3.56 \(\int \frac{e^{2 \text{csch}^{-1}(a x)}}{x^3} \, dx\)

Optimal. Leaf size=73 \[ -\frac{a \sqrt{\frac{1}{a^2 x^2}+1}}{4 x}-\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{2 a x^3}-\frac{1}{2 a^2 x^4}+\frac{1}{4} a^2 \text{csch}^{-1}(a x)-\frac{1}{2 x^2} \]

[Out]

-1/(2*a^2*x^4) - Sqrt[1 + 1/(a^2*x^2)]/(2*a*x^3) - 1/(2*x^2) - (a*Sqrt[1 + 1/(a^2*x^2)])/(4*x) + (a^2*ArcCsch[
a*x])/4

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Rubi [A]  time = 0.233842, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6338, 6742, 335, 279, 321, 215} \[ -\frac{a \sqrt{\frac{1}{a^2 x^2}+1}}{4 x}-\frac{\sqrt{\frac{1}{a^2 x^2}+1}}{2 a x^3}-\frac{1}{2 a^2 x^4}+\frac{1}{4} a^2 \text{csch}^{-1}(a x)-\frac{1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*a^2*x^4) - Sqrt[1 + 1/(a^2*x^2)]/(2*a*x^3) - 1/(2*x^2) - (a*Sqrt[1 + 1/(a^2*x^2)])/(4*x) + (a^2*ArcCsch[
a*x])/4

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 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 279

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 \frac{e^{2 \text{csch}^{-1}(a x)}}{x^3} \, dx &=\int \frac{\left (\sqrt{1+\frac{1}{a^2 x^2}}+\frac{1}{a x}\right )^2}{x^3} \, dx\\ &=\int \left (\frac{2}{a^2 x^5}+\frac{2 \sqrt{1+\frac{1}{a^2 x^2}}}{a x^4}+\frac{1}{x^3}\right ) \, dx\\ &=-\frac{1}{2 a^2 x^4}-\frac{1}{2 x^2}+\frac{2 \int \frac{\sqrt{1+\frac{1}{a^2 x^2}}}{x^4} \, dx}{a}\\ &=-\frac{1}{2 a^2 x^4}-\frac{1}{2 x^2}-\frac{2 \operatorname{Subst}\left (\int x^2 \sqrt{1+\frac{x^2}{a^2}} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{1}{2 a^2 x^4}-\frac{\sqrt{1+\frac{1}{a^2 x^2}}}{2 a x^3}-\frac{1}{2 x^2}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=-\frac{1}{2 a^2 x^4}-\frac{\sqrt{1+\frac{1}{a^2 x^2}}}{2 a x^3}-\frac{1}{2 x^2}-\frac{a \sqrt{1+\frac{1}{a^2 x^2}}}{4 x}+\frac{1}{4} a \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{2 a^2 x^4}-\frac{\sqrt{1+\frac{1}{a^2 x^2}}}{2 a x^3}-\frac{1}{2 x^2}-\frac{a \sqrt{1+\frac{1}{a^2 x^2}}}{4 x}+\frac{1}{4} a^2 \text{csch}^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.0486543, size = 73, normalized size = 1. \[ \left (-\frac{1}{2 a x^3}-\frac{a}{4 x}\right ) \sqrt{\frac{a^2 x^2+1}{a^2 x^2}}-\frac{1}{2 a^2 x^4}+\frac{1}{4} a^2 \sinh ^{-1}\left (\frac{1}{a x}\right )-\frac{1}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/(2*a^2*x^4) - 1/(2*x^2) + (-1/(2*a*x^3) - a/(4*x))*Sqrt[(1 + a^2*x^2)/(a^2*x^2)] + (a^2*ArcSinh[1/(a*x)])/4

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Maple [B]  time = 0.21, size = 176, normalized size = 2.4 \begin{align*} -{\frac{1}{2\,{a}^{2}{x}^{4}}}-{\frac{1}{2\,{x}^{2}}}+{\frac{a}{4\,{x}^{3}}\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}{x}^{2}}}} \left ( \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{-2}}{x}^{2}{a}^{2}-\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}\sqrt{{a}^{-2}}{x}^{4}{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}^{4}-2\, \left ({\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}} \right ) ^{3/2}\sqrt{{a}^{-2}} \right ){\frac{1}{\sqrt{{\frac{{a}^{2}{x}^{2}+1}{{a}^{2}}}}}}{\frac{1}{\sqrt{{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^3,x)

[Out]

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

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Maxima [B]  time = 1.01003, size = 188, normalized size = 2.58 \begin{align*} \frac{a^{3} \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} + 1\right ) - a^{3} \log \left (a x \sqrt{\frac{1}{a^{2} x^{2}} + 1} - 1\right ) - \frac{2 \,{\left (a^{6} x^{3}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{\frac{3}{2}} + a^{4} x \sqrt{\frac{1}{a^{2} x^{2}} + 1}\right )}}{a^{4} x^{4}{\left (\frac{1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{2} x^{2}{\left (\frac{1}{a^{2} x^{2}} + 1\right )} + 1}}{8 \, a} - \frac{1}{2 \, x^{2}} - \frac{1}{2 \, a^{2} x^{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))^2/x^3,x, algorithm="maxima")

[Out]

1/8*(a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) + 1) - a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) - 1) - 2*(a^6*x^3*(1/(a^2*x^2)
 + 1)^(3/2) + a^4*x*sqrt(1/(a^2*x^2) + 1))/(a^4*x^4*(1/(a^2*x^2) + 1)^2 - 2*a^2*x^2*(1/(a^2*x^2) + 1) + 1))/a
- 1/2/x^2 - 1/2/(a^2*x^4)

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Fricas [B]  time = 2.52876, size = 269, normalized size = 3.68 \begin{align*} \frac{a^{4} x^{4} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) - 2 \, a^{2} x^{2} -{\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt{\frac{a^{2} x^{2} + 1}{a^{2} x^{2}}} - 2}{4 \, a^{2} x^{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))^2/x^3,x, algorithm="fricas")

[Out]

1/4*(a^4*x^4*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x + 1) - a^4*x^4*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2))
- a*x - 1) - 2*a^2*x^2 - (a^3*x^3 + 2*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 2)/(a^2*x^4)

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Sympy [A]  time = 6.10566, size = 92, normalized size = 1.26 \begin{align*} \frac{a^{2} \operatorname{asinh}{\left (\frac{1}{a x} \right )}}{4} - \frac{a}{4 x \sqrt{1 + \frac{1}{a^{2} x^{2}}}} - \frac{1}{2 x^{2}} - \frac{3}{4 a x^{3} \sqrt{1 + \frac{1}{a^{2} x^{2}}}} - \frac{1}{2 a^{2} x^{4}} - \frac{1}{2 a^{3} x^{5} \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**3,x)

[Out]

a**2*asinh(1/(a*x))/4 - a/(4*x*sqrt(1 + 1/(a**2*x**2))) - 1/(2*x**2) - 3/(4*a*x**3*sqrt(1 + 1/(a**2*x**2))) -
1/(2*a**2*x**4) - 1/(2*a**3*x**5*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^3,x, algorithm="giac")

[Out]

Exception raised: TypeError