3.58 \(\int \frac {e^{2 \text {csch}^{-1}(a x)}}{x^5} \, dx\)

Optimal. Leaf size=96 \[ -\frac {1}{8} a^4 \text {csch}^{-1}(a x)-\frac {1}{3 a^2 x^6}-\frac {\sqrt {\frac {1}{a^2 x^2}+1}}{3 a x^5}-\frac {a \sqrt {\frac {1}{a^2 x^2}+1}}{12 x^3}+\frac {a^3 \sqrt {\frac {1}{a^2 x^2}+1}}{8 x}-\frac {1}{4 x^4} \]

[Out]

-1/3/a^2/x^6-1/4/x^4-1/8*a^4*arccsch(a*x)-1/3*(1+1/a^2/x^2)^(1/2)/a/x^5-1/12*a*(1+1/a^2/x^2)^(1/2)/x^3+1/8*a^3
*(1+1/a^2/x^2)^(1/2)/x

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {a^3 \sqrt {\frac {1}{a^2 x^2}+1}}{8 x}-\frac {a \sqrt {\frac {1}{a^2 x^2}+1}}{12 x^3}-\frac {\sqrt {\frac {1}{a^2 x^2}+1}}{3 a x^5}-\frac {1}{3 a^2 x^6}-\frac {1}{8} a^4 \text {csch}^{-1}(a x)-\frac {1}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCsch[a*x])/x^5,x]

[Out]

-1/(3*a^2*x^6) - Sqrt[1 + 1/(a^2*x^2)]/(3*a*x^5) - 1/(4*x^4) - (a*Sqrt[1 + 1/(a^2*x^2)])/(12*x^3) + (a^3*Sqrt[
1 + 1/(a^2*x^2)])/(8*x) - (a^4*ArcCsch[a*x])/8

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^5} \, dx &=\int \frac {\left (\sqrt {1+\frac {1}{a^2 x^2}}+\frac {1}{a x}\right )^2}{x^5} \, dx\\ &=\int \left (\frac {2}{a^2 x^7}+\frac {2 \sqrt {1+\frac {1}{a^2 x^2}}}{a x^6}+\frac {1}{x^5}\right ) \, dx\\ &=-\frac {1}{3 a^2 x^6}-\frac {1}{4 x^4}+\frac {2 \int \frac {\sqrt {1+\frac {1}{a^2 x^2}}}{x^6} \, dx}{a}\\ &=-\frac {1}{3 a^2 x^6}-\frac {1}{4 x^4}-\frac {2 \operatorname {Subst}\left (\int x^4 \sqrt {1+\frac {x^2}{a^2}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {1}{3 a^2 x^6}-\frac {\sqrt {1+\frac {1}{a^2 x^2}}}{3 a x^5}-\frac {1}{4 x^4}-\frac {\operatorname {Subst}\left (\int \frac {x^4}{\sqrt {1+\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{3 a}\\ &=-\frac {1}{3 a^2 x^6}-\frac {\sqrt {1+\frac {1}{a^2 x^2}}}{3 a x^5}-\frac {1}{4 x^4}-\frac {a \sqrt {1+\frac {1}{a^2 x^2}}}{12 x^3}+\frac {1}{4} a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {1}{3 a^2 x^6}-\frac {\sqrt {1+\frac {1}{a^2 x^2}}}{3 a x^5}-\frac {1}{4 x^4}-\frac {a \sqrt {1+\frac {1}{a^2 x^2}}}{12 x^3}+\frac {a^3 \sqrt {1+\frac {1}{a^2 x^2}}}{8 x}-\frac {1}{8} a^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {1}{3 a^2 x^6}-\frac {\sqrt {1+\frac {1}{a^2 x^2}}}{3 a x^5}-\frac {1}{4 x^4}-\frac {a \sqrt {1+\frac {1}{a^2 x^2}}}{12 x^3}+\frac {a^3 \sqrt {1+\frac {1}{a^2 x^2}}}{8 x}-\frac {1}{8} a^4 \text {csch}^{-1}(a x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 74, normalized size = 0.77 \[ \frac {\frac {\left (3 a^2 x^2+4\right ) \left (-2 a x \sqrt {\frac {1}{a^2 x^2}+1}+a^3 x^3 \sqrt {\frac {1}{a^2 x^2}+1}-2\right )}{x^6}-3 a^6 \sinh ^{-1}\left (\frac {1}{a x}\right )}{24 a^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(2*ArcCsch[a*x])/x^5,x]

[Out]

(((4 + 3*a^2*x^2)*(-2 - 2*a*Sqrt[1 + 1/(a^2*x^2)]*x + a^3*Sqrt[1 + 1/(a^2*x^2)]*x^3))/x^6 - 3*a^6*ArcSinh[1/(a
*x)])/(24*a^2)

________________________________________________________________________________________

fricas [A]  time = 2.31, size = 131, normalized size = 1.36 \[ -\frac {3 \, a^{6} x^{6} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - 3 \, a^{6} x^{6} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) + 6 \, a^{2} x^{2} - {\left (3 \, a^{5} x^{5} - 2 \, a^{3} x^{3} - 8 \, a x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} + 8}{24 \, a^{2} x^{6}} \]

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^5,x, algorithm="fricas")

[Out]

