3.71 \(\int \frac{1}{\sinh ^{-1}(a x)^4} \, dx\)
Optimal. Leaf size=76 \[ -\frac{\sqrt{a^2 x^2+1}}{6 a \sinh ^{-1}(a x)}-\frac{\sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}+\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{6 a}-\frac{x}{6 \sinh ^{-1}(a x)^2} \]
[Out]
-Sqrt[1 + a^2*x^2]/(3*a*ArcSinh[a*x]^3) - x/(6*ArcSinh[a*x]^2) - Sqrt[1 + a^2*x^2]/(6*a*ArcSinh[a*x]) + SinhIn
tegral[ArcSinh[a*x]]/(6*a)
________________________________________________________________________________________
Rubi [A] time = 0.149746, antiderivative size = 76, 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 =
{5655, 5774, 5779, 3298} \[ -\frac{\sqrt{a^2 x^2+1}}{6 a \sinh ^{-1}(a x)}-\frac{\sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}+\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{6 a}-\frac{x}{6 \sinh ^{-1}(a x)^2} \]
Antiderivative was successfully verified.
[In]
Int[ArcSinh[a*x]^(-4),x]
[Out]
-Sqrt[1 + a^2*x^2]/(3*a*ArcSinh[a*x]^3) - x/(6*ArcSinh[a*x]^2) - Sqrt[1 + a^2*x^2]/(6*a*ArcSinh[a*x]) + SinhIn
tegral[ArcSinh[a*x]]/(6*a)
Rule 5655
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^(n + 1
))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c^2*x^2], x], x] /; F
reeQ[{a, b, c}, x] && LtQ[n, -1]
Rule 5774
Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x
)^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && LtQ[n, -
1] && GtQ[d, 0]
Rule 5779
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c^
(m + 1), Subst[Int[(a + b*x)^n*Sinh[x]^m*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e,
n}, x] && EqQ[e, c^2*d] && IntegerQ[2*p] && GtQ[p, -1] && IGtQ[m, 0] && (IntegerQ[p] || GtQ[d, 0])
Rule 3298
Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]
Rubi steps
\begin{align*} \int \frac{1}{\sinh ^{-1}(a x)^4} \, dx &=-\frac{\sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}+\frac{1}{3} a \int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{6 \sinh ^{-1}(a x)^2}+\frac{1}{6} \int \frac{1}{\sinh ^{-1}(a x)^2} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{6 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{6 a \sinh ^{-1}(a x)}+\frac{1}{6} a \int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)} \, dx\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{6 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{6 a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{6 a}\\ &=-\frac{\sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{6 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{6 a \sinh ^{-1}(a x)}+\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{6 a}\\ \end{align*}
Mathematica [A] time = 0.0584567, size = 69, normalized size = 0.91 \[ -\frac{2 \sqrt{a^2 x^2+1}+\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2+\sinh ^{-1}(a x)^3 \left (-\text{Shi}\left (\sinh ^{-1}(a x)\right )\right )+a x \sinh ^{-1}(a x)}{6 a \sinh ^{-1}(a x)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[ArcSinh[a*x]^(-4),x]
[Out]
-(2*Sqrt[1 + a^2*x^2] + a*x*ArcSinh[a*x] + Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2 - ArcSinh[a*x]^3*SinhIntegral[ArcS
inh[a*x]])/(6*a*ArcSinh[a*x]^3)
________________________________________________________________________________________
Maple [A] time = 0.023, size = 61, normalized size = 0.8 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{ax}{6\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{1}{6\,{\it Arcsinh} \left ( ax \right ) }\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{6}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/arcsinh(a*x)^4,x)
