3.69 \(\int \frac{x^2}{\sinh ^{-1}(a x)^4} \, dx\)
Optimal. Leaf size=138 \[ -\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}-\frac{3 x^2 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)}-\frac{x^2 \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}-\frac{\sqrt{a^2 x^2+1}}{3 a^3 \sinh ^{-1}(a x)}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2} \]
[Out]
-(x^2*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^3) - x/(3*a^2*ArcSinh[a*x]^2) - x^3/(2*ArcSinh[a*x]^2) - Sqrt[1 + a
^2*x^2]/(3*a^3*ArcSinh[a*x]) - (3*x^2*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]) - SinhIntegral[ArcSinh[a*x]]/(24*a
^3) + (9*SinhIntegral[3*ArcSinh[a*x]])/(8*a^3)
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Rubi [A] time = 0.305834, antiderivative size = 138, normalized size of antiderivative = 1.,
number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used =
{5667, 5774, 5665, 3298, 5655, 5779} \[ -\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}-\frac{3 x^2 \sqrt{a^2 x^2+1}}{2 a \sinh ^{-1}(a x)}-\frac{x^2 \sqrt{a^2 x^2+1}}{3 a \sinh ^{-1}(a x)^3}-\frac{\sqrt{a^2 x^2+1}}{3 a^3 \sinh ^{-1}(a x)}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2} \]
Antiderivative was successfully verified.
[In]
Int[x^2/ArcSinh[a*x]^4,x]
[Out]
-(x^2*Sqrt[1 + a^2*x^2])/(3*a*ArcSinh[a*x]^3) - x/(3*a^2*ArcSinh[a*x]^2) - x^3/(2*ArcSinh[a*x]^2) - Sqrt[1 + a
^2*x^2]/(3*a^3*ArcSinh[a*x]) - (3*x^2*Sqrt[1 + a^2*x^2])/(2*a*ArcSinh[a*x]) - SinhIntegral[ArcSinh[a*x]]/(24*a
^3) + (9*SinhIntegral[3*ArcSinh[a*x]])/(8*a^3)
Rule 5667
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSinh[c*x])^(n +
1))/Sqrt[1 + c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSinh[c*x])^(n + 1))/Sqrt[1 + c
^2*x^2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]
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 5665
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 + c^2*x^2]*(a + b*ArcSi
nh[c*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]
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]
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 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])
Rubi steps
\begin{align*} \int \frac{x^2}{\sinh ^{-1}(a x)^4} \, dx &=-\frac{x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}+\frac{2 \int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3} \, dx}{3 a}+a \int \frac{x^3}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^3} \, dx\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2}+\frac{3}{2} \int \frac{x^2}{\sinh ^{-1}(a x)^2} \, dx+\frac{\int \frac{1}{\sinh ^{-1}(a x)^2} \, dx}{3 a^2}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{3 a^3 \sinh ^{-1}(a x)}-\frac{3 x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)}+\frac{3 \operatorname{Subst}\left (\int \left (-\frac{\sinh (x)}{4 x}+\frac{3 \sinh (3 x)}{4 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^3}+\frac{\int \frac{x}{\sqrt{1+a^2 x^2} \sinh ^{-1}(a x)} \, dx}{3 a}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{3 a^3 \sinh ^{-1}(a x)}-\frac{3 x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)}+\frac{\operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^3}-\frac{3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^3}+\frac{9 