3.131 \(\int x^3 \sin ^{-1}(a+b x)^2 \, dx\)

Optimal. Leaf size=343 \[ -\frac {a^4 \sin ^{-1}(a+b x)^2}{4 b^4}-\frac {2 a^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^4}+\frac {2 a^3 x}{b^3}-\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {3 a^2 \sin ^{-1}(a+b x)^2}{4 b^4}+\frac {3 a^2 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^3}{9 b^4}-\frac {(a+b x)^4}{32 b^4}-\frac {3 (a+b x)^2}{32 b^4}-\frac {2 a (a+b x)^2 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {4 a \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {3 \sin ^{-1}(a+b x)^2}{32 b^4}+\frac {(a+b x)^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b^4}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{16 b^4}+\frac {4 a x}{3 b^3}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2 \]

[Out]

4/3*a*x/b^3+2*a^3*x/b^3-3/32*(b*x+a)^2/b^4-3/4*a^2*(b*x+a)^2/b^4+2/9*a*(b*x+a)^3/b^4-1/32*(b*x+a)^4/b^4-3/32*a
rcsin(b*x+a)^2/b^4-3/4*a^2*arcsin(b*x+a)^2/b^4-1/4*a^4*arcsin(b*x+a)^2/b^4+1/4*x^4*arcsin(b*x+a)^2-4/3*a*arcsi
n(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^4-2*a^3*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^4+3/16*(b*x+a)*arcsin(b*x+a)*(1-(b*
x+a)^2)^(1/2)/b^4+3/2*a^2*(b*x+a)*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^4-2/3*a*(b*x+a)^2*arcsin(b*x+a)*(1-(b*x+
a)^2)^(1/2)/b^4+1/8*(b*x+a)^3*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)/b^4

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 343, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {4805, 4743, 4763, 4641, 4677, 8, 4707, 30} \[ \frac {2 a^3 x}{b^3}-\frac {3 a^2 (a+b x)^2}{4 b^4}-\frac {a^4 \sin ^{-1}(a+b x)^2}{4 b^4}-\frac {2 a^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^4}-\frac {3 a^2 \sin ^{-1}(a+b x)^2}{4 b^4}+\frac {3 a^2 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b^4}+\frac {2 a (a+b x)^3}{9 b^4}+\frac {4 a x}{3 b^3}-\frac {(a+b x)^4}{32 b^4}-\frac {3 (a+b x)^2}{32 b^4}-\frac {2 a (a+b x)^2 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {4 a \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {3 \sin ^{-1}(a+b x)^2}{32 b^4}+\frac {(a+b x)^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b^4}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{16 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSin[a + b*x]^2,x]

[Out]

(4*a*x)/(3*b^3) + (2*a^3*x)/b^3 - (3*(a + b*x)^2)/(32*b^4) - (3*a^2*(a + b*x)^2)/(4*b^4) + (2*a*(a + b*x)^3)/(
9*b^4) - (a + b*x)^4/(32*b^4) - (4*a*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(3*b^4) - (2*a^3*Sqrt[1 - (a + b*x
)^2]*ArcSin[a + b*x])/b^4 + (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(16*b^4) + (3*a^2*(a + b*x)*Sq
rt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(2*b^4) - (2*a*(a + b*x)^2*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(3*b^4)
+ ((a + b*x)^3*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(8*b^4) - (3*ArcSin[a + b*x]^2)/(32*b^4) - (3*a^2*ArcSi
n[a + b*x]^2)/(4*b^4) - (a^4*ArcSin[a + b*x]^2)/(4*b^4) + (x^4*ArcSin[a + b*x]^2)/4

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 4641

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

Rule 4677

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*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[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 4707

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

Rule 4743

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

Rule 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g},
x] && EqQ[c^2*d + e, 0] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && (m == 1 || p > 0 ||
(n == 1 && p > -1) || (m == 2 && p < -2))

