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

Optimal. Leaf size=65 \[ \frac {1}{8} a^3 \text {csch}^{-1}(a x)-\frac {a^2 \sqrt {\frac {1}{a^2 x^2}+1}}{8 x}-\frac {\sqrt {\frac {1}{a^2 x^2}+1}}{4 x^3}-\frac {1}{4 a x^4} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6336, 30, 335, 279, 321, 215} \[ -\frac {a^2 \sqrt {\frac {1}{a^2 x^2}+1}}{8 x}-\frac {\sqrt {\frac {1}{a^2 x^2}+1}}{4 x^3}+\frac {1}{8} a^3 \text {csch}^{-1}(a x)-\frac {1}{4 a x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 30

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

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

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Mathematica [A]  time = 0.05, size = 53, normalized size = 0.82 \[ \frac {a^4 x^4 \sinh ^{-1}\left (\frac {1}{a x}\right )-a x \sqrt {\frac {1}{a^2 x^2}+1} \left (a^2 x^2+2\right )-2}{8 a x^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2 - a*Sqrt[1 + 1/(a^2*x^2)]*x*(2 + a^2*x^2) + a^4*x^4*ArcSinh[1/(a*x)])/(8*a*x^4)

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fricas [B]  time = 0.68, size = 113, normalized size = 1.74 \[ \frac {a^{4} x^{4} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x + 1\right ) - a^{4} x^{4} \log \left (a x \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - a x - 1\right ) - {\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt {\frac {a^{2} x^{2} + 1}{a^{2} x^{2}}} - 2}{8 \, a x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 103, normalized size = 1.58 \[ \frac {a^{6} {\left | a \right |} \log \left (\sqrt {a^{2} x^{2} + 1} + 1\right ) \mathrm {sgn}\relax (x) - a^{6} {\left | a \right |} \log \left (\sqrt {a^{2} x^{2} + 1} - 1\right ) \mathrm {sgn}\relax (x) - \frac {2 \, {\left ({\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{6} {\left | a \right |} \mathrm {sgn}\relax (x) + \sqrt {a^{2} x^{2} + 1} a^{6} {\left | a \right |} \mathrm {sgn}\relax (x) + 2 \, a^{7}\right )}}{a^{4} x^{4}}}{16 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(a^6*abs(a)*log(sqrt(a^2*x^2 + 1) + 1)*sgn(x) - a^6*abs(a)*log(sqrt(a^2*x^2 + 1) - 1)*sgn(x) - 2*((a^2*x^
2 + 1)^(3/2)*a^6*abs(a)*sgn(x) + sqrt(a^2*x^2 + 1)*a^6*abs(a)*sgn(x) + 2*a^7)/(a^4*x^4))/a^4

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maple [B]  time = 0.06, size = 173, normalized size = 2.66 \[ \frac {\sqrt {\frac {a^{2} x^{2}+1}{a^{2} x^{2}}}\, a^{2} \left (\sqrt {\frac {1}{a^{2}}}\, \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} x^{2} a^{2}-\sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, x^{4} a^{2}+\ln \left (\frac {2 \sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}\, a^{2}+2}{a^{2} x}\right ) x^{4}-2 \left (\frac {a^{2} x^{2}+1}{a^{2}}\right )^{\frac {3}{2}} \sqrt {\frac {1}{a^{2}}}\right )}{8 x^{3} \sqrt {\frac {1}{a^{2}}}\, \sqrt {\frac {a^{2} x^{2}+1}{a^{2}}}}-\frac {1}{4 x^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/a/x+(1+1/x^2/a^2)^(1/2))/x^4,x)

[Out]

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

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maxima [B]  time = 0.32, size = 129, normalized size = 1.98 \[ \frac {1}{16} \, a^{3} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} + 1\right ) - \frac {1}{16} \, a^{3} \log \left (a x \sqrt {\frac {1}{a^{2} x^{2}} + 1} - 1\right ) - \frac {a^{6} x^{3} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{\frac {3}{2}} + a^{4} x \sqrt {\frac {1}{a^{2} x^{2}} + 1}}{8 \, {\left (a^{4} x^{4} {\left (\frac {1}{a^{2} x^{2}} + 1\right )}^{2} - 2 \, a^{2} x^{2} {\left (\frac {1}{a^{2} x^{2}} + 1\right )} + 1\right )}} - \frac {1}{4 \, a x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) + 1) - 1/16*a^3*log(a*x*sqrt(1/(a^2*x^2) + 1) - 1) - 1/8*(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) - 1/4/(a*x^4)

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mupad [B]  time = 2.54, size = 61, normalized size = 0.94 \[ \frac {\mathrm {asinh}\left (\frac {\sqrt {\frac {1}{a^2}}}{x}\right )}{8\,{\left (\frac {1}{a^2}\right )}^{3/2}}-\frac {\sqrt {\frac {1}{a^2\,x^2}+1}}{4\,x^3}-\frac {1}{4\,a\,x^4}-\frac {a^2\,\sqrt {\frac {1}{a^2\,x^2}+1}}{8\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/(a^2*x^2) + 1)^(1/2) + 1/(a*x))/x^4,x)

[Out]

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

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sympy [A]  time = 4.25, size = 83, normalized size = 1.28 \[ \frac {a^{3} \operatorname {asinh}{\left (\frac {1}{a x} \right )}}{8} - \frac {a^{2}}{8 x \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {3}{8 x^{3} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} - \frac {1}{4 a x^{4}} - \frac {1}{4 a^{2} x^{5} \sqrt {1 + \frac {1}{a^{2} x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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