-1/24*(3*a^6*x^6*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - a*x + 1) - 3*a^6*x^6*log(a*x*sqrt((a^2*x^2 + 1)/(a^2*
x^2)) - a*x - 1) + 6*a^2*x^2 - (3*a^5*x^5 - 2*a^3*x^3 - 8*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) + 8)/(a^2*x^6)

________________________________________________________________________________________

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^5,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

________________________________________________________________________________________

maple [B]  time = 0.05, size = 209, normalized size = 2.18 \[ -\frac {1}{3 a^{2} x^{6}}-\frac {1}{4 x^{4}}-\frac {a \sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, \left (3 \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} \sqrt {\frac {1}{a^{2}}}\, x^{4} a^{4}-3 \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, \sqrt {\frac {1}{a^{2}}}\, x^{6} a^{4}+3 \ln \left (\frac {2 \sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+2}{a^{2} x}\right ) x^{6} a^{2}-6 \sqrt {\frac {1}{a^{2}}}\, \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} x^{2} a^{2}+8 \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} \sqrt {\frac {1}{a^{2}}}\right )}{24 x^{5} \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, \sqrt {\frac {1}{a^{2}}}} \]

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^5,x)

[Out]

-1/3/a^2/x^6-1/4/x^4-1/24*a*((a^2*x^2+1)/a^2/x^2)^(1/2)/x^5*(3*((a^2*x^2+1)/a^2)^(3/2)*(1/a^2)^(1/2)*x^4*a^4-3
*((a^2*x^2+1)/a^2)^(1/2)*(1/a^2)^(1/2)*x^6*a^4+3*ln(2*((1/a^2)^(1/2)*((a^2*x^2+1)/a^2)^(1/2)*a^2+1)/a^2/x)*x^6
*a^2-6*(1/a^2)^(1/2)*((a^2*x^2+1)/a^2)^(3/2)*x^2*a^2+8*((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)

________________________________________________________________________________________

maxima [B]  time = 0.31, size = 180, normalized size = 1.88 \[ -\frac {3 \, a^{5} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right ) - 3 \, a^{5} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right ) - \frac {2 \, {\left (3 \, a^{10} x^{5} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 8 \, a^{8} x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, a^{6} x \sqrt {\frac {1}{a^{2} x^{2}} + 1}\right )}}{a^{6} x^{6} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{3} - 3 \, a^{4} x^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{2} + 3 \, a^{2} x^{2} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} - 1}}{48 \, a} - \frac {1}{4 \, x^{4}} - \frac {1}{3 \, a^{2} x^{6}} \]

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^5,x, algorithm="maxima")

[Out]

-1/48*(3*a^5*log(a*x*sqrt(1/(a^2*x^2) + 1) + 1) - 3*a^5*log(a*x*sqrt(1/(a^2*x^2) + 1) - 1) - 2*(3*a^10*x^5*(1/
(a^2*x^2) + 1)^(5/2) - 8*a^8*x^3*(1/(a^2*x^2) + 1)^(3/2) - 3*a^6*x*sqrt(1/(a^2*x^2) + 1))/(a^6*x^6*(1/(a^2*x^2
) + 1)^3 - 3*a^4*x^4*(1/(a^2*x^2) + 1)^2 + 3*a^2*x^2*(1/(a^2*x^2) + 1) - 1))/a - 1/4/x^4 - 1/3/(a^2*x^6)

________________________________________________________________________________________

mupad [B]  time = 2.37, size = 89, normalized size = 0.93 \[ \frac {a^3\,\sqrt {\frac {1}{a^2\,x^2}+1}}{8\,x}-\frac {1}{3\,a^2\,x^6}-\frac {a\,\sqrt {\frac {1}{a^2\,x^2}+1}}{12\,x^3}-\frac {1}{4\,x^4}-\frac {\sqrt {\frac {1}{a^2\,x^2}+1}}{3\,a\,x^5}-\frac {a^3\,\mathrm {asinh}\left (\frac {\sqrt {\frac {1}{a^2}}}{x}\right )}{8\,\sqrt {\frac {1}{a^2}}} \]

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^5,x)

[Out]

(a^3*(1/(a^2*x^2) + 1)^(1/2))/(8*x) - 1/(3*a^2*x^6) - (a*(1/(a^2*x^2) + 1)^(1/2))/(12*x^3) - 1/(4*x^4) - (1/(a
^2*x^2) + 1)^(1/2)/(3*a*x^5) - (a^3*asinh((1/a^2)^(1/2)/x))/(8*(1/a^2)^(1/2))

________________________________________________________________________________________

sympy [A]  time = 6.33, size = 114, normalized size = 1.19 \[ - \frac {a^{4} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{8} + \frac {a^{3}}{8 x \sqrt {1 + \frac {1}{a^{2} x^{2}}}} + \frac {a}{24 x^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 x^{4}} - \frac {5}{12 a x^{5} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{3 a^{2} x^{6}} - \frac {1}{3 a^{3} x^{7} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} \]

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**5,x)

[Out]

-a**4*asinh(1/(a*x))/8 + a**3/(8*x*sqrt(1 + 1/(a**2*x**2))) + a/(24*x**3*sqrt(1 + 1/(a**2*x**2))) - 1/(4*x**4)
- 5/(12*a*x**5*sqrt(1 + 1/(a**2*x**2))) - 1/(3*a**2*x**6) - 1/(3*a**3*x**7*sqrt(1 + 1/(a**2*x**2)))

________________________________________________________________________________________