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^3}\\ &=-\frac{x^2 \sqrt{1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac{x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac{x^3}{2 \sinh ^{-1}(a x)^2}-\frac{\sqrt{1+a^2 x^2}}{3 a^3 \sinh ^{-1}(a x)}-\frac{3 x^2 \sqrt{1+a^2 x^2}}{2 a \sinh ^{-1}(a x)}-\frac{\text{Shi}\left (\sinh ^{-1}(a x)\right )}{24 a^3}+\frac{9 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3}\\ \end{align*}
Mathematica [A] time = 0.290157, size = 99, normalized size = 0.72 \[ -\frac{\frac{4 \left (2 a^2 x^2 \sqrt{a^2 x^2+1}+a x \left (3 a^2 x^2+2\right ) \sinh ^{-1}(a x)+\sqrt{a^2 x^2+1} \left (9 a^2 x^2+2\right ) \sinh ^{-1}(a x)^2\right )}{\sinh ^{-1}(a x)^3}+\text{Shi}\left (\sinh ^{-1}(a x)\right )-27 \text{Shi}\left (3 \sinh ^{-1}(a x)\right )}{24 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[x^2/ArcSinh[a*x]^4,x]
[Out]
-((4*(2*a^2*x^2*Sqrt[1 + a^2*x^2] + a*x*(2 + 3*a^2*x^2)*ArcSinh[a*x] + Sqrt[1 + a^2*x^2]*(2 + 9*a^2*x^2)*ArcSi
nh[a*x]^2))/ArcSinh[a*x]^3 + SinhIntegral[ArcSinh[a*x]] - 27*SinhIntegral[3*ArcSinh[a*x]])/(24*a^3)
________________________________________________________________________________________
Maple [A] time = 0.031, size = 115, normalized size = 0.8 \begin{align*}{\frac{1}{{a}^{3}} \left ({\frac{1}{12\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{ax}{24\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}+{\frac{1}{24\,{\it Arcsinh} \left ( ax \right ) }\sqrt{{a}^{2}{x}^{2}+1}}-{\frac{{\it Shi} \left ({\it Arcsinh} \left ( ax \right ) \right ) }{24}}-{\frac{\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{12\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}}-{\frac{\sinh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}}}-{\frac{3\,\cosh \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8\,{\it Arcsinh} \left ( ax \right ) }}+{\frac{9\,{\it Shi} \left ( 3\,{\it Arcsinh} \left ( ax \right ) \right ) }{8}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2/arcsinh(a*x)^4,x)
[Out]
1/a^3*(1/12/arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)+1/24*a*x/arcsinh(a*x)^2+1/24/arcsinh(a*x)*(a^2*x^2+1)^(1/2)-1/24*
Shi(arcsinh(a*x))-1/12/arcsinh(a*x)^3*cosh(3*arcsinh(a*x))-1/8/arcsinh(a*x)^2*sinh(3*arcsinh(a*x))-3/8/arcsinh
(a*x)*cosh(3*arcsinh(a*x))+9/8*Shi(3*arcsinh(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(x^2/arcsinh(a*x)^4,x, algorithm="maxima")
[Out]
-1/6*(2*a^13*x^13 + 10*a^11*x^11 + 20*a^9*x^9 + 20*a^7*x^7 + 10*a^5*x^5 + 2*a^3*x^3 + 2*(a^8*x^8 + a^6*x^6)*(a
^2*x^2 + 1)^(5/2) + 2*(5*a^9*x^9 + 9*a^7*x^7 + 4*a^5*x^5)*(a^2*x^2 + 1)^2 + (9*a^13*x^13 + 45*a^11*x^11 + 90*a
^9*x^9 + 90*a^7*x^7 + 45*a^5*x^5 + 9*a^3*x^3 + (9*a^8*x^8 + 13*a^6*x^6 + 3*a^4*x^4 - a^2*x^2)*(a^2*x^2 + 1)^(5
/2) + (45*a^9*x^9 + 97*a^7*x^7 + 64*a^5*x^5 + 10*a^3*x^3 - 2*a*x)*(a^2*x^2 + 1)^2 + (90*a^10*x^10 + 258*a^8*x^
8 + 264*a^6*x^6 + 113*a^4*x^4 + 19*a^2*x^2 + 2)*(a^2*x^2 + 1)^(3/2) + 2*(45*a^11*x^11 + 161*a^9*x^9 + 219*a^7*
x^7 + 141*a^5*x^5 + 44*a^3*x^3 + 6*a*x)*(a^2*x^2 + 1) + (45*a^12*x^12 + 193*a^10*x^10 + 325*a^8*x^8 + 270*a^6*
