3.1.10 \(\int \frac {\sinh ^{-1}(a x)}{x^5} \, dx\) [10]

Optimal. Leaf size=56 \[ -\frac {a \sqrt {1+a^2 x^2}}{12 x^3}+\frac {a^3 \sqrt {1+a^2 x^2}}{6 x}-\frac {\sinh ^{-1}(a x)}{4 x^4} \]

[Out]

-1/4*arcsinh(a*x)/x^4-1/12*a*(a^2*x^2+1)^(1/2)/x^3+1/6*a^3*(a^2*x^2+1)^(1/2)/x

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Rubi [A]
time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5776, 277, 270} \begin {gather*} -\frac {a \sqrt {a^2 x^2+1}}{12 x^3}+\frac {a^3 \sqrt {a^2 x^2+1}}{6 x}-\frac {\sinh ^{-1}(a x)}{4 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]/x^5,x]

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*
c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 277

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^n)^(p + 1)/(a*(m + 1))), x]
- Dist[b*((m + n*(p + 1) + 1)/(a*(m + 1))), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\sinh ^{-1}(a x)}{x^5} \, dx &=-\frac {\sinh ^{-1}(a x)}{4 x^4}+\frac {1}{4} a \int \frac {1}{x^4 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1+a^2 x^2}}{12 x^3}-\frac {\sinh ^{-1}(a x)}{4 x^4}-\frac {1}{6} a^3 \int \frac {1}{x^2 \sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {a \sqrt {1+a^2 x^2}}{12 x^3}+\frac {a^3 \sqrt {1+a^2 x^2}}{6 x}-\frac {\sinh ^{-1}(a x)}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 40, normalized size = 0.71 \begin {gather*} \frac {a x \sqrt {1+a^2 x^2} \left (-1+2 a^2 x^2\right )-3 \sinh ^{-1}(a x)}{12 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]/x^5,x]

[Out]

(a*x*Sqrt[1 + a^2*x^2]*(-1 + 2*a^2*x^2) - 3*ArcSinh[a*x])/(12*x^4)

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Maple [A]
time = 0.19, size = 56, normalized size = 1.00

method	result	size
derivativedivides	\(a^{4} \left (-\frac {\arcsinh \left (a x \right )}{4 a^{4} x^{4}}-\frac {\sqrt {a^{2} x^{2}+1}}{12 a^{3} x^{3}}+\frac {\sqrt {a^{2} x^{2}+1}}{6 a x}\right )\)	\(56\)
default	\(a^{4} \left (-\frac {\arcsinh \left (a x \right )}{4 a^{4} x^{4}}-\frac {\sqrt {a^{2} x^{2}+1}}{12 a^{3} x^{3}}+\frac {\sqrt {a^{2} x^{2}+1}}{6 a x}\right )\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)/x^5,x,method=_RETURNVERBOSE)

[Out]

a^4*(-1/4/a^4/x^4*arcsinh(a*x)-1/12/a^3/x^3*(a^2*x^2+1)^(1/2)+1/6/a/x*(a^2*x^2+1)^(1/2))

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Maxima [A]
time = 0.26, size = 49, normalized size = 0.88 \begin {gather*} \frac {1}{12} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} a^{2}}{x} - \frac {\sqrt {a^{2} x^{2} + 1}}{x^{3}}\right )} a - \frac {\operatorname {arsinh}\left (a x\right )}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^5,x, algorithm="maxima")

[Out]

1/12*(2*sqrt(a^2*x^2 + 1)*a^2/x - sqrt(a^2*x^2 + 1)/x^3)*a - 1/4*arcsinh(a*x)/x^4

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Fricas [A]
time = 0.38, size = 49, normalized size = 0.88 \begin {gather*} \frac {{\left (2 \, a^{3} x^{3} - a x\right )} \sqrt {a^{2} x^{2} + 1} - 3 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{12 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^5,x, algorithm="fricas")

[Out]

1/12*((2*a^3*x^3 - a*x)*sqrt(a^2*x^2 + 1) - 3*log(a*x + sqrt(a^2*x^2 + 1)))/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}{\left (a x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)/x**5,x)

[Out]

Integral(asinh(a*x)/x**5, x)

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Giac [A]
time = 0.43, size = 77, normalized size = 1.38 \begin {gather*} \frac {{\left (3 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )} a^{3} {\left | a \right |}}{3 \, {\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} - \frac {\log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^5,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {asinh}\left (a\,x\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)/x^5,x)

[Out]

int(asinh(a*x)/x^5, x)

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