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

Optimal. Leaf size=163 \[ -\frac{3 x^3 \sqrt{a^2 x^2+1}}{128 a}+\frac{45 x \sqrt{a^2 x^2+1}}{256 a^3}-\frac{3 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{16 a}-\frac{9 x^2 \sinh ^{-1}(a x)}{32 a^2}+\frac{9 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 \sinh ^{-1}(a x)^3}{32 a^4}-\frac{45 \sinh ^{-1}(a x)}{256 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3+\frac{3}{32} x^4 \sinh ^{-1}(a x) \]

[Out]

(45*x*Sqrt[1 + a^2*x^2])/(256*a^3) - (3*x^3*Sqrt[1 + a^2*x^2])/(128*a) - (45*ArcSinh[a*x])/(256*a^4) - (9*x^2*
ArcSinh[a*x])/(32*a^2) + (3*x^4*ArcSinh[a*x])/32 + (9*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(32*a^3) - (3*x^3*Sq
rt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(16*a) - (3*ArcSinh[a*x]^3)/(32*a^4) + (x^4*ArcSinh[a*x]^3)/4

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Rubi [A]  time = 0.298368, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5661, 5758, 5675, 321, 215} \[ -\frac{3 x^3 \sqrt{a^2 x^2+1}}{128 a}+\frac{45 x \sqrt{a^2 x^2+1}}{256 a^3}-\frac{3 x^3 \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{16 a}-\frac{9 x^2 \sinh ^{-1}(a x)}{32 a^2}+\frac{9 x \sqrt{a^2 x^2+1} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 \sinh ^{-1}(a x)^3}{32 a^4}-\frac{45 \sinh ^{-1}(a x)}{256 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3+\frac{3}{32} x^4 \sinh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(45*x*Sqrt[1 + a^2*x^2])/(256*a^3) - (3*x^3*Sqrt[1 + a^2*x^2])/(128*a) - (45*ArcSinh[a*x])/(256*a^4) - (9*x^2*
ArcSinh[a*x])/(32*a^2) + (3*x^4*ArcSinh[a*x])/32 + (9*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(32*a^3) - (3*x^3*Sq
rt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(16*a) - (3*ArcSinh[a*x]^3)/(32*a^4) + (x^4*ArcSinh[a*x]^3)/4

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt
[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5758

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp
[(f*(f*x)^(m - 1)*Sqrt[d + e*x^2]*(a + b*ArcSinh[c*x])^n)/(e*m), x] + (-Dist[(f^2*(m - 1))/(c^2*m), Int[((f*x)
^(m - 2)*(a + b*ArcSinh[c*x])^n)/Sqrt[d + e*x^2], x], x] - Dist[(b*f*n*Sqrt[1 + c^2*x^2])/(c*m*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}, x] && EqQ[e, c^2*d] &&
GtQ[n, 0] && GtQ[m, 1] && IntegerQ[m]

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int x^3 \sinh ^{-1}(a x)^3 \, dx &=\frac{1}{4} x^4 \sinh ^{-1}(a x)^3-\frac{1}{4} (3 a) \int \frac{x^4 \sinh ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3+\frac{3}{8} \int x^3 \sinh ^{-1}(a x) \, dx+\frac{9 \int \frac{x^2 \sinh ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{16 a}\\ &=\frac{3}{32} x^4 \sinh ^{-1}(a x)+\frac{9 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{16 a}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3-\frac{9 \int \frac{\sinh ^{-1}(a x)^2}{\sqrt{1+a^2 x^2}} \, dx}{32 a^3}-\frac{9 \int x \sinh ^{-1}(a x) \, dx}{16 a^2}-\frac{1}{32} (3 a) \int \frac{x^4}{\sqrt{1+a^2 x^2}} \, dx\\ &=-\frac{3 x^3 \sqrt{1+a^2 x^2}}{128 a}-\frac{9 x^2 \sinh ^{-1}(a x)}{32 a^2}+\frac{3}{32} x^4 \sinh ^{-1}(a x)+\frac{9 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{16 a}-\frac{3 \sinh ^{-1}(a x)^3}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3+\frac{9 \int \frac{x^2}{\sqrt{1+a^2 x^2}} \, dx}{128 a}+\frac{9 \int \frac{x^2}{\sqrt{1+a^2 x^2}} \, dx}{32 a}\\ &=\frac{45 x \sqrt{1+a^2 x^2}}{256 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2}}{128 a}-\frac{9 x^2 \sinh ^{-1}(a x)}{32 a^2}+\frac{3}{32} x^4 \sinh ^{-1}(a x)+\frac{9 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{16 a}-\frac{3 \sinh ^{-1}(a x)^3}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3-\frac{9 \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{256 a^3}-\frac{9 \int \frac{1}{\sqrt{1+a^2 x^2}} \, dx}{64 a^3}\\ &=\frac{45 x \sqrt{1+a^2 x^2}}{256 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2}}{128 a}-\frac{45 \sinh ^{-1}(a x)}{256 a^4}-\frac{9 x^2 \sinh ^{-1}(a x)}{32 a^2}+\frac{3}{32} x^4 \sinh ^{-1}(a x)+\frac{9 x \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{32 a^3}-\frac{3 x^3 \sqrt{1+a^2 x^2} \sinh ^{-1}(a x)^2}{16 a}-\frac{3 \sinh ^{-1}(a x)^3}{32 a^4}+\frac{1}{4} x^4 \sinh ^{-1}(a x)^3\\ \end{align*}

