3.1.24 \(\int x^2 \sinh ^{-1}(a x)^3 \, dx\) [24]

Optimal. Leaf size=132 \[ \frac {14 \sqrt {1+a^2 x^2}}{9 a^3}-\frac {2 \left (1+a^2 x^2\right )^{3/2}}{27 a^3}-\frac {4 x \sinh ^{-1}(a x)}{3 a^2}+\frac {2}{9} x^3 \sinh ^{-1}(a x)+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a^3}-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3 \]

[Out]

-2/27*(a^2*x^2+1)^(3/2)/a^3-4/3*x*arcsinh(a*x)/a^2+2/9*x^3*arcsinh(a*x)+1/3*x^3*arcsinh(a*x)^3+14/9*(a^2*x^2+1
)^(1/2)/a^3+2/3*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/a^3-1/3*x^2*arcsinh(a*x)^2*(a^2*x^2+1)^(1/2)/a

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, Rules used = {5776, 5812, 5798, 5772, 267, 272, 45} \begin {gather*} -\frac {x^2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{3 a}-\frac {4 x \sinh ^{-1}(a x)}{3 a^2}-\frac {2 \left (a^2 x^2+1\right )^{3/2}}{27 a^3}+\frac {14 \sqrt {a^2 x^2+1}}{9 a^3}+\frac {2 \sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{3 a^3}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3+\frac {2}{9} x^3 \sinh ^{-1}(a x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*ArcSinh[a*x]^3,x]

[Out]

(14*Sqrt[1 + a^2*x^2])/(9*a^3) - (2*(1 + a^2*x^2)^(3/2))/(27*a^3) - (4*x*ArcSinh[a*x])/(3*a^2) + (2*x^3*ArcSin
h[a*x])/9 + (2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(3*a^3) - (x^2*Sqrt[1 + a^2*x^2]*ArcSinh[a*x]^2)/(3*a) + (x^3
*ArcSinh[a*x]^3)/3

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 267

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5772

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[x*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5776

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 5798

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

Rule 5812

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

Rubi steps

\begin {align*} \int x^2 \sinh ^{-1}(a x)^3 \, dx &=\frac {1}{3} x^3 \sinh ^{-1}(a x)^3-a \int \frac {x^3 \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3+\frac {2}{3} \int x^2 \sinh ^{-1}(a x) \, dx+\frac {2 \int \frac {x \sinh ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{3 a}\\ &=\frac {2}{9} x^3 \sinh ^{-1}(a x)+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a^3}-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3-\frac {4 \int \sinh ^{-1}(a x) \, dx}{3 a^2}-\frac {1}{9} (2 a) \int \frac {x^3}{\sqrt {1+a^2 x^2}} \, dx\\ &=-\frac {4 x \sinh ^{-1}(a x)}{3 a^2}+\frac {2}{9} x^3 \sinh ^{-1}(a x)+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a^3}-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3+\frac {4 \int \frac {x}{\sqrt {1+a^2 x^2}} \, dx}{3 a}-\frac {1}{9} a \text {Subst}\left (\int \frac {x}{\sqrt {1+a^2 x}} \, dx,x,x^2\right )\\ &=\frac {4 \sqrt {1+a^2 x^2}}{3 a^3}-\frac {4 x \sinh ^{-1}(a x)}{3 a^2}+\frac {2}{9} x^3 \sinh ^{-1}(a x)+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a^3}-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3-\frac {1}{9} a \text {Subst}\left (\int \left (-\frac {1}{a^2 \sqrt {1+a^2 x}}+\frac {\sqrt {1+a^2 x}}{a^2}\right ) \, dx,x,x^2\right )\\ &=\frac {14 \sqrt {1+a^2 x^2}}{9 a^3}-\frac {2 \left (1+a^2 x^2\right )^{3/2}}{27 a^3}-\frac {4 x \sinh ^{-1}(a x)}{3 a^2}+\frac {2}{9} x^3 \sinh ^{-1}(a x)+\frac {2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a^3}-\frac {x^2 \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{3 a}+\frac {1}{3} x^3 \sinh ^{-1}(a x)^3\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 93, normalized size = 0.70 \begin {gather*} \frac {-2 \left (-20+a^2 x^2\right ) \sqrt {1+a^2 x^2}+6 a x \left (-6+a^2 x^2\right ) \sinh ^{-1}(a x)-9 \left (-2+a^2 x^2\right ) \sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2+9 a^3 x^3 \sinh ^{-1}(a x)^3}{27 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcSinh[a*x]^3,x]

