3.1.61 \(\int \frac {x^3}{\sinh ^{-1}(a x)^3} \, dx\) [61]

Optimal. Leaf size=82 \[ -\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}-\frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a x)\right )}{a^4} \]

[Out]

-3/2*x^2/a^2/arcsinh(a*x)-2*x^4/arcsinh(a*x)-1/2*Shi(2*arcsinh(a*x))/a^4+Shi(4*arcsinh(a*x))/a^4-1/2*x^3*(a^2*
x^2+1)^(1/2)/a/arcsinh(a*x)^2

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Rubi [A]
time = 0.23, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5779, 5818, 5780, 5556, 3379, 12} \begin {gather*} -\frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a x)\right )}{a^4}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {x^3 \sqrt {a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac {2 x^4}{\sinh ^{-1}(a x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/ArcSinh[a*x]^3,x]

[Out]

-1/2*(x^3*Sqrt[1 + a^2*x^2])/(a*ArcSinh[a*x]^2) - (3*x^2)/(2*a^2*ArcSinh[a*x]) - (2*x^4)/ArcSinh[a*x] - SinhIn
tegral[2*ArcSinh[a*x]]/(2*a^4) + SinhIntegral[4*ArcSinh[a*x]]/a^4

Rule 12

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

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 5556

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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 5780

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/(b*c^(m + 1)), Subst[Int[x^n*Sinh
[-a/b + x/b]^m*Cosh[-a/b + x/b], x], x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

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]

Rubi steps

\begin {align*} \int \frac {x^3}{\sinh ^{-1}(a x)^3} \, dx &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}+\frac {3 \int \frac {x^2}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx}{2 a}+(2 a) \int \frac {x^4}{\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2} \, dx\\ &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}+8 \int \frac {x^3}{\sinh ^{-1}(a x)} \, dx+\frac {3 \int \frac {x}{\sinh ^{-1}(a x)} \, dx}{a^2}\\ &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}+\frac {3 \text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {8 \text {Subst}\left (\int \frac {\cosh (x) \sinh ^3(x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}\\ &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}+\frac {3 \text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {8 \text {Subst}\left (\int \left (-\frac {\sinh (2 x)}{4 x}+\frac {\sinh (4 x)}{8 x}\right ) \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}\\ &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}+\frac {\text {Subst}\left (\int \frac {\sinh (4 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}+\frac {3 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{2 a^4}-\frac {2 \text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a x)\right )}{a^4}\\ &=-\frac {x^3 \sqrt {1+a^2 x^2}}{2 a \sinh ^{-1}(a x)^2}-\frac {3 x^2}{2 a^2 \sinh ^{-1}(a x)}-\frac {2 x^4}{\sinh ^{-1}(a x)}-\frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{2 a^4}+\frac {\text {Shi}\left (4 \sinh ^{-1}(a x)\right )}{a^4}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 69, normalized size = 0.84 \begin {gather*} -\frac {\frac {a^2 x^2 \left (a x \sqrt {1+a^2 x^2}+\left (3+4 a^2 x^2\right ) \sinh ^{-1}(a x)\right )}{\sinh ^{-1}(a x)^2}+\text {Shi}\left (2 \sinh ^{-1}(a x)\right )-2 \text {Shi}\left (4 \sinh ^{-1}(a x)\right )}{2 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/ArcSinh[a*x]^3,x]

[Out]

-1/2*((a^2*x^2*(a*x*Sqrt[1 + a^2*x^2] + (3 + 4*a^2*x^2)*ArcSinh[a*x]))/ArcSinh[a*x]^2 + SinhIntegral[2*ArcSinh
[a*x]] - 2*SinhIntegral[4*ArcSinh[a*x]])/a^4

