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

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

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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]

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 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 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]]

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*}

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Mathematica [A]  time = 0.05, size = 73, normalized size = 1.00 \[ -\frac {1}{2 a^2 x^4}+\left (-\frac {1}{2 a x^3}-\frac {a}{4 x}\right ) \sqrt {\frac {a^2 x^2+1}{a^2 x^2}}+\frac {1}{4} a^2 \sinh ^{-1}\left (\frac {1}{a x}\right )-\frac {1}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*1/(a^2*x^4) - 1/(2*x^2) + (-1/2*1/(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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fricas [B]  time = 0.51, size = 121, normalized size = 1.66 \[ \frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) - 2 \, a^{2} x^{2} - {\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - 2}{4 \, a^{2} x^{4}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Error: Bad Argument Type

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maple [B]  time = 0.05, size = 176, normalized size = 2.41 \[ -\frac {1}{2 x^{2}}-\frac {1}{2 a^{2} x^{4}}+\frac {a \sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, \left (\sqrt {\frac {1}{a^{2}}}\, \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} x^{2} a^{2}-\sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, x^{4} a^{2}+\ln \left (\frac {2 \sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+2}{a^{2} x}\right ) x^{4}-2 \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} \sqrt {\frac {1}{a^{2}}}\right )}{4 x^{3} \sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 139, normalized size = 1.90 \[ \frac {a^{3} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right ) - a^{3} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (a^{6} x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + a^{4} x \sqrt {\frac {1}{a^{2} x^{2}} + 1}\right )}}{a^{4} x^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{2} x^{2} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} + 1}}{8 \, a} - \frac {1}{2 \, x^{2}} - \frac {1}{2 \, a^{2} x^{4}} \]

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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mupad [B]  time = 2.31, size = 68, normalized size = 0.93 \[ \frac {a\,\mathrm {asinh}\left (\frac {\sqrt {\frac {1}{a^2}}}{x}\right )}{4\,\sqrt {\frac {1}{a^2}}}-\frac {1}{2\,a^2\,x^4}-\frac {a\,\sqrt {\frac {1}{a^2\,x^2}+1}}{4\,x}-\frac {1}{2\,x^2}-\frac {\sqrt {\frac {1}{a^2\,x^2}+1}}{2\,a\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.61, size = 92, normalized size = 1.26 \[ \frac {a^{2} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{4} - \frac {a}{4 x \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{2 x^{2}} - \frac {3}{4 a x^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{2 a^{2} x^{4}} - \frac {1}{2 a^{3} x^{5} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} \]

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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