[Out]

(-2*(-20 + a^2*x^2)*Sqrt[1 + a^2*x^2] + 6*a*x*(-6 + a^2*x^2)*ArcSinh[a*x] - 9*(-2 + a^2*x^2)*Sqrt[1 + a^2*x^2]
*ArcSinh[a*x]^2 + 9*a^3*x^3*ArcSinh[a*x]^3)/(27*a^3)

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int x^{2} \arcsinh \left (a x \right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arcsinh(a*x)^3,x)

[Out]

int(x^2*arcsinh(a*x)^3,x)

________________________________________________________________________________________

Maxima [A]
time = 0.27, size = 116, normalized size = 0.88 \begin {gather*} \frac {1}{3} \, x^{3} \operatorname {arsinh}\left (a x\right )^{3} - \frac {1}{3} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \operatorname {arsinh}\left (a x\right )^{2} - \frac {2}{27} \, a {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2} - \frac {20 \, \sqrt {a^{2} x^{2} + 1}}{a^{2}}}{a^{2}} - \frac {3 \, {\left (a^{2} x^{3} - 6 \, x\right )} \operatorname {arsinh}\left (a x\right )}{a^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(a*x)^3,x, algorithm="maxima")

[Out]

1/3*x^3*arcsinh(a*x)^3 - 1/3*a*(sqrt(a^2*x^2 + 1)*x^2/a^2 - 2*sqrt(a^2*x^2 + 1)/a^4)*arcsinh(a*x)^2 - 2/27*a*(
(sqrt(a^2*x^2 + 1)*x^2 - 20*sqrt(a^2*x^2 + 1)/a^2)/a^2 - 3*(a^2*x^3 - 6*x)*arcsinh(a*x)/a^3)

________________________________________________________________________________________

Fricas [A]
time = 0.34, size = 124, normalized size = 0.94 \begin {gather*} \frac {9 \, a^{3} x^{3} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{3} - 9 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 2\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} + 6 \, {\left (a^{3} x^{3} - 6 \, a x\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right ) - 2 \, \sqrt {a^{2} x^{2} + 1} {\left (a^{2} x^{2} - 20\right )}}{27 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arcsinh(a*x)^3,x, algorithm="fricas")

[Out]

1/27*(9*a^3*x^3*log(a*x + sqrt(a^2*x^2 + 1))^3 - 9*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 2)*log(a*x + sqrt(a^2*x^2 + 1)
)^2 + 6*(a^3*x^3 - 6*a*x)*log(a*x + sqrt(a^2*x^2 + 1)) - 2*sqrt(a^2*x^2 + 1)*(a^2*x^2 - 20))/a^3

________________________________________________________________________________________

Sympy [A]
time = 0.30, size = 128, normalized size = 0.97 \begin {gather*} \begin {cases} \frac {x^{3} \operatorname {asinh}^{3}{\left (a x \right )}}{3} + \frac {2 x^{3} \operatorname {asinh}{\left (a x \right )}}{9} - \frac {x^{2} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{3 a} - \frac {2 x^{2} \sqrt {a^{2} x^{2} + 1}}{27 a} - \frac {4 x \operatorname {asinh}{\left (a x \right )}}{3 a^{2}} + \frac {2 \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}^{2}{\left (a x \right )}}{3 a^{3}} + \frac {40 \sqrt {a^{2} x^{2} + 1}}{27 a^{3}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*asinh(a*x)**3,x)

[Out]

Piecewise((x**3*asinh(a*x)**3/3 + 2*x**3*asinh(a*x)/9 - x**2*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(3*a) - 2*x**2*
sqrt(a**2*x**2 + 1)/(27*a) - 4*x*asinh(a*x)/(3*a**2) + 2*sqrt(a**2*x**2 + 1)*asinh(a*x)**2/(3*a**3) + 40*sqrt(
a**2*x**2 + 1)/(27*a**3), Ne(a, 0)), (0, True))

________________________________________________________________________________________

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^2*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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,{\mathrm {asinh}\left (a\,x\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*asinh(a*x)^3,x)

[Out]

int(x^2*asinh(a*x)^3, x)

________________________________________________________________________________________