[Out]
1/a*(-1/3/arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)-1/6*a*x/arcsinh(a*x)^2-1/6/arcsinh(a*x)*(a^2*x^2+1)^(1/2)+1/6*Shi(a
rcsinh(a*x)))
________________________________________________________________________________________
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/arcsinh(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 + 10*a^3*x^3 + 2*(a^6*x^6 + a^4*x^4)*(a^2*x^2 + 1)^(5
/2) + 2*(5*a^7*x^7 + 9*a^5*x^5 + 4*a^3*x^3)*(a^2*x^2 + 1)^2 + (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^6*x^6 + a^4*x^4 + 3*a^2*x^2 + 3)*(a^2*x^2 + 1)^(5/2) + (5*a^7*x^7 + 9*a^5*x^5 + 10*a^3*x^3 +
6*a*x)*(a^2*x^2 + 1)^2 + (10*a^8*x^8 + 26*a^6*x^6 + 22*a^4*x^4 + 3*a^2*x^2 - 3)*(a^2*x^2 + 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^2*x^2 + 1) + a*x + (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^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1))^2 + 4*(5*a^8*x^8 + 13*a^6*x^6
+ 11*a^4*x^4 + 3*a^2*x^2)*(a^2*x^2 + 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^2*x^2 + 1) + 2*a*x + (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^6*x^6 - a^2*x^2)*(a^2
*x^2 + 1)^(5/2) + (5*a^7*x^7 + 5*a^5*x^5 - 2*a^3*x^3 - 2*a*x)*(a^2*x^2 + 1)^2 + (10*a^8*x^8 + 20*a^6*x^6 + 10*
a^4*x^4 - a^2*x^2 - 1)*(a^2*x^2 + 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^2*x^
2 + 1) + a*x + (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^2*x^2 + 1))*log(a*x
+ sqrt(a^2*x^2 + 1)) + 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^2*x^2 +
1))/((a^11*x^10 + 5*a^9*x^8 + (a^2*x^2 + 1)^(5/2)*a^6*x^5 + 10*a^7*x^6 + 10*a^5*x^4 + 5*a^3*x^2 + 5*(a^7*x^6 +
a^5*x^4)*(a^2*x^2 + 1)^2 + 10*(a^8*x^7 + 2*a^6*x^5 + a^4*x^3)*(a^2*x^2 + 1)^(3/2) + 10*(a^9*x^8 + 3*a^7*x^6 +
3*a^5*x^4 + a^3*x^2)*(a^2*x^2 + 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^2*x^2 +
1) + a)*log(a*x + sqrt(a^2*x^2 + 1))^3) + integrate(1/6*(a^12*x^12 + 6*a^10*x^10 + 15*a^8*x^8 + 20*a^6*x^6 + 1
5*a^4*x^4 + (a^6*x^6 - a^4*x^4 - 9*a^2*x^2 - 15)*(a^2*x^2 + 1)^3 + 6*a^2*x^2 + (6*a^7*x^7 + a^5*x^5 - 31*a^3*x
^3 - 33*a*x)*(a^2*x^2 + 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^2*x^2 + 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^2*x^2 + 1)^(3/2) + (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^2*x^2 + 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^2*x^2 + 1) + 1)/((a^12*x^12 + 6*a^10*x^10 + 15*a^8*x^8 + (a^2*x^2 + 1)^3*a^6*x^6
+ 20*a^6*x^6 + 15*a^4*x^4 + 6*a^2*x^2 + 6*(a^7*x^7 + a^5*x^5)*(a^2*x^2 + 1)^(5/2) + 15*(a^8*x^8 + 2*a^6*x^6 +
a^4*x^4)*(a^2*x^2 + 1)^2 + 20*(a^9*x^9 + 3*a^7*x^7 + 3*a^5*x^5 + a^3*x^3)*(a^2*x^2 + 1)^(3/2) + 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^2*x^2 + 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^2*x^2 + 1) + 1)*log(a*x + sqrt(a^2*x^2 + 1))), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\operatorname{arsinh}\left (a x\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arcsinh(a*x)^4,x, algorithm="fricas")
[Out]
integral(arcsinh(a*x)^(-4), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{asinh}^{4}{\left (a x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/asinh(a*x)**4,x)
[Out]
Integral(asinh(a*x)**(-4), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{arsinh}\left (a x\right )^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/arcsinh(a*x)^4,x, algorithm="giac")
[Out]
integrate(arcsinh(a*x)^(-4), x)