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

Optimal. Leaf size=51 \[ -\frac {1}{5} a^4 \left (\frac {1}{a^2 x^2}+1\right )^{5/2}+\frac {1}{3} a^4 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}-\frac {1}{5 a x^5} \]

[Out]

1/3*a^4*(1+1/a^2/x^2)^(3/2)-1/5*a^4*(1+1/a^2/x^2)^(5/2)-1/5/a/x^5

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Rubi [A]  time = 0.04, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6336, 30, 266, 43} \[ -\frac {1}{5} a^4 \left (\frac {1}{a^2 x^2}+1\right )^{5/2}+\frac {1}{3} a^4 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}-\frac {1}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 + 1/(a^2*x^2))^(3/2))/3 - (a^4*(1 + 1/(a^2*x^2))^(5/2))/5 - 1/(5*a*x^5)

Rule 30

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 46, normalized size = 0.90 \[ \frac {a x \sqrt {\frac {1}{a^2 x^2}+1} \left (2 a^4 x^4-a^2 x^2-3\right )-3}{15 a x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 + a*Sqrt[1 + 1/(a^2*x^2)]*x*(-3 - a^2*x^2 + 2*a^4*x^4))/(15*a*x^5)

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fricas [A]  time = 1.02, size = 58, normalized size = 1.14 \[ \frac {2 \, a^{5} x^{5} + {\left (2 \, a^{5} x^{5} - a^{3} x^{3} - 3 \, a x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - 3}{15 \, a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))/x^5,x, algorithm="fricas")

[Out]

1/15*(2*a^5*x^5 + (2*a^5*x^5 - a^3*x^3 - 3*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 3)/(a*x^5)

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giac [B]  time = 0.20, size = 124, normalized size = 2.43 \[ \frac {4 \, {\left (15 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{6} a^{4} \mathrm {sgn}\relax (x) + 5 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{4} a^{4} \mathrm {sgn}\relax (x) + 5 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} a^{4} \mathrm {sgn}\relax (x) - a^{4} \mathrm {sgn}\relax (x)\right )}}{15 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{5}} - \frac {1}{5 \, a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))/x^5,x, algorithm="giac")

[Out]

4/15*(15*(x*abs(a) - sqrt(a^2*x^2 + 1))^6*a^4*sgn(x) + 5*(x*abs(a) - sqrt(a^2*x^2 + 1))^4*a^4*sgn(x) + 5*(x*ab
s(a) - sqrt(a^2*x^2 + 1))^2*a^4*sgn(x) - a^4*sgn(x))/((x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)^5 - 1/5/(a*x^5)

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maple [A]  time = 0.05, size = 52, normalized size = 1.02 \[ \frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, \left (a^{2} x^{2}+1\right ) \left (2 a^{2} x^{2}-3\right )}{15 x^{4}}-\frac {1}{5 a \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/x^2/a^2)^(1/2))/x^5,x)

[Out]

1/15*((a^2*x^2+1)/a^2/x^2)^(1/2)/x^4*(a^2*x^2+1)*(2*a^2*x^2-3)-1/5/a/x^5

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maxima [A]  time = 0.31, size = 41, normalized size = 0.80 \[ -\frac {1}{5} \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {5}{2}} + \frac {1}{3} \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - \frac {1}{5 \, a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))/x^5,x, algorithm="maxima")

[Out]

-1/5*a^4*(1/(a^2*x^2) + 1)^(5/2) + 1/3*a^4*(1/(a^2*x^2) + 1)^(3/2) - 1/5/(a*x^5)

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mupad [B]  time = 2.20, size = 61, normalized size = 1.20 \[ \frac {2\,a^4\,\sqrt {\frac {1}{a^2\,x^2}+1}}{15}-\frac {\frac {x\,\sqrt {\frac {1}{a^2\,x^2}+1}}{5}+\frac {1}{5\,a}}{x^5}-\frac {a^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{15\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/(a^2*x^2) + 1)^(1/2) + 1/(a*x))/x^5,x)

[Out]

(2*a^4*(1/(a^2*x^2) + 1)^(1/2))/15 - ((x*(1/(a^2*x^2) + 1)^(1/2))/5 + 1/(5*a))/x^5 - (a^2*(1/(a^2*x^2) + 1)^(1
/2))/(15*x^2)

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sympy [A]  time = 2.70, size = 65, normalized size = 1.27 \[ \frac {2 a^{3} \sqrt {a^{2} x^{2} + 1}}{15 x} - \frac {a \sqrt {a^{2} x^{2} + 1}}{15 x^{3}} - \frac {\sqrt {a^{2} x^{2} + 1}}{5 a x^{5}} - \frac {1}{5 a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))/x**5,x)

[Out]

2*a**3*sqrt(a**2*x**2 + 1)/(15*x) - a*sqrt(a**2*x**2 + 1)/(15*x**3) - sqrt(a**2*x**2 + 1)/(5*a*x**5) - 1/(5*a*
x**5)

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