3.4.13 \(\int \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)^3 \, dx\) [313]

Optimal. Leaf size=135 \[ \frac {3 (a+b x)^2}{8 b}-\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)}{4 b}+\frac {3 \text {ArcSin}(a+b x)^2}{8 b}-\frac {3 (a+b x)^2 \text {ArcSin}(a+b x)^2}{4 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)^3}{2 b}+\frac {\text {ArcSin}(a+b x)^4}{8 b} \]

[Out]

3/8*(b*x+a)^2/b+3/8*arcsin(b*x+a)^2/b-3/4*(b*x+a)^2*arcsin(b*x+a)^2/b+1/8*arcsin(b*x+a)^4/b-3/4*(b*x+a)*arcsin
(b*x+a)*(1-(b*x+a)^2)^(1/2)/b+1/2*(b*x+a)*arcsin(b*x+a)^3*(1-(b*x+a)^2)^(1/2)/b

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Rubi [A]
time = 0.13, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4891, 4741, 4737, 4723, 4795, 30} \begin {gather*} \frac {\text {ArcSin}(a+b x)^4}{8 b}+\frac {(a+b x) \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)^3}{2 b}-\frac {3 (a+b x)^2 \text {ArcSin}(a+b x)^2}{4 b}+\frac {3 \text {ArcSin}(a+b x)^2}{8 b}-\frac {3 (a+b x) \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)}{4 b}+\frac {3 (a+b x)^2}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^3,x]

[Out]

(3*(a + b*x)^2)/(8*b) - (3*(a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x])/(4*b) + (3*ArcSin[a + b*x]^2)/(8*b
) - (3*(a + b*x)^2*ArcSin[a + b*x]^2)/(4*b) + ((a + b*x)*Sqrt[1 - (a + b*x)^2]*ArcSin[a + b*x]^3)/(2*b) + ArcS
in[a + b*x]^4/(8*b)

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 4737

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

Rule 4741

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

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]

Rule 4891

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-C/d^2 + (C/d^2)*x^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B
, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 133, normalized size = 0.99 \begin {gather*} \frac {3 b x (2 a+b x)-6 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)-3 \left (-1+2 a^2+4 a b x+2 b^2 x^2\right ) \text {ArcSin}(a+b x)^2+4 (a+b x) \sqrt {1-a^2-2 a b x-b^2 x^2} \text {ArcSin}(a+b x)^3+\text {ArcSin}(a+b x)^4}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^3,x]

[Out]

(3*b*x*(2*a + b*x) - 6*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x] - 3*(-1 + 2*a^2 + 4*a*b*x +
2*b^2*x^2)*ArcSin[a + b*x]^2 + 4*(a + b*x)*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]*ArcSin[a + b*x]^3 + ArcSin[a + b
*x]^4)/(8*b)

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Maple [A]
time = 0.75, size = 215, normalized size = 1.59

method	result	size
default	\(\frac {4 \arcsin \left (b x +a \right )^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x -6 \arcsin \left (b x +a \right )^{2} b^{2} x^{2}+4 \arcsin \left (b x +a \right )^{3} \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a -12 \arcsin \left (b x +a \right )^{2} a b x +\arcsin \left (b x +a \right )^{4}-6 \arcsin \left (b x +a \right )^{2} a^{2}-6 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, b x +3 b^{2} x^{2}-6 \arcsin \left (b x +a \right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +6 a b x +3 \arcsin \left (b x +a \right )^{2}+3 a^{2}}{8 b}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8*(4*arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*b*x-6*arcsin(b*x+a)^2*b^2*x^2+4*arcsin(b*x+a)^3*(-b^2*x^
2-2*a*b*x-a^2+1)^(1/2)*a-12*arcsin(b*x+a)^2*a*b*x+arcsin(b*x+a)^4-6*arcsin(b*x+a)^2*a^2-6*arcsin(b*x+a)*(-b^2*
x^2-2*a*b*x-a^2+1)^(1/2)*b*x+3*b^2*x^2-6*arcsin(b*x+a)*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)*a+6*a*b*x+3*arcsin(b*x+a
)^2+3*a^2)/b

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a)^3, x)

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Fricas [A]
time = 3.69, size = 110, normalized size = 0.81 \begin {gather*} \frac {3 \, b^{2} x^{2} + \arcsin \left (b x + a\right )^{4} + 6 \, a b x - 3 \, {\left (2 \, b^{2} x^{2} + 4 \, a b x + 2 \, a^{2} - 1\right )} \arcsin \left (b x + a\right )^{2} + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (2 \, {\left (b x + a\right )} \arcsin \left (b x + a\right )^{3} - 3 \, {\left (b x + a\right )} \arcsin \left (b x + a\right )\right )}}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2),x, algorithm="fricas")

[Out]

1/8*(3*b^2*x^2 + arcsin(b*x + a)^4 + 6*a*b*x - 3*(2*b^2*x^2 + 4*a*b*x + 2*a^2 - 1)*arcsin(b*x + a)^2 + 2*sqrt(
-b^2*x^2 - 2*a*b*x - a^2 + 1)*(2*(b*x + a)*arcsin(b*x + a)^3 - 3*(b*x + a)*arcsin(b*x + a)))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- \left (a + b x - 1\right ) \left (a + b x + 1\right )} \operatorname {asin}^{3}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**3*(-b**2*x**2-2*a*b*x-a**2+1)**(1/2),x)

[Out]

Integral(sqrt(-(a + b*x - 1)*(a + b*x + 1))*asin(a + b*x)**3, x)

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Giac [A]
time = 0.48, size = 162, normalized size = 1.20 \begin {gather*} \frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )^{3}}{2 \, b} + \frac {\arcsin \left (b x + a\right )^{4}}{8 \, b} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \arcsin \left (b x + a\right )^{2}}{4 \, b} - \frac {3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (b x + a\right )} \arcsin \left (b x + a\right )}{4 \, b} - \frac {3 \, \arcsin \left (b x + a\right )^{2}}{8 \, b} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )}}{8 \, b} + \frac {3}{16 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a)^3/b + 1/8*arcsin(b*x + a)^4/b - 3/4*(b^2*x^2
+ 2*a*b*x + a^2 - 1)*arcsin(b*x + a)^2/b - 3/4*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*(b*x + a)*arcsin(b*x + a)/b
- 3/8*arcsin(b*x + a)^2/b + 3/8*(b^2*x^2 + 2*a*b*x + a^2 - 1)/b + 3/16/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^3*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2),x)

[Out]

int(asin(a + b*x)^3*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2), x)

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