3.69 \(\int \frac{1}{\cosh ^{-1}(a x)^4} \, dx\)
Optimal. Leaf size=86 \[ \frac{\text{Chi}\left (\cosh ^{-1}(a x)\right )}{6 a}-\frac{x}{6 \cosh ^{-1}(a x)^2}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{6 a \cosh ^{-1}(a x)}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{3 a \cosh ^{-1}(a x)^3} \]
[Out]
-(Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(3*a*ArcCosh[a*x]^3) - x/(6*ArcCosh[a*x]^2) - (Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(
6*a*ArcCosh[a*x]) + CoshIntegral[ArcCosh[a*x]]/(6*a)
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Rubi [A] time = 0.377686, antiderivative size = 86, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used =
{5656, 5775, 5781, 3301} \[ \frac{\text{Chi}\left (\cosh ^{-1}(a x)\right )}{6 a}-\frac{x}{6 \cosh ^{-1}(a x)^2}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{6 a \cosh ^{-1}(a x)}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{3 a \cosh ^{-1}(a x)^3} \]
Antiderivative was successfully verified.
[In]
Int[ArcCosh[a*x]^(-4),x]
[Out]
-(Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(3*a*ArcCosh[a*x]^3) - x/(6*ArcCosh[a*x]^2) - (Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(
6*a*ArcCosh[a*x]) + CoshIntegral[ArcCosh[a*x]]/(6*a)
Rule 5656
Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(a + b*ArcCosh[c
*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCosh[c*x])^(n + 1))/(Sqrt[-1 + c*x]*Sqr
t[1 + c*x]), x], x] /; FreeQ[{a, b, c}, x] && LtQ[n, -1]
Rule 5775
Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[((f*x)^m*(a + b*ArcCosh[c*x])^(n + 1))/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] - Dist[(f
*m)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), Int[(f*x)^(m - 1)*(a + b*ArcCosh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d1
, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && LtQ[n, -1] && GtQ[d1, 0] && LtQ[d2, 0]
Rule 5781
Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x_))^(p_
.), x_Symbol] :> Dist[(-(d1*d2))^p/c^(m + 1), Subst[Int[(a + b*x)^n*Cosh[x]^m*Sinh[x]^(2*p + 1), x], x, ArcCos
h[c*x]], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && IntegerQ[p
+ 1/2] && GtQ[p, -1] && IGtQ[m, 0] && (GtQ[d1, 0] && LtQ[d2, 0])
Rule 3301
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]
Rubi steps
\begin{align*} \int \frac{1}{\cosh ^{-1}(a x)^4} \, dx &=-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{3 a \cosh ^{-1}(a x)^3}+\frac{1}{3} a \int \frac{x}{\sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3} \, dx\\ &=-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{3 a \cosh ^{-1}(a x)^3}-\frac{x}{6 \cosh ^{-1}(a x)^2}+\frac{1}{6} \int \frac{1}{\cosh ^{-1}(a x)^2} \, dx\\ &=-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{3 a \cosh ^{-1}(a x)^3}-\frac{x}{6 \cosh ^{-1}(a x)^2}-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{6 a \cosh ^{-1}(a x)}+\frac{1}{6} a \int \frac{x}{\sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)} \, dx\\ &=-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{3 a \cosh ^{-1}(a x)^3}-\frac{x}{6 \cosh ^{-1}(a x)^2}-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{6 a \cosh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\cosh (x)}{x} \, dx,x,\cosh ^{-1}(a x)\right )}{6 a}\\ &=-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{3 a \cosh ^{-1}(a x)^3}-\frac{x}{6 \cosh ^{-1}(a x)^2}-\frac{\sqrt{-1+a x} \sqrt{1+a x}}{6 a \cosh ^{-1}(a x)}+\frac{\text{Chi}\left (\cosh ^{-1}(a x)\right )}{6 a}\\ \end{align*}
Mathematica [A] time = 0.179326, size = 116, normalized size = 1.35 \[ \frac{\sqrt{\frac{a x-1}{a x+1}} \sqrt{a x+1} \cosh ^{-1}(a x)^3 \text{Chi}\left (\cosh ^{-1}(a x)\right )-2 (a x-1) \sqrt{a x+1}-(a x-1) \sqrt{a x+1} \cosh ^{-1}(a x)^2-a x \sqrt{a x-1} \cosh ^{-1}(a x)}{6 a \sqrt{a x-1} \cosh ^{-1}(a x)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[ArcCosh[a*x]^(-4),x]
[Out]
(-2*(-1 + a*x)*Sqrt[1 + a*x] - a*x*Sqrt[-1 + a*x]*ArcCosh[a*x] - (-1 + a*x)*Sqrt[1 + a*x]*ArcCosh[a*x]^2 + Sqr
t[(-1 + a*x)/(1 + a*x)]*Sqrt[1 + a*x]*ArcCosh[a*x]^3*CoshIntegral[ArcCosh[a*x]])/(6*a*Sqrt[-1 + a*x]*ArcCosh[a
*x]^3)
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Maple [A] time = 0.027, size = 67, normalized size = 0.8 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{3\, \left ({\rm arccosh} \left (ax\right ) \right ) ^{3}}\sqrt{ax-1}\sqrt{ax+1}}-{\frac{ax}{6\, \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}}}-{\frac{1}{6\,{\rm arccosh} \left (ax\right )}\sqrt{ax-1}\sqrt{ax+1}}+{\frac{{\it Chi} \left ({\rm arccosh} \left (ax\right ) \right ) }{6}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/arccosh(a*x)^4,x)
[Out]
1/a*(-1/3/arccosh(a*x)^3*(a*x-1)^(1/2)*(a*x+1)^(1/2)-1/6*a*x/arccosh(a*x)^2-1/6/arccosh(a*x)*(a*x-1)^(1/2)*(a*
x+1)^(1/2)+1/6*Chi(arccosh(a*x)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arccosh(a*x)^4,x, algorithm="maxima")
[Out]
-1/6*(2*a^11*x^11 - 10*a^9*x^9 + 20*a^7*x^7 - 20*a^5*x^5 + 2*(a^6*x^6 - a^4*x^4)*(a*x + 1)^(5/2)*(a*x - 1)^(5/
2) + 10*a^3*x^3 + 2*(5*a^7*x^7 - 9*a^5*x^5 + 4*a^3*x^3)*(a*x + 1)^2*(a*x - 1)^2 + 4*(5*a^8*x^8 - 13*a^6*x^6 +
11*a^4*x^4 - 3*a^2*x^2)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + 4*(5*a^9*x^9 - 17*a^7*x^7 + 21*a^5*x^5 - 11*a^3*x^3
+ 2*a*x)*(a*x + 1)*(a*x - 1) + (a^11*x^11 - 5*a^9*x^9 + 10*a^7*x^7 - 10*a^5*x^5 + (a^6*x^6 - a^4*x^4 + 3*a^2*x
^2 - 3)*(a*x + 1)^(5/2)*(a*x - 1)^(5/2) + 5*a^3*x^3 + (5*a^7*x^7 - 9*a^5*x^5 + 10*a^3*x^3 - 6*a*x)*(a*x + 1)^2
*(a*x - 1)^2 + (10*a^8*x^8 - 26*a^6*x^6 + 22*a^4*x^4 - 3*a^2*x^2 - 3)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + 2*(5*a
^9*x^9 - 17*a^7*x^7 + 18*a^5*x^5 - 5*a^3*x^3 - a*x)*(a*x + 1)*(a*x - 1) + (5*a^10*x^10 - 21*a^8*x^8 + 31*a^6*x
^6 - 20*a^4*x^4 + 6*a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(a*x - 1) - a*x)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))^2 +
