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

Optimal. Leaf size=40 \[ -\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{2 x}-\frac{1}{2 a x^2}-\frac{1}{2} a \text{csch}^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.028842, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6336, 30, 335, 195, 215} \[ -\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{2 x}-\frac{1}{2 a x^2}-\frac{1}{2} a \text{csch}^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCsch[a*x]/x^2,x]

[Out]

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

Rule 6336

Int[E^ArcCsch[(a_.)*(x_)^(p_.)]*(x_)^(m_.), x_Symbol] :> Dist[1/a, Int[x^(m - p), x], x] + Int[x^m*Sqrt[1 + 1/
(a^2*x^(2*p))], x] /; FreeQ[{a, m, p}, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 335

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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

Mathematica [A]  time = 0.0272924, size = 43, normalized size = 1.08 \[ -\frac{a x \sqrt{\frac{1}{a^2 x^2}+1}+a^2 x^2 \sinh ^{-1}\left (\frac{1}{a x}\right )+1}{2 a x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCsch[a*x]/x^2,x]

[Out]

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

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Maple [B]  time = 0.188, size = 145, normalized size = 3.6 \begin{align*} -{\frac{1}{2\,x}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ({a}^{2} \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{-2}}-\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}\sqrt{{a}^{-2}}{x}^{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 ){x}^{2} \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}{\frac{1}{\sqrt{{a}^{-2}}}}}-{\frac{1}{2\,a{x}^{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))/x^2,x)

[Out]

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

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Maxima [B]  time = 1.0079, size = 116, normalized size = 2.9 \begin{align*} -\frac{a^{2} x \sqrt{\frac{1}{a^{2} x^{2}} + 1}}{2 \,{\left (a^{2} x^{2}{\left (\frac{1}{a^{2} x^{2}} + 1\right )} - 1\right )}} - \frac{1}{4} \, a \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right ) + \frac{1}{4} \, a \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right ) - \frac{1}{2 \, a 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))/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.94851, size = 232, normalized size = 5.8 \begin{align*} -\frac{a^{2} x^{2} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - a^{2} x^{2} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) + a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} + 1}{2 \, a 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))/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.07544, size = 36, normalized size = 0.9 \begin{align*} - \frac{a \operatorname{asinh}{\left (\frac{1}{a x} \right )}}{2} - \frac{\sqrt{1 + \frac{1}{a^{2} x^{2}}}}{2 x} - \frac{1}{2 a 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))/x**2,x)

[Out]

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

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Giac [B]  time = 1.13513, size = 111, normalized size = 2.78 \begin{align*} -\frac{a^{4}{\left | a \right |} \log \left (\sqrt{a^{2} x^{2} + 1} + 1\right ) \mathrm{sgn}\left (x\right ) - a^{4}{\left | a \right |} \log \left (\sqrt{a^{2} x^{2} + 1} - 1\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \,{\left (\sqrt{a^{2} x^{2} + 1} a^{4}{\left | a \right |} \mathrm{sgn}\left (x\right ) + a^{5}\right )}}{a^{2} x^{2}}}{4 \, a^{4}} \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))/x^2,x, algorithm="giac")

[Out]

-1/4*(a^4*abs(a)*log(sqrt(a^2*x^2 + 1) + 1)*sgn(x) - a^4*abs(a)*log(sqrt(a^2*x^2 + 1) - 1)*sgn(x) + 2*(sqrt(a^
2*x^2 + 1)*a^4*abs(a)*sgn(x) + a^5)/(a^2*x^2))/a^4