3.2.40 \(\int \text {ArcSin}(a+b x)^3 \, dx\) [140]

Optimal. Leaf size=82 \[ -\frac {6 \sqrt {1-(a+b x)^2}}{b}-\frac {6 (a+b x) \text {ArcSin}(a+b x)}{b}+\frac {3 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)^2}{b}+\frac {(a+b x) \text {ArcSin}(a+b x)^3}{b} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4887, 4715, 4767, 267} \begin {gather*} \frac {(a+b x) \text {ArcSin}(a+b x)^3}{b}+\frac {3 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)^2}{b}-\frac {6 (a+b x) \text {ArcSin}(a+b x)}{b}-\frac {6 \sqrt {1-(a+b x)^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^3,x]

[Out]

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

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 4715

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

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 4887

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcSin[x])^n, x],
x, c + d*x], x] /; FreeQ[{a, b, c, d, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 74, normalized size = 0.90 \begin {gather*} \frac {-6 \sqrt {1-(a+b x)^2}-6 (a+b x) \text {ArcSin}(a+b x)+3 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)^2+(a+b x) \text {ArcSin}(a+b x)^3}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^3,x]

[Out]

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

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Maple [A]
time = 0.06, size = 71, normalized size = 0.87

method	result	size
derivativedivides	\(\frac {\arcsin \left (b x +a \right )^{3} \left (b x +a \right )+3 \arcsin \left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}-6 \sqrt {1-\left (b x +a \right )^{2}}-6 \left (b x +a \right ) \arcsin \left (b x +a \right )}{b}\)	\(71\)
default	\(\frac {\arcsin \left (b x +a \right )^{3} \left (b x +a \right )+3 \arcsin \left (b x +a \right )^{2} \sqrt {1-\left (b x +a \right )^{2}}-6 \sqrt {1-\left (b x +a \right )^{2}}-6 \left (b x +a \right ) \arcsin \left (b x +a \right )}{b}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

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

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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,x, algorithm="maxima")

[Out]

x*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^3 + 3*b*integrate(sqrt(b*x + a + 1)*sqrt(-b*x - a + 1
)*x*arctan2(b*x + a, sqrt(b*x + a + 1)*sqrt(-b*x - a + 1))^2/(b^2*x^2 + 2*a*b*x + a^2 - 1), x)

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Fricas [A]
time = 1.97, size = 66, normalized size = 0.80 \begin {gather*} \frac {{\left (b x + a\right )} \arcsin \left (b x + a\right )^{3} - 6 \, {\left (b x + a\right )} \arcsin \left (b x + a\right ) + 3 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (\arcsin \left (b x + a\right )^{2} - 2\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.15, size = 109, normalized size = 1.33 \begin {gather*} \begin {cases} \frac {a \operatorname {asin}^{3}{\left (a + b x \right )}}{b} - \frac {6 a \operatorname {asin}{\left (a + b x \right )}}{b} + x \operatorname {asin}^{3}{\left (a + b x \right )} - 6 x \operatorname {asin}{\left (a + b x \right )} + \frac {3 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}^{2}{\left (a + b x \right )}}{b} - \frac {6 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}{b} & \text {for}\: b \neq 0 \\x \operatorname {asin}^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*asin(a + b*x)**3/b - 6*a*asin(a + b*x)/b + x*asin(a + b*x)**3 - 6*x*asin(a + b*x) + 3*sqrt(-a**2
- 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)**2/b - 6*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)/b, Ne(b, 0)), (x*asin(
a)**3, True))

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Giac [A]
time = 0.38, size = 78, normalized size = 0.95 \begin {gather*} \frac {{\left (b x + a\right )} \arcsin \left (b x + a\right )^{3}}{b} + \frac {3 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} \arcsin \left (b x + a\right )^{2}}{b} - \frac {6 \, {\left (b x + a\right )} \arcsin \left (b x + a\right )}{b} - \frac {6 \, \sqrt {-{\left (b x + a\right )}^{2} + 1}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.25, size = 59, normalized size = 0.72 \begin {gather*} \frac {\left (3\,{\mathrm {asin}\left (a+b\,x\right )}^2-6\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{b}-\frac {\left (6\,\mathrm {asin}\left (a+b\,x\right )-{\mathrm {asin}\left (a+b\,x\right )}^3\right )\,\left (a+b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^3,x)

[Out]

((3*asin(a + b*x)^2 - 6)*(1 - (a + b*x)^2)^(1/2))/b - ((6*asin(a + b*x) - asin(a + b*x)^3)*(a + b*x))/b

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