3.21 \(\int \frac{\sinh ^{-1}(a x)^2}{x^5} \, dx\)

Optimal. Leaf size=85 \[ -\frac{a^2}{12 x^2}+\frac{a^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{3 x}-\frac{a \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{6 x^3}-\frac{1}{3} a^4 \log (x)-\frac{\sinh ^{-1}(a x)^2}{4 x^4} \]

[Out]

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

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Rubi [A]  time = 0.137922, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5747, 5723, 29, 30} \[ -\frac{a^2}{12 x^2}+\frac{a^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{3 x}-\frac{a \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{6 x^3}-\frac{1}{3} a^4 \log (x)-\frac{\sinh ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^2/x^5,x]

[Out]

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

Rule 5661

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]

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

Rule 5723

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] - Dist[(b*c*n*d^IntPart[p]*(d + e
*x^2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Arc
Sinh[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 30

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

Rubi steps

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

Mathematica [A]  time = 0.0592826, size = 64, normalized size = 0.75 \[ -\frac{a^2 x^2+4 a^4 x^4 \log (x)-2 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-1\right ) \sinh ^{-1}(a x)+3 \sinh ^{-1}(a x)^2}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a*x]^2/x^5,x]

[Out]

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

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Maple [A]  time = 0.106, size = 99, normalized size = 1.2 \begin{align*}{\frac{{a}^{4}{\it Arcsinh} \left ( ax \right ) }{3}}+{\frac{{a}^{3}{\it Arcsinh} \left ( ax \right ) }{3\,x}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{a{\it Arcsinh} \left ( ax \right ) }{6\,{x}^{3}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{{a}^{2}}{12\,{x}^{2}}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}{4\,{x}^{4}}}-{\frac{{a}^{4}}{3}\ln \left ( \left ( ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) ^{2}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^2/x^5,x)

[Out]

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

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Maxima [A]  time = 1.22541, size = 96, normalized size = 1.13 \begin{align*} -\frac{1}{12} \,{\left (4 \, a^{2} \log \left (x\right ) + \frac{1}{x^{2}}\right )} a^{2} + \frac{1}{6} \,{\left (\frac{2 \, \sqrt{a^{2} x^{2} + 1} a^{2}}{x} - \frac{\sqrt{a^{2} x^{2} + 1}}{x^{3}}\right )} a \operatorname{arsinh}\left (a x\right ) - \frac{\operatorname{arsinh}\left (a x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.16209, size = 194, normalized size = 2.28 \begin{align*} -\frac{4 \, a^{4} x^{4} \log \left (x\right ) + a^{2} x^{2} - 2 \,{\left (2 \, a^{3} x^{3} - a x\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) + 3 \, \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}^{2}{\left (a x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**2/x**5,x)

[Out]

Integral(asinh(a*x)**2/x**5, x)

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Giac [B]  time = 1.59414, size = 200, normalized size = 2.35 \begin{align*} -\frac{1}{12} \,{\left (2 \, a^{3} \log \left (x^{2}\right ) - 4 \, a^{3} \log \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} + 1}\right ) - \frac{8 \,{\left (3 \,{\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}\right )}^{2} - 1\right )} a^{2}{\left | a \right |} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )}{{\left ({\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} + 1}\right )}^{2} - 1\right )}^{3}} - \frac{2 \, a^{3} x^{2} - a}{x^{2}}\right )} a - \frac{\log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x^5,x, algorithm="giac")

[Out]

-1/12*(2*a^3*log(x^2) - 4*a^3*log(-x*abs(a) + sqrt(a^2*x^2 + 1)) - 8*(3*(x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)*
a^2*abs(a)*log(a*x + sqrt(a^2*x^2 + 1))/((x*abs(a) - sqrt(a^2*x^2 + 1))^2 - 1)^3 - (2*a^3*x^2 - a)/x^2)*a - 1/
4*log(a*x + sqrt(a^2*x^2 + 1))^2/x^4