3.63 \(\int \frac {x}{\sinh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=63 \[ \frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}-\frac {x \sqrt {a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac {1}{2 a^2 \sinh ^{-1}(a x)}-\frac {x^2}{\sinh ^{-1}(a x)} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {5667, 5774, 5669, 5448, 12, 3298, 5675} \[ \frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}-\frac {x \sqrt {a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}-\frac {1}{2 a^2 \sinh ^{-1}(a x)}-\frac {x^2}{\sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 5448

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 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 5669

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

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 62, normalized size = 0.98 \[ \frac {\text {Shi}\left (2 \sinh ^{-1}(a x)\right )}{a^2}-\frac {x \sqrt {a^2 x^2+1}}{2 a \sinh ^{-1}(a x)^2}+\frac {-2 a^2 x^2-1}{2 a^2 \sinh ^{-1}(a x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x}{\operatorname {arsinh}\left (a x\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x/arcsinh(a*x)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\operatorname {arsinh}\left (a x\right )^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/arcsinh(a*x)^3, x)

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maple [A]  time = 0.17, size = 43, normalized size = 0.68 \[ \frac {-\frac {\sinh \left (2 \arcsinh \left (a x \right )\right )}{4 \arcsinh \left (a x \right )^{2}}-\frac {\cosh \left (2 \arcsinh \left (a x \right )\right )}{2 \arcsinh \left (a x \right )}+\Shi \left (2 \arcsinh \left (a x \right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {a^{8} x^{8} + 3 \, a^{6} x^{6} + 3 \, a^{4} x^{4} + a^{2} x^{2} + {\left (a^{5} x^{5} + a^{3} x^{3}\right )} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} + {\left (3 \, a^{6} x^{6} + 5 \, a^{4} x^{4} + 2 \, a^{2} x^{2}\right )} {\left (a^{2} x^{2} + 1\right )} + {\left (2 \, a^{8} x^{8} + 6 \, a^{6} x^{6} + 6 \, a^{4} x^{4} + 2 \, a^{2} x^{2} + 2 \, {\left (a^{5} x^{5} + a^{3} x^{3}\right )} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} + {\left (6 \, a^{6} x^{6} + 10 \, a^{4} x^{4} + 5 \, a^{2} x^{2} + 1\right )} {\left (a^{2} x^{2} + 1\right )} + {\left (6 \, a^{7} x^{7} + 14 \, a^{5} x^{5} + 11 \, a^{3} x^{3} + 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1}\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) + {\left (3 \, a^{7} x^{7} + 7 \, a^{5} x^{5} + 5 \, a^{3} x^{3} + a x\right )} \sqrt {a^{2} x^{2} + 1}}{2 \, {\left (a^{8} x^{6} + 3 \, a^{6} x^{4} + {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} a^{5} x^{3} + 3 \, a^{4} x^{2} + 3 \, {\left (a^{6} x^{4} + a^{4} x^{2}\right )} {\left (a^{2} x^{2} + 1\right )} + a^{2} + 3 \, {\left (a^{7} x^{5} + 2 \, a^{5} x^{3} + a^{3} x\right )} \sqrt {a^{2} x^{2} + 1}\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}} + \int \frac {4 \, a^{9} x^{9} + 16 \, a^{7} x^{7} + 4 \, {\left (a^{2} x^{2} + 1\right )}^{2} a^{5} x^{5} + 24 \, a^{5} x^{5} + 16 \, a^{3} x^{3} + {\left (16 \, a^{6} x^{6} + 16 \, a^{4} x^{4} - 3\right )} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} + 24 \, {\left (a^{7} x^{7} + 2 \, a^{5} x^{5} + a^{3} x^{3}\right )} {\left (a^{2} x^{2} + 1\right )} + 4 \, a x + {\left (16 \, a^{8} x^{8} + 48 \, a^{6} x^{6} + 48 \, a^{4} x^{4} + 19 \, a^{2} x^{2} + 3\right )} \sqrt {a^{2} x^{2} + 1}}{2 \, {\left (a^{9} x^{8} + 4 \, a^{7} x^{6} + {\left (a^{2} x^{2} + 1\right )}^{2} a^{5} x^{4} + 6 \, a^{5} x^{4} + 4 \, a^{3} x^{2} + 4 \, {\left (a^{6} x^{5} + a^{4} x^{3}\right )} {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} + 6 \, {\left (a^{7} x^{6} + 2 \, a^{5} x^{4} + a^{3} x^{2}\right )} {\left (a^{2} x^{2} + 1\right )} + 4 \, {\left (a^{8} x^{7} + 3 \, a^{6} x^{5} + 3 \, a^{4} x^{3} + a^{2} x\right )} \sqrt {a^{2} x^{2} + 1} + a\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^8*x^8 + 3*a^6*x^6 + 3*a^4*x^4 + a^2*x^2 + (a^5*x^5 + a^3*x^3)*(a^2*x^2 + 1)^(3/2) + (3*a^6*x^6 + 5*a^4
*x^4 + 2*a^2*x^2)*(a^2*x^2 + 1) + (2*a^8*x^8 + 6*a^6*x^6 + 6*a^4*x^4 + 2*a^2*x^2 + 2*(a^5*x^5 + a^3*x^3)*(a^2*
x^2 + 1)^(3/2) + (6*a^6*x^6 + 10*a^4*x^4 + 5*a^2*x^2 + 1)*(a^2*x^2 + 1) + (6*a^7*x^7 + 14*a^5*x^5 + 11*a^3*x^3
+ 3*a*x)*sqrt(a^2*x^2 + 1))*log(a*x + sqrt(a^2*x^2 + 1)) + (3*a^7*x^7 + 7*a^5*x^5 + 5*a^3*x^3 + a*x)*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*
(4*a^9*x^9 + 16*a^7*x^7 + 4*(a^2*x^2 + 1)^2*a^5*x^5 + 24*a^5*x^5 + 16*a^3*x^3 + (16*a^6*x^6 + 16*a^4*x^4 - 3)*
(a^2*x^2 + 1)^(3/2) + 24*(a^7*x^7 + 2*a^5*x^5 + a^3*x^3)*(a^2*x^2 + 1) + 4*a*x + (16*a^8*x^8 + 48*a^6*x^6 + 48
*a^4*x^4 + 19*a^2*x^2 + 3)*sqrt(a^2*x^2 + 1))/((a^9*x^8 + 4*a^7*x^6 + (a^2*x^2 + 1)^2*a^5*x^4 + 6*a^5*x^4 + 4*
a^3*x^2 + 4*(a^6*x^5 + a^4*x^3)*(a^2*x^2 + 1)^(3/2) + 6*(a^7*x^6 + 2*a^5*x^4 + a^3*x^2)*(a^2*x^2 + 1) + 4*(a^8
*x^7 + 3*a^6*x^5 + 3*a^4*x^3 + a^2*x)*sqrt(a^2*x^2 + 1) + a)*log(a*x + sqrt(a^2*x^2 + 1))), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{{\mathrm {asinh}\left (a\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x/asinh(a*x)^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/asinh(a*x)**3, x)

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