3.2.34 \(\int \text {ArcSin}(a+b x)^2 \, dx\) [134]

Optimal. Leaf size=47 \[ -2 x+\frac {2 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)}{b}+\frac {(a+b x) \text {ArcSin}(a+b x)^2}{b} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4887, 4715, 4767, 8} \begin {gather*} \frac {(a+b x) \text {ArcSin}(a+b x)^2}{b}+\frac {2 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)}{b}-2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^2,x]

[Out]

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

Rule 8

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

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

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Mathematica [A]
time = 0.02, size = 49, normalized size = 1.04 \begin {gather*} \frac {-2 (a+b x)+2 \sqrt {1-(a+b x)^2} \text {ArcSin}(a+b x)+(a+b x) \text {ArcSin}(a+b x)^2}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^2,x]

[Out]

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

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Maple [A]
time = 0.07, size = 48, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{b}\)	\(48\)
default	\(\frac {\arcsin \left (b x +a \right )^{2} \left (b x +a \right )-2 b x -2 a +2 \arcsin \left (b x +a \right ) \sqrt {1-\left (b x +a \right )^{2}}}{b}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(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))

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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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.97, size = 53, normalized size = 1.13 \begin {gather*} \frac {{\left (b x + a\right )} \arcsin \left (b x + a\right )^{2} - 2 \, b x + 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.11, size = 63, normalized size = 1.34 \begin {gather*} \begin {cases} \frac {a \operatorname {asin}^{2}{\left (a + b x \right )}}{b} + x \operatorname {asin}^{2}{\left (a + b x \right )} - 2 x + \frac {2 \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname {asin}{\left (a + b x \right )}}{b} & \text {for}\: b \neq 0 \\x \operatorname {asin}^{2}{\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)**2,x)

[Out]

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

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Giac [A]
time = 0.40, size = 52, normalized size = 1.11 \begin {gather*} \frac {{\left (b x + a\right )} \arcsin \left (b x + a\right )^{2}}{b} + \frac {2 \, \sqrt {-{\left (b x + a\right )}^{2} + 1} \arcsin \left (b x + a\right )}{b} - \frac {2 \, {\left (b x + a\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.25, size = 44, normalized size = 0.94 \begin {gather*} \frac {\left ({\mathrm {asin}\left (a+b\,x\right )}^2-2\right )\,\left (a+b\,x\right )}{b}+\frac {2\,\mathrm {asin}\left (a+b\,x\right )\,\sqrt {1-{\left (a+b\,x\right )}^2}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a + b*x)^2,x)

[Out]

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

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