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int x^3 \sin ^{-1}(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \sin ^{-1}(x)^2 \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2-\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2-\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {a^4 \sin ^{-1}(x)}{b^4 \sqrt {1-x^2}}-\frac {4 a^3 x \sin ^{-1}(x)}{b^4 \sqrt {1-x^2}}+\frac {6 a^2 x^2 \sin ^{-1}(x)}{b^4 \sqrt {1-x^2}}-\frac {4 a x^3 \sin ^{-1}(x)}{b^4 \sqrt {1-x^2}}+\frac {x^4 \sin ^{-1}(x)}{b^4 \sqrt {1-x^2}}\right ) \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int \frac {x^4 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b^4}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {x^3 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {x^2 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{b^4}+\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \frac {x \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{b^4}-\frac {a^4 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b^4}\\ &=-\frac {2 a^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^4}+\frac {3 a^2 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b^4}-\frac {2 a (a+b x)^2 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}+\frac {(a+b x)^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b^4}-\frac {a^4 \sin ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2-\frac {\operatorname {Subst}\left (\int x^3 \, dx,x,a+b x\right )}{8 b^4}-\frac {3 \operatorname {Subst}\left (\int \frac {x^2 \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{8 b^4}+\frac {(2 a) \operatorname {Subst}\left (\int x^2 \, dx,x,a+b x\right )}{3 b^4}+\frac {(4 a) \operatorname {Subst}\left (\int \frac {x \sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{3 b^4}-\frac {\left (3 a^2\right ) \operatorname {Subst}(\int x \, dx,x,a+b x)}{2 b^4}-\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{2 b^4}+\frac {\left (2 a^3\right ) \operatorname {Subst}(\int 1 \, dx,x,a+b x)}{b^4}\\ &=\frac {2 a^3 x}{b^3}-\frac {3 a^2 (a+b x)^2}{4 b^4}+\frac {2 a (a+b x)^3}{9 b^4}-\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {2 a^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{16 b^4}+\frac {3 a^2 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b^4}-\frac {2 a (a+b x)^2 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}+\frac {(a+b x)^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b^4}-\frac {3 a^2 \sin ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sin ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2-\frac {3 \operatorname {Subst}(\int x \, dx,x,a+b x)}{16 b^4}-\frac {3 \operatorname {Subst}\left (\int \frac {\sin ^{-1}(x)}{\sqrt {1-x^2}} \, dx,x,a+b x\right )}{16 b^4}+\frac {(4 a) \operatorname {Subst}(\int 1 \, dx,x,a+b x)}{3 b^4}\\ &=\frac {4 a x}{3 b^3}+\frac {2 a^3 x}{b^3}-\frac {3 (a+b x)^2}{32 b^4}-\frac {3 a^2 (a+b x)^2}{4 b^4}+\frac {2 a (a+b x)^3}{9 b^4}-\frac {(a+b x)^4}{32 b^4}-\frac {4 a \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}-\frac {2 a^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^4}+\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{16 b^4}+\frac {3 a^2 (a+b x) \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{2 b^4}-\frac {2 a (a+b x)^2 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{3 b^4}+\frac {(a+b x)^3 \sqrt {1-(a+b x)^2} \sin ^{-1}(a+b x)}{8 b^4}-\frac {3 \sin ^{-1}(a+b x)^2}{32 b^4}-\frac {3 a^2 \sin ^{-1}(a+b x)^2}{4 b^4}-\frac {a^4 \sin ^{-1}(a+b x)^2}{4 b^4}+\frac {1}{4} x^4 \sin ^{-1}(a+b x)^2\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.23, size = 148, normalized size = 0.43 \[ \frac {-9 \left (8 a^4+24 a^2-8 b^4 x^4+3\right ) \sin ^{-1}(a+b x)^2+b x \left (300 a^3-78 a^2 b x+a \left (28 b^2 x^2+330\right )-9 b x \left (b^2 x^2+3\right )\right )-6 \sqrt {-a^2-2 a b x-b^2 x^2+1} \left (50 a^3-26 a^2 b x+14 a b^2 x^2+55 a-6 b^3 x^3-9 b x\right ) \sin ^{-1}(a+b x)}{288 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSin[a + b*x]^2,x]

[Out]

