3.57 \(\int \frac {e^{2 \text {csch}^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=58 \[ -\frac {2}{5 a^2 x^5}-\frac {2}{5} a^3 \left (\frac {1}{a^2 x^2}+1\right )^{5/2}+\frac {2}{3} a^3 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}-\frac {1}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6338, 6742, 266, 43} \[ -\frac {2}{5} a^3 \left (\frac {1}{a^2 x^2}+1\right )^{5/2}+\frac {2}{3} a^3 \left (\frac {1}{a^2 x^2}+1\right )^{3/2}-\frac {2}{5 a^2 x^5}-\frac {1}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcCsch[a*x])/x^4,x]

[Out]

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

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

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Mathematica [A]  time = 0.06, size = 54, normalized size = 0.93 \[ -\frac {5 a^2 x^2+2 a x \sqrt {\frac {1}{a^2 x^2}+1} \left (-2 a^4 x^4+a^2 x^2+3\right )+6}{15 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcCsch[a*x])/x^4,x]

[Out]

-1/15*(6 + 5*a^2*x^2 + 2*a*Sqrt[1 + 1/(a^2*x^2)]*x*(3 + a^2*x^2 - 2*a^4*x^4))/(a^2*x^5)

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fricas [A]  time = 0.73, size = 67, normalized size = 1.16 \[ \frac {4 \, a^{5} x^{5} - 5 \, a^{2} x^{2} + 2 \, {\left (2 \, a^{5} x^{5} - a^{3} x^{3} - 3 \, a x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - 6}{15 \, a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x, algorithm="fricas")

[Out]

1/15*(4*a^5*x^5 - 5*a^2*x^2 + 2*(2*a^5*x^5 - a^3*x^3 - 3*a*x)*sqrt((a^2*x^2 + 1)/(a^2*x^2)) - 6)/(a^2*x^5)

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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^4,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 [A]  time = 0.05, size = 73, normalized size = 1.26 \[ \frac {-\frac {a^{2}}{3 x^{3}}-\frac {1}{5 x^{5}}}{a^{2}}+\frac {2 \sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, \left (a^{2} x^{2}+1\right ) \left (2 a^{2} x^{2}-3\right )}{15 a \,x^{4}}-\frac {1}{5 x^{5} a^{2}} \]

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^4,x)

[Out]

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

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maxima [A]  time = 0.30, size = 52, normalized size = 0.90 \[ -\frac {2 \, {\left (3 \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {5}{2}} - 5 \, a^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}}\right )}}{15 \, a} - \frac {1}{3 \, x^{3}} - \frac {2}{5 \, a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a^2/x^2)^(1/2))^2/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.26, size = 67, normalized size = 1.16 \[ \frac {4\,a^3\,\sqrt {\frac {1}{a^2\,x^2}+1}}{15}-\frac {\frac {2\,a\,x\,\sqrt {\frac {1}{a^2\,x^2}+1}}{15}+\frac {1}{3}}{x^3}-\frac {\frac {2}{5\,a^2}+\frac {2\,x\,\sqrt {\frac {1}{a^2\,x^2}+1}}{5\,a}}{x^5} \]

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^4,x)

[Out]

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

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sympy [A]  time = 3.07, size = 76, normalized size = 1.31 \[ \frac {4 a^{2} \sqrt {a^{2} x^{2} + 1}}{15 x} - \frac {2 \sqrt {a^{2} x^{2} + 1}}{15 x^{3}} - \frac {1}{3 x^{3}} - \frac {2 \sqrt {a^{2} x^{2} + 1}}{5 a^{2} x^{5}} - \frac {2}{5 a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/a/x+(1+1/a**2/x**2)**(1/2))**2/x**4,x)

[Out]

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

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