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Maple [A]
time = 1.45, size = 82, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (2 \arcsinh \left (a x \right )\right )}{2}-\frac {\sinh \left (4 \arcsinh \left (a x \right )\right )}{16 \arcsinh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}+\hyperbolicSineIntegral \left (4 \arcsinh \left (a x \right )\right )}{a^{4}}\)	\(82\)
default	\(\frac {\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{8 \arcsinh \left (a x \right )^{2}}+\frac {\cosh \left (2 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}-\frac {\hyperbolicSineIntegral \left (2 \arcsinh \left (a x \right )\right )}{2}-\frac {\sinh \left (4 \arcsinh \left (a x \right )\right )}{16 \arcsinh \left (a x \right )^{2}}-\frac {\cosh \left (4 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )}+\hyperbolicSineIntegral \left (4 \arcsinh \left (a x \right )\right )}{a^{4}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/arcsinh(a*x)^3,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/8/arcsinh(a*x)^2*sinh(2*arcsinh(a*x))+1/4/arcsinh(a*x)*cosh(2*arcsinh(a*x))-1/2*Shi(2*arcsinh(a*x))-1
/16/arcsinh(a*x)^2*sinh(4*arcsinh(a*x))-1/4/arcsinh(a*x)*cosh(4*arcsinh(a*x))+Shi(4*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^3/arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^8*x^10 + 3*a^6*x^8 + 3*a^4*x^6 + a^2*x^4 + (a^5*x^7 + a^3*x^5)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^8 + 5*a^
4*x^6 + 2*a^2*x^4)*(a^2*x^2 + 1) + (4*a^8*x^10 + 12*a^6*x^8 + 12*a^4*x^6 + 4*a^2*x^4 + 2*(2*a^5*x^7 + 3*a^3*x^
5 + a*x^3)*(a^2*x^2 + 1)^(3/2) + 3*(4*a^6*x^8 + 8*a^4*x^6 + 5*a^2*x^4 + x^2)*(a^2*x^2 + 1) + (12*a^7*x^9 + 30*
a^5*x^7 + 25*a^3*x^5 + 7*a*x^3)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^9 + 7*a^5*x^7 + 5*a
^3*x^5 + a*x^3)*sqrt(a^2*x^2 + 1))/((a^8*x^6 + 3*a^6*x^4 + (a^2*x^2 + 1)^(3/2)*a^5*x^3 + 3*a^4*x^2 + 3*(a^6*x^
4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 3*(a^7*x^5 + 2*a^5*x^3 + a^3*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 +
1))^2) + integrate(1/2*(16*a^10*x^11 + 64*a^8*x^9 + 96*a^6*x^7 + 64*a^4*x^5 + 16*a^2*x^3 + 4*(4*a^6*x^7 + 3*a
^4*x^5)*(a^2*x^2 + 1)^2 + (64*a^7*x^8 + 100*a^5*x^6 + 42*a^3*x^4 + 3*a*x^2)*(a^2*x^2 + 1)^(3/2) + 6*(16*a^8*x^
9 + 38*a^6*x^7 + 30*a^4*x^5 + 9*a^2*x^3 + x)*(a^2*x^2 + 1) + (64*a^9*x^10 + 204*a^7*x^8 + 234*a^5*x^6 + 115*a^
3*x^4 + 21*a*x^2)*sqrt(a^2*x^2 + 1))/((a^10*x^8 + 4*a^8*x^6 + (a^2*x^2 + 1)^2*a^6*x^4 + 6*a^6*x^4 + 4*a^4*x^2
+ 4*(a^7*x^5 + a^5*x^3)*(a^2*x^2 + 1)^(3/2) + 6*(a^8*x^6 + 2*a^6*x^4 + a^4*x^2)*(a^2*x^2 + 1) + a^2 + 4*(a^9*x
^7 + 3*a^7*x^5 + 3*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^3/arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

integral(x^3/arcsinh(a*x)^3, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\operatorname {asinh}^{3}{\left (a x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/asinh(a*x)**3,x)

[Out]

Integral(x**3/asinh(a*x)**3, x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/arcsinh(a*x)^3,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3}{{\mathrm {asinh}\left (a\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/asinh(a*x)^3,x)

[Out]

int(x^3/asinh(a*x)^3, x)

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