\(\int \frac {\text {arccosh}(a x)}{x^4} \, dx\) [9]

Optimal result

Integrand size = 8, antiderivative size = 65 \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\frac {a \sqrt {-1+a x} \sqrt {1+a x}}{6 x^2}-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{6} a^3 \arctan \left (\sqrt {-1+a x} \sqrt {1+a x}\right ) \]

[Out]

-1/3*arccosh(a*x)/x^3+1/6*a^3*arctan((a*x-1)^(1/2)*(a*x+1)^(1/2))+1/6*a*(a*x-1)^(1/2)*(a*x+1)^(1/2)/x^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {5883, 105, 12, 94, 211} \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\frac {1}{6} a^3 \arctan \left (\sqrt {a x-1} \sqrt {a x+1}\right )-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {a \sqrt {a x-1} \sqrt {a x+1}}{6 x^2} \]

[In]

Int[ArcCosh[a*x]/x^4,x]

[Out]

(a*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(6*x^2) - ArcCosh[a*x]/(3*x^3) + (a^3*ArcTan[Sqrt[-1 + a*x]*Sqrt[1 + a*x]])/6

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 5883

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)/(Sqrt
[1 + c*x]*Sqrt[-1 + c*x])), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{3} a \int \frac {1}{x^3 \sqrt {-1+a x} \sqrt {1+a x}} \, dx \\ & = \frac {a \sqrt {-1+a x} \sqrt {1+a x}}{6 x^2}-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{6} a \int \frac {a^2}{x \sqrt {-1+a x} \sqrt {1+a x}} \, dx \\ & = \frac {a \sqrt {-1+a x} \sqrt {1+a x}}{6 x^2}-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{6} a^3 \int \frac {1}{x \sqrt {-1+a x} \sqrt {1+a x}} \, dx \\ & = \frac {a \sqrt {-1+a x} \sqrt {1+a x}}{6 x^2}-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{6} a^4 \text {Subst}\left (\int \frac {1}{a+a x^2} \, dx,x,\sqrt {-1+a x} \sqrt {1+a x}\right ) \\ & = \frac {a \sqrt {-1+a x} \sqrt {1+a x}}{6 x^2}-\frac {\text {arccosh}(a x)}{3 x^3}+\frac {1}{6} a^3 \arctan \left (\sqrt {-1+a x} \sqrt {1+a x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20 \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\frac {-2 \text {arccosh}(a x)+\frac {a x \left (-1+a^2 x^2+a^2 x^2 \sqrt {-1+a^2 x^2} \arctan \left (\sqrt {-1+a^2 x^2}\right )\right )}{\sqrt {-1+a x} \sqrt {1+a x}}}{6 x^3} \]

[In]

Integrate[ArcCosh[a*x]/x^4,x]

[Out]

(-2*ArcCosh[a*x] + (a*x*(-1 + a^2*x^2 + a^2*x^2*Sqrt[-1 + a^2*x^2]*ArcTan[Sqrt[-1 + a^2*x^2]]))/(Sqrt[-1 + a*x
]*Sqrt[1 + a*x]))/(6*x^3)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.15

method	result	size
parts	\(-\frac {\operatorname {arccosh}\left (a x \right )}{3 x^{3}}-\frac {a \sqrt {a x -1}\, \sqrt {a x +1}\, \left (\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) a^{2} x^{2}-\sqrt {a^{2} x^{2}-1}\right )}{6 \sqrt {a^{2} x^{2}-1}\, x^{2}}\)	\(75\)
derivativedivides	\(a^{3} \left (-\frac {\operatorname {arccosh}\left (a x \right )}{3 a^{3} x^{3}}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) a^{2} x^{2}-\sqrt {a^{2} x^{2}-1}\right )}{6 a^{2} x^{2} \sqrt {a^{2} x^{2}-1}}\right )\)	\(84\)
default	\(a^{3} \left (-\frac {\operatorname {arccosh}\left (a x \right )}{3 a^{3} x^{3}}-\frac {\sqrt {a x -1}\, \sqrt {a x +1}\, \left (\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) a^{2} x^{2}-\sqrt {a^{2} x^{2}-1}\right )}{6 a^{2} x^{2} \sqrt {a^{2} x^{2}-1}}\right )\)	\(84\)

[In]

int(arccosh(a*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*arccosh(a*x)/x^3-1/6*a*(a*x-1)^(1/2)*(a*x+1)^(1/2)*(arctan(1/(a^2*x^2-1)^(1/2))*a^2*x^2-(a^2*x^2-1)^(1/2)
)/(a^2*x^2-1)^(1/2)/x^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.38 \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\frac {2 \, a^{3} x^{3} \arctan \left (-a x + \sqrt {a^{2} x^{2} - 1}\right ) + 2 \, x^{3} \log \left (-a x + \sqrt {a^{2} x^{2} - 1}\right ) + \sqrt {a^{2} x^{2} - 1} a x + 2 \, {\left (x^{3} - 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{6 \, x^{3}} \]

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="fricas")

[Out]

1/6*(2*a^3*x^3*arctan(-a*x + sqrt(a^2*x^2 - 1)) + 2*x^3*log(-a*x + sqrt(a^2*x^2 - 1)) + sqrt(a^2*x^2 - 1)*a*x
+ 2*(x^3 - 1)*log(a*x + sqrt(a^2*x^2 - 1)))/x^3

Sympy [F]

\[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\int \frac {\operatorname {acosh}{\left (a x \right )}}{x^{4}}\, dx \]

[In]

integrate(acosh(a*x)/x**4,x)

[Out]

Integral(acosh(a*x)/x**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.66 \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=-\frac {1}{6} \, {\left (a^{2} \arcsin \left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\sqrt {a^{2} x^{2} - 1}}{x^{2}}\right )} a - \frac {\operatorname {arcosh}\left (a x\right )}{3 \, x^{3}} \]

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="maxima")

[Out]

-1/6*(a^2*arcsin(1/(a*abs(x))) - sqrt(a^2*x^2 - 1)/x^2)*a - 1/3*arccosh(a*x)/x^3

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\frac {a^{4} \arctan \left (\sqrt {a^{2} x^{2} - 1}\right ) + \frac {\sqrt {a^{2} x^{2} - 1} a^{2}}{x^{2}}}{6 \, a} - \frac {\log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{3 \, x^{3}} \]

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\text {arccosh}(a x)}{x^4} \, dx=\int \frac {\mathrm {acosh}\left (a\,x\right )}{x^4} \,d x \]

[In]

int(acosh(a*x)/x^4,x)

[Out]

int(acosh(a*x)/x^4, x)