2*(5*a^10*x^10 - 21*a^8*x^8 + 34*a^6*x^6 - 26*a^4*x^4 + 9*a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(a*x - 1) - 2*a*x +
(a^11*x^11 - 5*a^9*x^9 + 10*a^7*x^7 - 10*a^5*x^5 + (a^6*x^6 - a^2*x^2)*(a*x + 1)^(5/2)*(a*x - 1)^(5/2) + 5*a^3
*x^3 + (5*a^7*x^7 - 5*a^5*x^5 - 2*a^3*x^3 + 2*a*x)*(a*x + 1)^2*(a*x - 1)^2 + (10*a^8*x^8 - 20*a^6*x^6 + 10*a^4
*x^4 + a^2*x^2 - 1)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + 2*(5*a^9*x^9 - 15*a^7*x^7 + 16*a^5*x^5 - 7*a^3*x^3 + a*x
)*(a*x + 1)*(a*x - 1) + (5*a^10*x^10 - 20*a^8*x^8 + 31*a^6*x^6 - 23*a^4*x^4 + 8*a^2*x^2 - 1)*sqrt(a*x + 1)*sqr
t(a*x - 1) - a*x)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1)))/((a^11*x^10 - 5*a^9*x^8 + (a*x + 1)^(5/2)*(a*x - 1)^
(5/2)*a^6*x^5 + 10*a^7*x^6 - 10*a^5*x^4 + 5*(a^7*x^6 - a^5*x^4)*(a*x + 1)^2*(a*x - 1)^2 + 5*a^3*x^2 + 10*(a^8*
x^7 - 2*a^6*x^5 + a^4*x^3)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + 10*(a^9*x^8 - 3*a^7*x^6 + 3*a^5*x^4 - a^3*x^2)*(a
*x + 1)*(a*x - 1) + 5*(a^10*x^9 - 4*a^8*x^7 + 6*a^6*x^5 - 4*a^4*x^3 + a^2*x)*sqrt(a*x + 1)*sqrt(a*x - 1) - a)*
log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))^3) + integrate(1/6*(a^12*x^12 - 6*a^10*x^10 + 15*a^8*x^8 - 20*a^6*x^6 +
15*a^4*x^4 + (a^6*x^6 + a^4*x^4 - 9*a^2*x^2 + 15)*(a*x + 1)^3*(a*x - 1)^3 + (6*a^7*x^7 - a^5*x^5 - 31*a^3*x^3
+ 33*a*x)*(a*x + 1)^(5/2)*(a*x - 1)^(5/2) + (15*a^8*x^8 - 20*a^6*x^6 - 19*a^4*x^4 + 3*a^2*x^2 + 21)*(a*x + 1)
^2*(a*x - 1)^2 + (20*a^9*x^9 - 50*a^7*x^7 + 54*a^5*x^5 - 59*a^3*x^3 + 35*a*x)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2)
- 6*a^2*x^2 + (15*a^10*x^10 - 55*a^8*x^8 + 101*a^6*x^6 - 90*a^4*x^4 + 22*a^2*x^2 + 7)*(a*x + 1)*(a*x - 1) + (6
*a^11*x^11 - 29*a^9*x^9 + 65*a^7*x^7 - 66*a^5*x^5 + 23*a^3*x^3 + a*x)*sqrt(a*x + 1)*sqrt(a*x - 1) + 1)/((a^12*
x^12 - 6*a^10*x^10 + (a*x + 1)^3*(a*x - 1)^3*a^6*x^6 + 15*a^8*x^8 - 20*a^6*x^6 + 15*a^4*x^4 + 6*(a^7*x^7 - a^5
*x^5)*(a*x + 1)^(5/2)*(a*x - 1)^(5/2) + 15*(a^8*x^8 - 2*a^6*x^6 + a^4*x^4)*(a*x + 1)^2*(a*x - 1)^2 + 20*(a^9*x
^9 - 3*a^7*x^7 + 3*a^5*x^5 - a^3*x^3)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) - 6*a^2*x^2 + 15*(a^10*x^10 - 4*a^8*x^8
+ 6*a^6*x^6 - 4*a^4*x^4 + a^2*x^2)*(a*x + 1)*(a*x - 1) + 6*(a^11*x^11 - 5*a^9*x^9 + 10*a^7*x^7 - 10*a^5*x^5 +
5*a^3*x^3 - a*x)*sqrt(a*x + 1)*sqrt(a*x - 1) + 1)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\operatorname{arcosh}\left (a x\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arccosh(a*x)^4,x, algorithm="fricas")
[Out]
integral(arccosh(a*x)^(-4), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{acosh}^{4}{\left (a x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/acosh(a*x)**4,x)
[Out]
Integral(acosh(a*x)**(-4), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{arcosh}\left (a x\right )^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arccosh(a*x)^4,x, algorithm="giac")
[Out]
integrate(arccosh(a*x)^(-4), x)