\(\int x^4 \arcsin (a x)^2 \, dx\) [12]

Optimal result

Integrand size = 10, antiderivative size = 120 \[ \int x^4 \arcsin (a x)^2 \, dx=-\frac {16 x}{75 a^4}-\frac {8 x^3}{225 a^2}-\frac {2 x^5}{125}+\frac {16 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^5}+\frac {8 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^3}+\frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}+\frac {1}{5} x^5 \arcsin (a x)^2 \]

[Out]

-16/75*x/a^4-8/225*x^3/a^2-2/125*x^5+1/5*x^5*arcsin(a*x)^2+16/75*arcsin(a*x)*(-a^2*x^2+1)^(1/2)/a^5+8/75*x^2*a
rcsin(a*x)*(-a^2*x^2+1)^(1/2)/a^3+2/25*x^4*arcsin(a*x)*(-a^2*x^2+1)^(1/2)/a

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4723, 4795, 4767, 8, 30} \[ \int x^4 \arcsin (a x)^2 \, dx=-\frac {16 x}{75 a^4}+\frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}-\frac {8 x^3}{225 a^2}+\frac {16 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^5}+\frac {8 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^3}+\frac {1}{5} x^5 \arcsin (a x)^2-\frac {2 x^5}{125} \]

[In]

Int[x^4*ArcSin[a*x]^2,x]

[Out]

(-16*x)/(75*a^4) - (8*x^3)/(225*a^2) - (2*x^5)/125 + (16*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(75*a^5) + (8*x^2*Sqrt
[1 - a^2*x^2]*ArcSin[a*x])/(75*a^3) + (2*x^4*Sqrt[1 - a^2*x^2]*ArcSin[a*x])/(25*a) + (x^5*ArcSin[a*x]^2)/5

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

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

Rule 4723

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSi
n[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSin[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 4767

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

Rule 4795

Int[((a_.) + ArcSin[(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*ArcSin[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*ArcSin[c*x])^n, x], x] + Dist[b*f*(n/(c*(m + 2*p + 1)))*S
imp[(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*ArcSin[c*x])^(n - 1), x],
x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && IGtQ[m, 1] && NeQ[m + 2*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5} x^5 \arcsin (a x)^2-\frac {1}{5} (2 a) \int \frac {x^5 \arcsin (a x)}{\sqrt {1-a^2 x^2}} \, dx \\ & = \frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}+\frac {1}{5} x^5 \arcsin (a x)^2-\frac {2 \int x^4 \, dx}{25}-\frac {8 \int \frac {x^3 \arcsin (a x)}{\sqrt {1-a^2 x^2}} \, dx}{25 a} \\ & = -\frac {2 x^5}{125}+\frac {8 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^3}+\frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}+\frac {1}{5} x^5 \arcsin (a x)^2-\frac {16 \int \frac {x \arcsin (a x)}{\sqrt {1-a^2 x^2}} \, dx}{75 a^3}-\frac {8 \int x^2 \, dx}{75 a^2} \\ & = -\frac {8 x^3}{225 a^2}-\frac {2 x^5}{125}+\frac {16 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^5}+\frac {8 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^3}+\frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}+\frac {1}{5} x^5 \arcsin (a x)^2-\frac {16 \int 1 \, dx}{75 a^4} \\ & = -\frac {16 x}{75 a^4}-\frac {8 x^3}{225 a^2}-\frac {2 x^5}{125}+\frac {16 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^5}+\frac {8 x^2 \sqrt {1-a^2 x^2} \arcsin (a x)}{75 a^3}+\frac {2 x^4 \sqrt {1-a^2 x^2} \arcsin (a x)}{25 a}+\frac {1}{5} x^5 \arcsin (a x)^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.68 \[ \int x^4 \arcsin (a x)^2 \, dx=\frac {-2 a x \left (120+20 a^2 x^2+9 a^4 x^4\right )+30 \sqrt {1-a^2 x^2} \left (8+4 a^2 x^2+3 a^4 x^4\right ) \arcsin (a x)+225 a^5 x^5 \arcsin (a x)^2}{1125 a^5} \]

[In]

Integrate[x^4*ArcSin[a*x]^2,x]

[Out]

(-2*a*x*(120 + 20*a^2*x^2 + 9*a^4*x^4) + 30*Sqrt[1 - a^2*x^2]*(8 + 4*a^2*x^2 + 3*a^4*x^4)*ArcSin[a*x] + 225*a^
5*x^5*ArcSin[a*x]^2)/(1125*a^5)

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63

method	result	size
derivativedivides	\(\frac {\frac {a^{5} x^{5} \arcsin \left (a x \right )^{2}}{5}+\frac {2 \arcsin \left (a x \right ) \left (3 a^{4} x^{4}+4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{75}-\frac {2 a^{5} x^{5}}{125}-\frac {8 a^{3} x^{3}}{225}-\frac {16 a x}{75}}{a^{5}}\)	\(76\)
default	\(\frac {\frac {a^{5} x^{5} \arcsin \left (a x \right )^{2}}{5}+\frac {2 \arcsin \left (a x \right ) \left (3 a^{4} x^{4}+4 a^{2} x^{2}+8\right ) \sqrt {-a^{2} x^{2}+1}}{75}-\frac {2 a^{5} x^{5}}{125}-\frac {8 a^{3} x^{3}}{225}-\frac {16 a x}{75}}{a^{5}}\)	\(76\)

