3.1.69 \(\int \frac {x^2}{\sinh ^{-1}(a x)^4} \, dx\) [69]

Optimal. Leaf size=138 \[ -\frac {x^2 \sqrt {1+a^2 x^2}}{3 a \sinh ^{-1}(a x)^3}-\frac {x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac {x^3}{2 \sinh ^{-1}(a x)^2}-\frac {\sqrt {1+a^2 x^2}}{3 a^3 \sinh ^{-1}(a x)}-\frac {3 x^2 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)}-\frac {\text {Shi}\left (\sinh ^{-1}(a x)\right )}{24 a^3}+\frac {9 \text {Shi}\left (3 \sinh ^{-1}(a x)\right )}{8 a^3} \]

[Out]

-1/3*x/a^2/arcsinh(a*x)^2-1/2*x^3/arcsinh(a*x)^2-1/24*Shi(arcsinh(a*x))/a^3+9/8*Shi(3*arcsinh(a*x))/a^3-1/3*x^
2*(a^2*x^2+1)^(1/2)/a/arcsinh(a*x)^3-1/3*(a^2*x^2+1)^(1/2)/a^3/arcsinh(a*x)-3/2*x^2*(a^2*x^2+1)^(1/2)/a/arcsin
h(a*x)

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Rubi [A]
time = 0.21, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5779, 5818, 5778, 3379, 5773, 5819} \begin {gather*} -\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 {x}{3 a^2 \sinh ^{-1}(a x)^2}-\frac {\sqrt {a^2 x^2+1}}{3 a^3 \sinh ^{-1}(a x)}-\frac {x^3}{2 \sinh ^{-1}(a x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/ArcSinh[a*x]^4,x]

[Out]

-1/3*(x^2*Sqrt[1 + a^2*x^2])/(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 3379

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 5773

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 5778

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^2*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[x^(n + 1), Si
nh[-a/b + x/b]^(m - 1)*(m + (m + 1)*Sinh[-a/b + x/b]^2), x], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}
, x] && IGtQ[m, 0] && GeQ[n, -2] && LtQ[n, -1]

Rule 5779

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 5818

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[((f*x)^m/(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] - Dist[f*(m/
(b*c*(n + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], 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]

Rule 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(1/(b*
c^(m + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Subst[Int[x^n*Sinh[-a/b + x/b]^m*Cosh[-a/b + x/b]^(2*p + 1),
x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IGtQ[2*p + 2, 0] && IGtQ[m,
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 \text {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 {\text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{3 a^3}-\frac {3 \text {Subst}\left (\int \frac {\sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{8 a^3}+\frac {9 \text {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*}

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Mathematica [A]
time = 0.23, size = 99, normalized size = 0.72 \begin {gather*} -\frac {\frac {4 \left (2 a^2 x^2 \sqrt {1+a^2 x^2}+a x \left (2+3 a^2 x^2\right ) \sinh ^{-1}(a x)+\sqrt {1+a^2 x^2} \left (2+9 a^2 x^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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/ArcSinh[a*x]^4,x]

[Out]

-1/24*((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)*
ArcSinh[a*x]^2))/ArcSinh[a*x]^3 + SinhIntegral[ArcSinh[a*x]] - 27*SinhIntegral[3*ArcSinh[a*x]])/a^3

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Maple [A]
time = 1.23, size = 115, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {\sqrt {a^{2} x^{2}+1}}{12 \arcsinh \left (a x \right )^{3}}+\frac {a x}{24 \arcsinh \left (a x \right )^{2}}+\frac {\sqrt {a^{2} x^{2}+1}}{24 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right )}{24}-\frac {\cosh \left (3 \arcsinh \left (a x \right )\right )}{12 \arcsinh \left (a x \right )^{3}}-\frac {\sinh \left (3 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )^{2}}-\frac {3 \cosh \left (3 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )}+\frac {9 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right )}{8}}{a^{3}}\)	\(115\)
default	\(\frac {\frac {\sqrt {a^{2} x^{2}+1}}{12 \arcsinh \left (a x \right )^{3}}+\frac {a x}{24 \arcsinh \left (a x \right )^{2}}+\frac {\sqrt {a^{2} x^{2}+1}}{24 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (\arcsinh \left (a x \right )\right )}{24}-\frac {\cosh \left (3 \arcsinh \left (a x \right )\right )}{12 \arcsinh \left (a x \right )^{3}}-\frac {\sinh \left (3 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )^{2}}-\frac {3 \cosh \left (3 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )}+\frac {9 \hyperbolicSineIntegral \left (3 \arcsinh \left (a x \right )\right )}{8}}{a^{3}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/arcsinh(a*x)^4,x,method=_RETURNVERBOSE)

[Out]

1/a^3*(1/12/arcsinh(a*x)^3*(a^2*x^2+1)^(1/2)+1/24/arcsinh(a*x)^2*a*x+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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\operatorname {asinh}^{4}{\left (a x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\mathrm {asinh}\left (a\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/asinh(a*x)^4,x)

[Out]

int(x^2/asinh(a*x)^4, x)

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