x^6 + 112*a^4*x^4 + 19*a^2*x^2)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1))^2 + 4*(5*a^10*x^10 + 13*a^8*x^
8 + 11*a^6*x^6 + 3*a^4*x^4)*(a^2*x^2 + 1)^(3/2) + 4*(5*a^11*x^11 + 17*a^9*x^9 + 21*a^7*x^7 + 11*a^5*x^5 + 2*a^
3*x^3)*(a^2*x^2 + 1) + (3*a^13*x^13 + 15*a^11*x^11 + 30*a^9*x^9 + 30*a^7*x^7 + 15*a^5*x^5 + 3*a^3*x^3 + (3*a^8
*x^8 + 4*a^6*x^6 + a^4*x^4)*(a^2*x^2 + 1)^(5/2) + (15*a^9*x^9 + 31*a^7*x^7 + 20*a^5*x^5 + 4*a^3*x^3)*(a^2*x^2
+ 1)^2 + (30*a^10*x^10 + 84*a^8*x^8 + 84*a^6*x^6 + 35*a^4*x^4 + 5*a^2*x^2)*(a^2*x^2 + 1)^(3/2) + 2*(15*a^11*x^
11 + 53*a^9*x^9 + 71*a^7*x^7 + 44*a^5*x^5 + 12*a^3*x^3 + a*x)*(a^2*x^2 + 1) + (15*a^12*x^12 + 64*a^10*x^10 + 1
07*a^8*x^8 + 87*a^6*x^6 + 34*a^4*x^4 + 5*a^2*x^2)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + 2*(5*a^12*
x^12 + 21*a^10*x^10 + 34*a^8*x^8 + 26*a^6*x^6 + 9*a^4*x^4 + a^2*x^2)*sqrt(a^2*x^2 + 1))/((a^13*x^10 + 5*a^11*x
^8 + (a^2*x^2 + 1)^(5/2)*a^8*x^5 + 10*a^9*x^6 + 10*a^7*x^4 + 5*a^5*x^2 + 5*(a^9*x^6 + a^7*x^4)*(a^2*x^2 + 1)^2
+ a^3 + 10*(a^10*x^7 + 2*a^8*x^5 + a^6*x^3)*(a^2*x^2 + 1)^(3/2) + 10*(a^11*x^8 + 3*a^9*x^6 + 3*a^7*x^4 + a^5*
x^2)*(a^2*x^2 + 1) + 5*(a^12*x^9 + 4*a^10*x^7 + 6*a^8*x^5 + 4*a^6*x^3 + a^4*x)*sqrt(a^2*x^2 + 1))*log(a*x + sq
rt(a^2*x^2 + 1))^3) + integrate(1/6*(27*a^14*x^14 + 162*a^12*x^12 + 405*a^10*x^10 + 540*a^8*x^8 + 405*a^6*x^6
+ 162*a^4*x^4 + (27*a^8*x^8 + 13*a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2)*(a^2*x^2 + 1)^3 + 27*a^2*x^2 + (162*a^9*x^9
+ 227*a^7*x^7 + 63*a^5*x^5 - 3*a^3*x^3 + 6*a*x)*(a^2*x^2 + 1)^(5/2) + (405*a^10*x^10 + 940*a^8*x^8 + 687*a^6*x
^6 + 143*a^4*x^4 - 21*a^2*x^2 - 12)*(a^2*x^2 + 1)^2 + (540*a^11*x^11 + 1750*a^9*x^9 + 2058*a^7*x^7 + 1017*a^5*
x^5 + 145*a^3*x^3 - 24*a*x)*(a^2*x^2 + 1)^(3/2) + (405*a^12*x^12 + 1685*a^10*x^10 + 2727*a^8*x^8 + 2118*a^6*x^
6 + 782*a^4*x^4 + 123*a^2*x^2 + 12)*(a^2*x^2 + 1) + (162*a^13*x^13 + 823*a^11*x^11 + 1695*a^9*x^9 + 1790*a^7*x
^7 + 1015*a^5*x^5 + 297*a^3*x^3 + 38*a*x)*sqrt(a^2*x^2 + 1))/((a^14*x^12 + 6*a^12*x^10 + 15*a^10*x^8 + (a^2*x^
2 + 1)^3*a^8*x^6 + 20*a^8*x^6 + 15*a^6*x^4 + 6*a^4*x^2 + 6*(a^9*x^7 + a^7*x^5)*(a^2*x^2 + 1)^(5/2) + 15*(a^10*
x^8 + 2*a^8*x^6 + a^6*x^4)*(a^2*x^2 + 1)^2 + 20*(a^11*x^9 + 3*a^9*x^7 + 3*a^7*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/
2) + 15*(a^12*x^10 + 4*a^10*x^8 + 6*a^8*x^6 + 4*a^6*x^4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 6*(a^13*x^11 + 5*a^11
*x^9 + 10*a^9*x^7 + 10*a^7*x^5 + 5*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 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{x^{2}}{\operatorname{arsinh}\left (a x\right )^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/arcsinh(a*x)^4,x, algorithm="fricas")
[Out]
integral(x^2/arcsinh(a*x)^4, x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{asinh}^{4}{\left (a x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2/asinh(a*x)**4,x)
[Out]
Integral(x**2/asinh(a*x)**4, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{\operatorname{arsinh}\left (a x\right )^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2/arcsinh(a*x)^4,x, algorithm="giac")
[Out]
integrate(x^2/arcsinh(a*x)^4, x)