(b*x*(300*a^3 - 78*a^2*b*x - 9*b*x*(3 + b^2*x^2) + a*(330 + 28*b^2*x^2)) - 6*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]
*(55*a + 50*a^3 - 9*b*x - 26*a^2*b*x + 14*a*b^2*x^2 - 6*b^3*x^3)*ArcSin[a + b*x] - 9*(3 + 24*a^2 + 8*a^4 - 8*b
^4*x^4)*ArcSin[a + b*x]^2)/(288*b^4)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 147, normalized size = 0.43 \[ -\frac {9 \, b^{4} x^{4} - 28 \, a b^{3} x^{3} + 3 \, {\left (26 \, a^{2} + 9\right )} b^{2} x^{2} - 30 \, {\left (10 \, a^{3} + 11 \, a\right )} b x - 9 \, {\left (8 \, b^{4} x^{4} - 8 \, a^{4} - 24 \, a^{2} - 3\right )} \arcsin \left (b x + a\right )^{2} - 6 \, {\left (6 \, b^{3} x^{3} - 14 \, a b^{2} x^{2} - 50 \, a^{3} + {\left (26 \, a^{2} + 9\right )} b x - 55 \, a\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )}{288 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/288*(9*b^4*x^4 - 28*a*b^3*x^3 + 3*(26*a^2 + 9)*b^2*x^2 - 30*(10*a^3 + 11*a)*b*x - 9*(8*b^4*x^4 - 8*a^4 - 24
*a^2 - 3)*arcsin(b*x + a)^2 - 6*(6*b^3*x^3 - 14*a*b^2*x^2 - 50*a^3 + (26*a^2 + 9)*b*x - 55*a)*sqrt(-b^2*x^2 -
2*a*b*x - a^2 + 1)*arcsin(b*x + a))/b^4

________________________________________________________________________________________

giac [A]  time = 2.80, size = 440, normalized size = 1.28 \[ -\frac {{\left (b x + a\right )} a^{3} \arcsin \left (b x + a\right )^{2}}{b^{4}} - \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )} a \arcsin \left (b x + a\right )^{2}}{b^{4}} + \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a^{2} \arcsin \left (b x + a\right )^{2}}{2 \, b^{4}} + \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} a^{2} \arcsin \left (b x + a\right )}{2 \, b^{4}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a^{3} \arcsin \left (b x + a\right )}{b^{4}} + \frac {2 \, {\left (b x + a\right )} a^{3}}{b^{4}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )}^{2} \arcsin \left (b x + a\right )^{2}}{4 \, b^{4}} - \frac {{\left (b x + a\right )} a \arcsin \left (b x + a\right )^{2}}{b^{4}} + \frac {3 \, a^{2} \arcsin \left (b x + a\right )^{2}}{4 \, b^{4}} - \frac {{\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} {\left (b x + a\right )} \arcsin \left (b x + a\right )}{8 \, b^{4}} + \frac {2 \, {\left (-{\left (b x + a\right )}^{2} + 1\right )}^{\frac {3}{2}} a \arcsin \left (b x + a\right )}{3 \, b^{4}} + \frac {2 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} {\left (b x + a\right )} a}{9 \, b^{4}} - \frac {3 \, {\left ({\left (b x + a\right )}^{2} - 1\right )} a^{2}}{4 \, b^{4}} + \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )^{2}}{2 \, b^{4}} + \frac {5 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )}{16 \, b^{4}} - \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} a \arcsin \left (b x + a\right )}{b^{4}} - \frac {{\left ({\left (b x + a\right )}^{2} - 1\right )}^{2}}{32 \, b^{4}} + \frac {14 \, {\left (b x + a\right )} a}{9 \, b^{4}} - \frac {3 \, a^{2}}{8 \, b^{4}} + \frac {5 \, \arcsin \left (b x + a\right )^{2}}{32 \, b^{4}} - \frac {5 \, {\left ({\left (b x + a\right )}^{2} - 1\right )}}{32 \, b^{4}} - \frac {17}{256 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(b*x+a)^2,x, algorithm="giac")

[Out]

