3.191 \(\int (c e+d e x) (a+b \sin ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=105 \[ \frac{e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}-\frac{e \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac{b^2 e (c+d x)^2}{4 d} \]

[Out]

-(b^2*e*(c + d*x)^2)/(4*d) + (b*e*(c + d*x)*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(2*d) - (e*(a + b*A
rcSin[c + d*x])^2)/(4*d) + (e*(c + d*x)^2*(a + b*ArcSin[c + d*x])^2)/(2*d)

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Rubi [A]  time = 0.145095, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {4805, 12, 4627, 4707, 4641, 30} \[ \frac{e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d}+\frac{b e \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d}-\frac{e \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac{b^2 e (c+d x)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)*(a + b*ArcSin[c + d*x])^2,x]

[Out]

-(b^2*e*(c + d*x)^2)/(4*d) + (b*e*(c + d*x)*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(2*d) - (e*(a + b*A
rcSin[c + d*x])^2)/(4*d) + (e*(c + d*x)^2*(a + b*ArcSin[c + d*x])^2)/(2*d)

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4627

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 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 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 30

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

Rubi steps

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

Mathematica [A]  time = 0.0735132, size = 86, normalized size = 0.82 \[ -\frac{e \left (-2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^2-2 b \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )+\left (a+b \sin ^{-1}(c+d x)\right )^2+b^2 (c+d x)^2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)*(a + b*ArcSin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.03, size = 146, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({\frac{e \left ( dx+c \right ) ^{2}{a}^{2}}{2}}+e{b}^{2} \left ({\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{2} \left ( \left ( dx+c \right ) ^{2}-1 \right ) }{2}}+{\frac{\arcsin \left ( dx+c \right ) }{2} \left ( \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}+\arcsin \left ( dx+c \right ) \right ) }-{\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{4}}-{\frac{ \left ( dx+c \right ) ^{2}}{4}} \right ) +2\,eab \left ( 1/2\,\arcsin \left ( dx+c \right ) \left ( dx+c \right ) ^{2}+1/4\, \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}-1/4\,\arcsin \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2*e*(d*x+c)^2*a^2+e*b^2*(1/2*arcsin(d*x+c)^2*((d*x+c)^2-1)+1/2*arcsin(d*x+c)*((d*x+c)*(1-(d*x+c)^2)^(1/
2)+arcsin(d*x+c))-1/4*arcsin(d*x+c)^2-1/4*(d*x+c)^2)+2*e*a*b*(1/2*arcsin(d*x+c)*(d*x+c)^2+1/4*(d*x+c)*(1-(d*x+
c)^2)^(1/2)-1/4*arcsin(d*x+c)))

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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((d*e*x+c*e)*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.4353, size = 421, normalized size = 4.01 \begin{align*} \frac{{\left (2 \, a^{2} - b^{2}\right )} d^{2} e x^{2} + 2 \,{\left (2 \, a^{2} - b^{2}\right )} c d e x +{\left (2 \, b^{2} d^{2} e x^{2} + 4 \, b^{2} c d e x +{\left (2 \, b^{2} c^{2} - b^{2}\right )} e\right )} \arcsin \left (d x + c\right )^{2} + 2 \,{\left (2 \, a b d^{2} e x^{2} + 4 \, a b c d e x +{\left (2 \, a b c^{2} - a b\right )} e\right )} \arcsin \left (d x + c\right ) + 2 \,{\left (a b d e x + a b c e +{\left (b^{2} d e x + b^{2} c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*a^2 - b^2)*d^2*e*x^2 + 2*(2*a^2 - b^2)*c*d*e*x + (2*b^2*d^2*e*x^2 + 4*b^2*c*d*e*x + (2*b^2*c^2 - b^2)*
e)*arcsin(d*x + c)^2 + 2*(2*a*b*d^2*e*x^2 + 4*a*b*c*d*e*x + (2*a*b*c^2 - a*b)*e)*arcsin(d*x + c) + 2*(a*b*d*e*
x + a*b*c*e + (b^2*d*e*x + b^2*c*e)*arcsin(d*x + c))*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1))/d

