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

Optimal. Leaf size=130 \[ -\frac{a^2 \sin ^{-1}(a+b x)^2}{2 b^2}-\frac{(a+b x)^2}{4 b^2}+\frac{\sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{2 b^2}-\frac{\sin ^{-1}(a+b x)^2}{4 b^2}-\frac{2 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^2+\frac{2 a x}{b} \]

[Out]

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

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Rubi [A]  time = 0.248699, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {4805, 4743, 4763, 4641, 4677, 8, 4707, 30} \[ -\frac{a^2 \sin ^{-1}(a+b x)^2}{2 b^2}-\frac{(a+b x)^2}{4 b^2}+\frac{\sqrt{1-(a+b x)^2} (a+b x) \sin ^{-1}(a+b x)}{2 b^2}-\frac{\sin ^{-1}(a+b x)^2}{4 b^2}-\frac{2 a \sqrt{1-(a+b x)^2} \sin ^{-1}(a+b x)}{b^2}+\frac{1}{2} x^2 \sin ^{-1}(a+b x)^2+\frac{2 a x}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*ArcSin[a + b*x]^2,x]

[Out]

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

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 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 8

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

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 30

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

Rubi steps

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

Mathematica [A]  time = 0.0901196, size = 83, normalized size = 0.64 \[ \frac{\left (-2 a^2+2 b^2 x^2-1\right ) \sin ^{-1}(a+b x)^2-2 (3 a-b x) \sqrt{-a^2-2 a b x-b^2 x^2+1} \sin ^{-1}(a+b x)+b x (6 a-b x)}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x*ArcSin[a + b*x]^2,x]

[Out]

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

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Maple [A]  time = 0.042, size = 124, normalized size = 1. \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( -1+ \left ( bx+a \right ) ^{2} \right ) }{2}}+{\frac{\arcsin \left ( bx+a \right ) }{2} \left ( \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}}+\arcsin \left ( bx+a \right ) \right ) }-{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( bx+a \right ) ^{2}}{4}}-a \left ( \left ( \arcsin \left ( bx+a \right ) \right ) ^{2} \left ( bx+a \right ) -2\,bx-2\,a+2\,\sqrt{1- \left ( bx+a \right ) ^{2}}\arcsin \left ( bx+a \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^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))-1/4*ar
csin(b*x+a)^2-1/4*(b*x+a)^2-a*(arcsin(b*x+a)^2*(b*x+a)-2*b*x-2*a+2*(1-(b*x+a)^2)^(1/2)*arcsin(b*x+a)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.81804, size = 190, normalized size = 1.46 \begin{align*} -\frac{b^{2} x^{2} - 6 \, a b x -{\left (2 \, b^{2} x^{2} - 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 (b x - 3 \, a\right )} \arcsin \left (b x + a\right )}{4 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.0149, size = 138, normalized size = 1.06 \begin{align*} \begin{cases} - \frac{a^{2} \operatorname{asin}^{2}{\left (a + b x \right )}}{2 b^{2}} + \frac{3 a x}{2 b} - \frac{3 a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{2 b^{2}} + \frac{x^{2} \operatorname{asin}^{2}{\left (a + b x \right )}}{2} - \frac{x^{2}}{4} + \frac{x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} \operatorname{asin}{\left (a + b x \right )}}{2 b} - \frac{\operatorname{asin}^{2}{\left (a + b x \right )}}{4 b^{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \operatorname{asin}^{2}{\left (a \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a**2*asin(a + b*x)**2/(2*b**2) + 3*a*x/(2*b) - 3*a*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a +
b*x)/(2*b**2) + x**2*asin(a + b*x)**2/2 - x**2/4 + x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1)*asin(a + b*x)/(2*b)
- asin(a + b*x)**2/(4*b**2), Ne(b, 0)), (x**2*asin(a)**2/2, True))

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Giac [A]  time = 1.22604, size = 188, normalized size = 1.45 \begin{align*} -\frac{{\left (b x + a\right )} a \arcsin \left (b x + a\right )^{2}}{b^{2}} + \frac{{\left ({\left (b x + a\right )}^{2} - 1\right )} \arcsin \left (b x + a\right )^{2}}{2 \, b^{2}} + \frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left (b x + a\right )} \arcsin \left (b x + a\right )}{2 \, b^{2}} - \frac{2 \, \sqrt{-{\left (b x + a\right )}^{2} + 1} a \arcsin \left (b x + a\right )}{b^{2}} + \frac{2 \,{\left (b x + a\right )} a}{b^{2}} + \frac{\arcsin \left (b x + a\right )^{2}}{4 \, b^{2}} - \frac{{\left (b x + a\right )}^{2} - 1}{4 \, b^{2}} - \frac{1}{8 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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