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

Optimal. Leaf size=95 \[ -\frac {1}{3} a^4 \log (x)+\frac {a^3 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{3 x}+\frac {a^2}{12 x^2}-\frac {\cosh ^{-1}(a x)^2}{4 x^4}+\frac {a \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{6 x^3} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5662, 5748, 5724, 29, 30} \[ \frac {a^2}{12 x^2}-\frac {1}{3} a^4 \log (x)+\frac {a^3 \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{3 x}+\frac {a \sqrt {a x-1} \sqrt {a x+1} \cosh ^{-1}(a x)}{6 x^3}-\frac {\cosh ^{-1}(a x)^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2/(12*x^2) + (a*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x])/(6*x^3) + (a^3*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCo
sh[a*x])/(3*x) - ArcCosh[a*x]^2/(4*x^4) - (a^4*Log[x])/3

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]

Rule 5662

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

Rule 5724

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x
_))^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d
1*d2*f*(m + 1)), x] + Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(f*(m
+ 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/2)*(a + b*ArcCosh
[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2,
0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1] && IntegerQ[p + 1/2]

Rule 5748

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_)*((d2_) + (e2_.)*(x_
))^(p_), x_Symbol] :> Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d1*
d2*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m + 1)), Int[(f*x)^(m + 2)*(d1 + e1*x)^p*(d2 + e2*x)^p*(a
+ b*ArcCosh[c*x])^n, x], x] + Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p
])/(f*(m + 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/2)*(a + b
*ArcCosh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d1, e1, d2, e2, f, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 +
c*d2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegerQ[m] && IntegerQ[p + 1/2]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 69, normalized size = 0.73 \[ \frac {-4 a^4 x^4 \log (x)+a^2 x^2+2 a x \sqrt {a x-1} \sqrt {a x+1} \left (2 a^2 x^2+1\right ) \cosh ^{-1}(a x)-3 \cosh ^{-1}(a x)^2}{12 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 85, normalized size = 0.89 \[ -\frac {4 \, a^{4} x^{4} \log \relax (x) - a^{2} x^{2} - 2 \, {\left (2 \, a^{3} x^{3} + a x\right )} \sqrt {a^{2} x^{2} - 1} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) + 3 \, \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )^{2}}{12 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(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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giac [A]  time = 0.49, size = 147, normalized size = 1.55 \[ -\frac {1}{12} \, {\left (2 \, a^{3} \log \left (x^{2}\right ) - 4 \, a^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) - \frac {8 \, {\left (3 \, {\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )} a^{2} {\left | a \right |} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{{\left ({\left (x {\left | a \right |} - \sqrt {a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{3}} - \frac {2 \, a^{3} x^{2} + a}{x^{2}}\right )} a - \frac {\log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 109, normalized size = 1.15 \[ \frac {a^{4} \mathrm {arccosh}\left (a x \right )}{3}+\frac {a^{3} \mathrm {arccosh}\left (a x \right ) \sqrt {a x -1}\, \sqrt {a x +1}}{3 x}+\frac {a \,\mathrm {arccosh}\left (a x \right ) \sqrt {a x -1}\, \sqrt {a x +1}}{6 x^{3}}+\frac {a^{2}}{12 x^{2}}-\frac {\mathrm {arccosh}\left (a x \right )^{2}}{4 x^{4}}-\frac {a^{4} \ln \left (1+\left (a x +\sqrt {a x -1}\, \sqrt {a x +1}\right )^{2}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 72, normalized size = 0.76 \[ -\frac {1}{12} \, {\left (4 \, a^{2} \log \relax (x) - \frac {1}{x^{2}}\right )} a^{2} + \frac {1}{6} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} - 1} a^{2}}{x} + \frac {\sqrt {a^{2} x^{2} - 1}}{x^{3}}\right )} a \operatorname {arcosh}\left (a x\right ) - \frac {\operatorname {arcosh}\left (a x\right )^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(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*arccosh(a*x) - 1/
4*arccosh(a*x)^2/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {acosh}\left (a\,x\right )}^2}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(a*x)^2/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acosh}^{2}{\left (a x \right )}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(a*x)**2/x**5, x)

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