\(\int x^3 \text {arcsinh}(a x)^2 \, dx\) [13]

Optimal result

Integrand size = 10, antiderivative size = 96 \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=-\frac {3 x^2}{32 a^2}+\frac {x^4}{32}+\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{16 a^3}-\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a}-\frac {3 \text {arcsinh}(a x)^2}{32 a^4}+\frac {1}{4} x^4 \text {arcsinh}(a x)^2 \]

[Out]

-3/32*x^2/a^2+1/32*x^4-3/32*arcsinh(a*x)^2/a^4+1/4*x^4*arcsinh(a*x)^2+3/16*x*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a^
3-1/8*x^3*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/a

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5776, 5812, 5783, 30} \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=-\frac {3 \text {arcsinh}(a x)^2}{32 a^4}-\frac {x^3 \sqrt {a^2 x^2+1} \text {arcsinh}(a x)}{8 a}-\frac {3 x^2}{32 a^2}+\frac {3 x \sqrt {a^2 x^2+1} \text {arcsinh}(a x)}{16 a^3}+\frac {1}{4} x^4 \text {arcsinh}(a x)^2+\frac {x^4}{32} \]

[In]

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

[Out]

(-3*x^2)/(32*a^2) + x^4/32 + (3*x*Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(16*a^3) - (x^3*Sqrt[1 + a^2*x^2]*ArcSinh[a*
x])/(8*a) - (3*ArcSinh[a*x]^2)/(32*a^4) + (x^4*ArcSinh[a*x]^2)/4

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -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*} \text {integral}& = \frac {1}{4} x^4 \text {arcsinh}(a x)^2-\frac {1}{2} a \int \frac {x^4 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx \\ & = -\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a}+\frac {1}{4} x^4 \text {arcsinh}(a x)^2+\frac {\int x^3 \, dx}{8}+\frac {3 \int \frac {x^2 \text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 a} \\ & = \frac {x^4}{32}+\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{16 a^3}-\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a}+\frac {1}{4} x^4 \text {arcsinh}(a x)^2-\frac {3 \int \frac {\text {arcsinh}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{16 a^3}-\frac {3 \int x \, dx}{16 a^2} \\ & = -\frac {3 x^2}{32 a^2}+\frac {x^4}{32}+\frac {3 x \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{16 a^3}-\frac {x^3 \sqrt {1+a^2 x^2} \text {arcsinh}(a x)}{8 a}-\frac {3 \text {arcsinh}(a x)^2}{32 a^4}+\frac {1}{4} x^4 \text {arcsinh}(a x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.75 \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\frac {a^2 x^2 \left (-3+a^2 x^2\right )-2 a x \sqrt {1+a^2 x^2} \left (-3+2 a^2 x^2\right ) \text {arcsinh}(a x)+\left (-3+8 a^4 x^4\right ) \text {arcsinh}(a x)^2}{32 a^4} \]

[In]

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

[Out]

(a^2*x^2*(-3 + a^2*x^2) - 2*a*x*Sqrt[1 + a^2*x^2]*(-3 + 2*a^2*x^2)*ArcSinh[a*x] + (-3 + 8*a^4*x^4)*ArcSinh[a*x
]^2)/(32*a^4)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {a^{4} x^{4} \operatorname {arcsinh}\left (a x \right )^{2}}{4}-\frac {a^{3} x^{3} \operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}}{8}+\frac {3 \,\operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x}{16}-\frac {3 \operatorname {arcsinh}\left (a x \right )^{2}}{32}+\frac {a^{4} x^{4}}{32}-\frac {3 a^{2} x^{2}}{32}-\frac {3}{32}}{a^{4}}\)	\(87\)
default	\(\frac {\frac {a^{4} x^{4} \operatorname {arcsinh}\left (a x \right )^{2}}{4}-\frac {a^{3} x^{3} \operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}}{8}+\frac {3 \,\operatorname {arcsinh}\left (a x \right ) \sqrt {a^{2} x^{2}+1}\, a x}{16}-\frac {3 \operatorname {arcsinh}\left (a x \right )^{2}}{32}+\frac {a^{4} x^{4}}{32}-\frac {3 a^{2} x^{2}}{32}-\frac {3}{32}}{a^{4}}\)	\(87\)