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Sympy [A]  time = 1.27237, size = 335, normalized size = 3.19 \begin{align*} \begin{cases} a^{2} c e x + \frac{a^{2} d e x^{2}}{2} + \frac{a b c^{2} e \operatorname{asin}{\left (c + d x \right )}}{d} + 2 a b c e x \operatorname{asin}{\left (c + d x \right )} + \frac{a b c e \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{2 d} + a b d e x^{2} \operatorname{asin}{\left (c + d x \right )} + \frac{a b e x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{2} - \frac{a b e \operatorname{asin}{\left (c + d x \right )}}{2 d} + \frac{b^{2} c^{2} e \operatorname{asin}^{2}{\left (c + d x \right )}}{2 d} + b^{2} c e x \operatorname{asin}^{2}{\left (c + d x \right )} - \frac{b^{2} c e x}{2} + \frac{b^{2} c e \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname{asin}{\left (c + d x \right )}}{2 d} + \frac{b^{2} d e x^{2} \operatorname{asin}^{2}{\left (c + d x \right )}}{2} - \frac{b^{2} d e x^{2}}{4} + \frac{b^{2} e x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1} \operatorname{asin}{\left (c + d x \right )}}{2} - \frac{b^{2} e \operatorname{asin}^{2}{\left (c + d x \right )}}{4 d} & \text{for}\: d \neq 0 \\c e x \left (a + b \operatorname{asin}{\left (c \right )}\right )^{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)*(a+b*asin(d*x+c))**2,x)

[Out]

Piecewise((a**2*c*e*x + a**2*d*e*x**2/2 + a*b*c**2*e*asin(c + d*x)/d + 2*a*b*c*e*x*asin(c + d*x) + a*b*c*e*sqr
t(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(2*d) + a*b*d*e*x**2*asin(c + d*x) + a*b*e*x*sqrt(-c**2 - 2*c*d*x - d**2*x*
*2 + 1)/2 - a*b*e*asin(c + d*x)/(2*d) + b**2*c**2*e*asin(c + d*x)**2/(2*d) + b**2*c*e*x*asin(c + d*x)**2 - b**
2*c*e*x/2 + b**2*c*e*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(2*d) + b**2*d*e*x**2*asin(c + d*x)**
2/2 - b**2*d*e*x**2/4 + b**2*e*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/2 - b**2*e*asin(c + d*x)*
*2/(4*d), Ne(d, 0)), (c*e*x*(a + b*asin(c))**2, True))

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Giac [B]  time = 1.26747, size = 261, normalized size = 2.49 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} \arcsin \left (d x + c\right )^{2} e}{2 \, d} + \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right ) e}{2 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a b \arcsin \left (d x + c\right ) e}{d} + \frac{b^{2} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} a b e}{2 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} e}{2 \, d} - \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} e}{4 \, d} + \frac{a b \arcsin \left (d x + c\right ) e}{2 \, d} - \frac{b^{2} e}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((d*x + c)^2 - 1)*b^2*arcsin(d*x + c)^2*e/d + 1/2*sqrt(-(d*x + c)^2 + 1)*(d*x + c)*b^2*arcsin(d*x + c)*e/d
+ ((d*x + c)^2 - 1)*a*b*arcsin(d*x + c)*e/d + 1/4*b^2*arcsin(d*x + c)^2*e/d + 1/2*sqrt(-(d*x + c)^2 + 1)*(d*x
+ c)*a*b*e/d + 1/2*((d*x + c)^2 - 1)*a^2*e/d - 1/4*((d*x + c)^2 - 1)*b^2*e/d + 1/2*a*b*arcsin(d*x + c)*e/d -
1/8*b^2*e/d