Mathematica [A]  time = 0.0690697, size = 110, normalized size = 0.67 \[ \frac{3 a x \left (15-2 a^2 x^2\right ) \sqrt{a^2 x^2+1}+8 \left (8 a^4 x^4-3\right ) \sinh ^{-1}(a x)^3-24 a x \sqrt{a^2 x^2+1} \left (2 a^2 x^2-3\right ) \sinh ^{-1}(a x)^2+3 \left (8 a^4 x^4-24 a^2 x^2-15\right ) \sinh ^{-1}(a x)}{256 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x*(15 - 2*a^2*x^2)*Sqrt[1 + a^2*x^2] + 3*(-15 - 24*a^2*x^2 + 8*a^4*x^4)*ArcSinh[a*x] - 24*a*x*Sqrt[1 + a^
2*x^2]*(-3 + 2*a^2*x^2)*ArcSinh[a*x]^2 + 8*(-3 + 8*a^4*x^4)*ArcSinh[a*x]^3)/(256*a^4)

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Maple [A]  time = 0.034, size = 168, normalized size = 1. \begin{align*}{\frac{1}{{a}^{4}} \left ({\frac{{a}^{2}{x}^{2} \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{ \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3} \left ({a}^{2}{x}^{2}+1 \right ) }{4}}-{\frac{3\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}ax}{16} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{15\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{2}ax}{32}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{5\, \left ({\it Arcsinh} \left ( ax \right ) \right ) ^{3}}{32}}+{\frac{3\,{\it Arcsinh} \left ( ax \right ){a}^{2}{x}^{2} \left ({a}^{2}{x}^{2}+1 \right ) }{32}}-{\frac{3\,ax}{128} \left ({a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{51\,ax}{256}\sqrt{{a}^{2}{x}^{2}+1}}+{\frac{51\,{\it Arcsinh} \left ( ax \right ) }{256}}-{\frac{ \left ( 3\,{a}^{2}{x}^{2}+3 \right ){\it Arcsinh} \left ( ax \right ) }{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^4*(1/4*a^2*x^2*arcsinh(a*x)^3*(a^2*x^2+1)-1/4*arcsinh(a*x)^3*(a^2*x^2+1)-3/16*arcsinh(a*x)^2*a*x*(a^2*x^2+
1)^(3/2)+15/32*arcsinh(a*x)^2*a*x*(a^2*x^2+1)^(1/2)+5/32*arcsinh(a*x)^3+3/32*arcsinh(a*x)*a^2*x^2*(a^2*x^2+1)-
3/128*a*x*(a^2*x^2+1)^(3/2)+51/256*a*x*(a^2*x^2+1)^(1/2)+51/256*arcsinh(a*x)-3/8*(a^2*x^2+1)*arcsinh(a*x))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{4} \, x^{4} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} - \int \frac{3 \,{\left (a^{3} x^{6} + \sqrt{a^{2} x^{2} + 1} a^{2} x^{5} + a x^{4}\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2}}{4 \,{\left (a^{3} x^{3} + a x +{\left (a^{2} x^{2} + 1\right )}^{\frac{3}{2}}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^4*log(a*x + sqrt(a^2*x^2 + 1))^3 - integrate(3/4*(a^3*x^6 + sqrt(a^2*x^2 + 1)*a^2*x^5 + a*x^4)*log(a*x +
sqrt(a^2*x^2 + 1))^2/(a^3*x^3 + a*x + (a^2*x^2 + 1)^(3/2)), x)

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Fricas [A]  time = 2.14171, size = 327, normalized size = 2.01 \begin{align*} \frac{8 \,{\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{3} - 24 \,{\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt{a^{2} x^{2} + 1} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right )^{2} + 3 \,{\left (8 \, a^{4} x^{4} - 24 \, a^{2} x^{2} - 15\right )} \log \left (a x + \sqrt{a^{2} x^{2} + 1}\right ) - 3 \,{\left (2 \, a^{3} x^{3} - 15 \, a x\right )} \sqrt{a^{2} x^{2} + 1}}{256 \, a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(8*(8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 + 1))^3 - 24*(2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^2 + 1)*log(a*x + s
qrt(a^2*x^2 + 1))^2 + 3*(8*a^4*x^4 - 24*a^2*x^2 - 15)*log(a*x + sqrt(a^2*x^2 + 1)) - 3*(2*a^3*x^3 - 15*a*x)*sq
rt(a^2*x^2 + 1))/a^4

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Sympy [A]  time = 4.26919, size = 160, normalized size = 0.98 \begin{align*} \begin{cases} \frac{x^{4} \operatorname{asinh}^{3}{\left (a x \right )}}{4} + \frac{3 x^{4} \operatorname{asinh}{\left (a x \right )}}{32} - \frac{3 x^{3} \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a x \right )}}{16 a} - \frac{3 x^{3} \sqrt{a^{2} x^{2} + 1}}{128 a} - \frac{9 x^{2} \operatorname{asinh}{\left (a x \right )}}{32 a^{2}} + \frac{9 x \sqrt{a^{2} x^{2} + 1} \operatorname{asinh}^{2}{\left (a x \right )}}{32 a^{3}} + \frac{45 x \sqrt{a^{2} x^{2} + 1}}{256 a^{3}} - \frac{3 \operatorname{asinh}^{3}{\left (a x \right )}}{32 a^{4}} - \frac{45 \operatorname{asinh}{\left (a x \right )}}{256 a^{4}} & \text{for}\: a \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x**4*asinh(a*x)**3/4 + 3*x**4*asinh(a*x)/32 - 3*x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(16*a) - 3*x
**3*sqrt(a**2*x**2 + 1)/(128*a) - 9*x**2*asinh(a*x)/(32*a**2) + 9*x*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(32*a**3
) + 45*x*sqrt(a**2*x**2 + 1)/(256*a**3) - 3*asinh(a*x)**3/(32*a**4) - 45*asinh(a*x)/(256*a**4), Ne(a, 0)), (0,
True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arsinh}\left (a x\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arcsinh(a*x)^3, x)