[In]

int(x^4*arcsin(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^5*(1/5*a^5*x^5*arcsin(a*x)^2+2/75*arcsin(a*x)*(3*a^4*x^4+4*a^2*x^2+8)*(-a^2*x^2+1)^(1/2)-2/125*a^5*x^5-8/2
25*a^3*x^3-16/75*a*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int x^4 \arcsin (a x)^2 \, dx=\frac {225 \, a^{5} x^{5} \arcsin \left (a x\right )^{2} - 18 \, a^{5} x^{5} - 40 \, a^{3} x^{3} + 30 \, {\left (3 \, a^{4} x^{4} + 4 \, a^{2} x^{2} + 8\right )} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right ) - 240 \, a x}{1125 \, a^{5}} \]

[In]

integrate(x^4*arcsin(a*x)^2,x, algorithm="fricas")

[Out]

1/1125*(225*a^5*x^5*arcsin(a*x)^2 - 18*a^5*x^5 - 40*a^3*x^3 + 30*(3*a^4*x^4 + 4*a^2*x^2 + 8)*sqrt(-a^2*x^2 + 1
)*arcsin(a*x) - 240*a*x)/a^5

Sympy [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.95 \[ \int x^4 \arcsin (a x)^2 \, dx=\begin {cases} \frac {x^{5} \operatorname {asin}^{2}{\left (a x \right )}}{5} - \frac {2 x^{5}}{125} + \frac {2 x^{4} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{25 a} - \frac {8 x^{3}}{225 a^{2}} + \frac {8 x^{2} \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{75 a^{3}} - \frac {16 x}{75 a^{4}} + \frac {16 \sqrt {- a^{2} x^{2} + 1} \operatorname {asin}{\left (a x \right )}}{75 a^{5}} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} \]

[In]

integrate(x**4*asin(a*x)**2,x)

[Out]

Piecewise((x**5*asin(a*x)**2/5 - 2*x**5/125 + 2*x**4*sqrt(-a**2*x**2 + 1)*asin(a*x)/(25*a) - 8*x**3/(225*a**2)
+ 8*x**2*sqrt(-a**2*x**2 + 1)*asin(a*x)/(75*a**3) - 16*x/(75*a**4) + 16*sqrt(-a**2*x**2 + 1)*asin(a*x)/(75*a*
*5), Ne(a, 0)), (0, True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.85 \[ \int x^4 \arcsin (a x)^2 \, dx=\frac {1}{5} \, x^{5} \arcsin \left (a x\right )^{2} + \frac {2}{75} \, {\left (\frac {3 \, \sqrt {-a^{2} x^{2} + 1} x^{4}}{a^{2}} + \frac {4 \, \sqrt {-a^{2} x^{2} + 1} x^{2}}{a^{4}} + \frac {8 \, \sqrt {-a^{2} x^{2} + 1}}{a^{6}}\right )} a \arcsin \left (a x\right ) - \frac {2 \, {\left (9 \, a^{4} x^{5} + 20 \, a^{2} x^{3} + 120 \, x\right )}}{1125 \, a^{4}} \]

[In]

integrate(x^4*arcsin(a*x)^2,x, algorithm="maxima")

[Out]

1/5*x^5*arcsin(a*x)^2 + 2/75*(3*sqrt(-a^2*x^2 + 1)*x^4/a^2 + 4*sqrt(-a^2*x^2 + 1)*x^2/a^4 + 8*sqrt(-a^2*x^2 +
1)/a^6)*a*arcsin(a*x) - 2/1125*(9*a^4*x^5 + 20*a^2*x^3 + 120*x)/a^4

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.41 \[ \int x^4 \arcsin (a x)^2 \, dx=\frac {{\left (a^{2} x^{2} - 1\right )}^{2} x \arcsin \left (a x\right )^{2}}{5 \, a^{4}} + \frac {2 \, {\left (a^{2} x^{2} - 1\right )} x \arcsin \left (a x\right )^{2}}{5 \, a^{4}} - \frac {2 \, {\left (a^{2} x^{2} - 1\right )}^{2} x}{125 \, a^{4}} + \frac {x \arcsin \left (a x\right )^{2}}{5 \, a^{4}} + \frac {2 \, {\left (a^{2} x^{2} - 1\right )}^{2} \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )}{25 \, a^{5}} - \frac {76 \, {\left (a^{2} x^{2} - 1\right )} x}{1125 \, a^{4}} - \frac {4 \, {\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} \arcsin \left (a x\right )}{15 \, a^{5}} - \frac {298 \, x}{1125 \, a^{4}} + \frac {2 \, \sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )}{5 \, a^{5}} \]

[In]

integrate(x^4*arcsin(a*x)^2,x, algorithm="giac")

[Out]

1/5*(a^2*x^2 - 1)^2*x*arcsin(a*x)^2/a^4 + 2/5*(a^2*x^2 - 1)*x*arcsin(a*x)^2/a^4 - 2/125*(a^2*x^2 - 1)^2*x/a^4
+ 1/5*x*arcsin(a*x)^2/a^4 + 2/25*(a^2*x^2 - 1)^2*sqrt(-a^2*x^2 + 1)*arcsin(a*x)/a^5 - 76/1125*(a^2*x^2 - 1)*x/
a^4 - 4/15*(-a^2*x^2 + 1)^(3/2)*arcsin(a*x)/a^5 - 298/1125*x/a^4 + 2/5*sqrt(-a^2*x^2 + 1)*arcsin(a*x)/a^5

Mupad [F(-1)]

Timed out. \[ \int x^4 \arcsin (a x)^2 \, dx=\int x^4\,{\mathrm {asin}\left (a\,x\right )}^2 \,d x \]

[In]

int(x^4*asin(a*x)^2,x)

[Out]

int(x^4*asin(a*x)^2, x)