-(b*x + a)*a^3*arcsin(b*x + a)^2/b^4 - ((b*x + a)^2 - 1)*(b*x + a)*a*arcsin(b*x + a)^2/b^4 + 3/2*((b*x + a)^2
- 1)*a^2*arcsin(b*x + a)^2/b^4 + 3/2*sqrt(-(b*x + a)^2 + 1)*(b*x + a)*a^2*arcsin(b*x + a)/b^4 - 2*sqrt(-(b*x +
a)^2 + 1)*a^3*arcsin(b*x + a)/b^4 + 2*(b*x + a)*a^3/b^4 + 1/4*((b*x + a)^2 - 1)^2*arcsin(b*x + a)^2/b^4 - (b*
x + a)*a*arcsin(b*x + a)^2/b^4 + 3/4*a^2*arcsin(b*x + a)^2/b^4 - 1/8*(-(b*x + a)^2 + 1)^(3/2)*(b*x + a)*arcsin
(b*x + a)/b^4 + 2/3*(-(b*x + a)^2 + 1)^(3/2)*a*arcsin(b*x + a)/b^4 + 2/9*((b*x + a)^2 - 1)*(b*x + a)*a/b^4 - 3
/4*((b*x + a)^2 - 1)*a^2/b^4 + 1/2*((b*x + a)^2 - 1)*arcsin(b*x + a)^2/b^4 + 5/16*sqrt(-(b*x + a)^2 + 1)*(b*x
+ a)*arcsin(b*x + a)/b^4 - 2*sqrt(-(b*x + a)^2 + 1)*a*arcsin(b*x + a)/b^4 - 1/32*((b*x + a)^2 - 1)^2/b^4 + 14/
9*(b*x + a)*a/b^4 - 3/8*a^2/b^4 + 5/32*arcsin(b*x + a)^2/b^4 - 5/32*((b*x + a)^2 - 1)/b^4 - 17/256/b^4

________________________________________________________________________________________

maple [A]  time = 0.13, size = 435, normalized size = 1.27 \[ \frac {\frac {\arcsin \left (b x +a \right )^{2} \left (-1+\left (b x +a \right )^{2}\right )^{2}}{4}-\frac {\arcsin \left (b x +a \right ) \left (-2 \left (b x +a \right )^{3} \sqrt {1-\left (b x +a \right )^{2}}+5 \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+3 \arcsin \left (b x +a \right )\right )}{16}-\frac {5 \arcsin \left (b x +a \right )^{2}}{32}-\frac {\left (-1+\left (b x +a \right )^{2}\right )^{2}}{32}-\frac {5 \left (b x +a \right )^{2}}{32}-\frac {3}{32}+\frac {3 a^{2} \left (2 \arcsin \left (b x +a \right )^{2} \left (b x +a \right )^{2}+2 \sqrt {1-\left (b x +a \right )^{2}}\, \arcsin \left (b x +a \right ) \left (b x +a \right )-\arcsin \left (b x +a \right )^{2}-\left (b x +a \right )^{2}\right )}{4}-\frac {a \left (9 \arcsin \left (b x +a \right )^{2} \left (b x +a \right )^{3}+6 \sqrt {1-\left (b x +a \right )^{2}}\, \arcsin \left (b x +a \right ) \left (b x +a \right )^{2}-27 \arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 \left (b x +a \right )^{3}-42 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+42 b x +42 a \right )}{9}-a^{3} \left (\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}\right )+\frac {\arcsin \left (b x +a \right )^{2} \left (-1+\left (b x +a \right )^{2}\right )}{2}+\frac {\arcsin \left (b x +a \right ) \left (\left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}+\arcsin \left (b x +a \right )\right )}{2}-3 a \left (\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsin(b*x+a)^2,x)

[Out]

