a^2 x^2}+1}+a x\right )-4 \sinh ^{-1}\left (\frac{1}{a x}\right )+4 \log (x)}{2 a^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*x*(4*Sqrt[1 + 1/(a^2*x^2)] + a*x) - 4*ArcSinh[1/(a*x)] + 4*Log[x])/(2*a^2)

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Maple [B]  time = 0.179, size = 120, normalized size = 2.8 \begin{align*}{\frac{{x}^{2}}{2}}+2\,{\frac{\ln \left ( x \right ) }{{a}^{2}}}+2\,{\frac{x}{{a}^{3}\sqrt{{a}^{-2}}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( \sqrt{{a}^{-2}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}{a}^{2}-\ln \left ( 2\,{\frac{1}{{a}^{2}x} \left ( \sqrt{{a}^{-2}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}{a}^{2}+1 \right ) } \right ) \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,x)

[Out]

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

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Maxima [A]  time = 0.964158, size = 101, normalized size = 2.35 \begin{align*} \frac{1}{2} \, x^{2} + \frac{2 \, x \sqrt{\frac{1}{a^{2} x^{2}} + 1} - \frac{\log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right )}{a} + \frac{\log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right )}{a}}{a} + \frac{2 \, \log \left (x\right )}{a^{2}} \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,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.49821, size = 235, normalized size = 5.47 \begin{align*} \frac{a^{2} x^{2} + 4 \, a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - 4 \, \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) + 4 \, \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) + 4 \, \log \left (x\right )}{2 \, a^{2}} \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,x, algorithm="fricas")

[Out]

1/2*(a^2*x^2 + 4*a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 4*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x + 1) + 4*lo
g(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x - 1) + 4*log(x))/a^2

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Sympy [A]  time = 4.36319, size = 63, normalized size = 1.47 \begin{align*} \frac{x^{2}}{2} + \frac{2 x}{a \sqrt{1 + \frac{1}{a^{2} x^{2}}}} + \frac{2 \log{\left (x \right )}}{a^{2}} - \frac{2 \operatorname{asinh}{\left (\frac{1}{a x} \right )}}{a^{2}} + \frac{2}{a^{3} x \sqrt{1 + \frac{1}{a^{2} x^{2}}}} \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,x)

[Out]

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

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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,x, algorithm="giac")

[Out]

Exception raised: TypeError