[In]

int(x^3*arcsinh(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(1/4*a^4*x^4*arcsinh(a*x)^2-1/8*a^3*x^3*arcsinh(a*x)*(a^2*x^2+1)^(1/2)+3/16*arcsinh(a*x)*(a^2*x^2+1)^(1/
2)*a*x-3/32*arcsinh(a*x)^2+1/32*a^4*x^4-3/32*a^2*x^2-3/32)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.96 \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\frac {a^{4} x^{4} - 3 \, a^{2} x^{2} + {\left (8 \, a^{4} x^{4} - 3\right )} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2} - 2 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \sqrt {a^{2} x^{2} + 1} \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )}{32 \, a^{4}} \]

[In]

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

[Out]

1/32*(a^4*x^4 - 3*a^2*x^2 + (8*a^4*x^4 - 3)*log(a*x + sqrt(a^2*x^2 + 1))^2 - 2*(2*a^3*x^3 - 3*a*x)*sqrt(a^2*x^
2 + 1)*log(a*x + sqrt(a^2*x^2 + 1)))/a^4

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.94 \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\begin {cases} \frac {x^{4} \operatorname {asinh}^{2}{\left (a x \right )}}{4} + \frac {x^{4}}{32} - \frac {x^{3} \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{8 a} - \frac {3 x^{2}}{32 a^{2}} + \frac {3 x \sqrt {a^{2} x^{2} + 1} \operatorname {asinh}{\left (a x \right )}}{16 a^{3}} - \frac {3 \operatorname {asinh}^{2}{\left (a x \right )}}{32 a^{4}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((x**4*asinh(a*x)**2/4 + x**4/32 - x**3*sqrt(a**2*x**2 + 1)*asinh(a*x)/(8*a) - 3*x**2/(32*a**2) + 3*x
*sqrt(a**2*x**2 + 1)*asinh(a*x)/(16*a**3) - 3*asinh(a*x)**2/(32*a**4), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14 \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\frac {1}{4} \, x^{4} \operatorname {arsinh}\left (a x\right )^{2} + \frac {1}{32} \, {\left (\frac {x^{4}}{a^{2}} - \frac {3 \, x^{2}}{a^{4}} + \frac {3 \, \log \left (a x + \sqrt {a^{2} x^{2} + 1}\right )^{2}}{a^{6}}\right )} a^{2} - \frac {1}{16} \, {\left (\frac {2 \, \sqrt {a^{2} x^{2} + 1} x^{3}}{a^{2}} - \frac {3 \, \sqrt {a^{2} x^{2} + 1} x}{a^{4}} + \frac {3 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} a \operatorname {arsinh}\left (a x\right ) \]

[In]

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

[Out]

1/4*x^4*arcsinh(a*x)^2 + 1/32*(x^4/a^2 - 3*x^2/a^4 + 3*log(a*x + sqrt(a^2*x^2 + 1))^2/a^6)*a^2 - 1/16*(2*sqrt(
a^2*x^2 + 1)*x^3/a^2 - 3*sqrt(a^2*x^2 + 1)*x/a^4 + 3*arcsinh(a*x)/a^5)*a*arcsinh(a*x)

Giac [F(-2)]

Exception generated. \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(x^3*arcsinh(a*x)^2,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(-1)]

Timed out. \[ \int x^3 \text {arcsinh}(a x)^2 \, dx=\int x^3\,{\mathrm {asinh}\left (a\,x\right )}^2 \,d x \]

[In]

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

[Out]

int(x^3*asinh(a*x)^2, x)