1/b^4*(1/4*arcsin(b*x+a)^2*(-1+(b*x+a)^2)^2-1/16*arcsin(b*x+a)*(-2*(b*x+a)^3*(1-(b*x+a)^2)^(1/2)+5*(b*x+a)*(1-
(b*x+a)^2)^(1/2)+3*arcsin(b*x+a))-5/32*arcsin(b*x+a)^2-1/32*(-1+(b*x+a)^2)^2-5/32*(b*x+a)^2-3/32+3/4*a^2*(2*ar
csin(b*x+a)^2*(b*x+a)^2+2*(1-(b*x+a)^2)^(1/2)*arcsin(b*x+a)*(b*x+a)-arcsin(b*x+a)^2-(b*x+a)^2)-1/9*a*(9*arcsin
(b*x+a)^2*(b*x+a)^3+6*(1-(b*x+a)^2)^(1/2)*arcsin(b*x+a)*(b*x+a)^2-27*arcsin(b*x+a)^2*(b*x+a)-2*(b*x+a)^3-42*ar
csin(b*x+a)*(1-(b*x+a)^2)^(1/2)+42*b*x+42*a)-a^3*(arcsin(b*x+a)^2*(b*x+a)-2*b*x-2*a+2*arcsin(b*x+a)*(1-(b*x+a)
^2)^(1/2))+1/2*arcsin(b*x+a)^2*(-1+(b*x+a)^2)+1/2*arcsin(b*x+a)*((b*x+a)*(1-(b*x+a)^2)^(1/2)+arcsin(b*x+a))-3*
a*(arcsin(b*x+a)^2*(b*x+a)-2*b*x-2*a+2*arcsin(b*x+a)*(1-(b*x+a)^2)^(1/2)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, x^{4} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )^{2} + b \int \frac {\sqrt {b x + a + 1} \sqrt {-b x - a + 1} x^{4} \arctan \left (b x + a, \sqrt {b x + a + 1} \sqrt {-b x - a + 1}\right )}{2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsin(b*x+a)^2,x, algorithm="maxima")

[Out]

1/4*x^4*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2 + b*integrate(1/2*sqrt(b*x + a + 1)*sqrt(-b*x
- a + 1)*x^4*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))/(b^2*x^2 + 2*a*b*x + a^2 - 1), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^3\,{\mathrm {asin}\left (a+b\,x\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*asin(a + b*x)^2,x)

[Out]

int(x^3*asin(a + b*x)^2, x)

________________________________________________________________________________________

sympy [A]  time = 2.91, size = 366, normalized size = 1.07 \[ \begin {cases} - \frac {a^{4} \operatorname {asin}^{2}{\left (a + b x \right )}}{4 b^{4}} + \frac {25 a^{3} x}{24 b^{3}} - \frac {25 a^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{24 b^{4}} - \frac {13 a^{2} x^{2}}{48 b^{2}} + \frac {13 a^{2} x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{24 b^{3}} - \frac {3 a^{2} \operatorname {asin}^{2}{\left (a + b x \right )}}{4 b^{4}} + \frac {7 a x^{3}}{72 b} - \frac {7 a x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{24 b^{2}} + \frac {55 a x}{48 b^{3}} - \frac {55 a \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{48 b^{4}} + \frac {x^{4} \operatorname {asin}^{2}{\left (a + b x \right )}}{4} - \frac {x^{4}}{32} + \frac {x^{3} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{8 b} - \frac {3 x^{2}}{32 b^{2}} + \frac {3 x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{16 b^{3}} - \frac {3 \operatorname {asin}^{2}{\left (a + b x \right )}}{32 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {asin}^{2}{\relax (a )}}{4} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asin(b*x+a)**2,x)

[Out]

Piecewise((-a**4*asin(a + b*x)**2/(4*b**4) + 25*a**3*x/(24*b**3) - 25*a**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 +
1)*asin(a + b*x)/(24*b**4) - 13*a**2*x**2/(48*b**2) + 13*a**2*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a +
b*x)/(24*b**3) - 3*a**2*asin(a + b*x)**2/(4*b**4) + 7*a*x**3/(72*b) - 7*a*x**2*sqrt(-a**2 - 2*a*b*x - b**2*x*
*2 + 1)*asin(a + b*x)/(24*b**2) + 55*a*x/(48*b**3) - 55*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/
(48*b**4) + x**4*asin(a + b*x)**2/4 - x**4/32 + x**3*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(8*b)
- 3*x**2/(32*b**2) + 3*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(16*b**3) - 3*asin(a + b*x)**2/(
32*b**4), Ne(b, 0)), (x**4*asin(a)**2/4, True))

________________________________________________________________________________________