3.189 \(\int (c e+d e x)^3 (a+b \sin ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=176 \[ \frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac{b e^3 \sqrt{1-(c+d x)^2} (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{8 d}+\frac{3 b e^3 \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )}{16 d}-\frac{3 e^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{32 d}-\frac{b^2 e^3 (c+d x)^4}{32 d}-\frac{3 b^2 e^3 (c+d x)^2}{32 d} \]

[Out]

(-3*b^2*e^3*(c + d*x)^2)/(32*d) - (b^2*e^3*(c + d*x)^4)/(32*d) + (3*b*e^3*(c + d*x)*Sqrt[1 - (c + d*x)^2]*(a +
b*ArcSin[c + d*x]))/(16*d) + (b*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(8*d) - (3*e^3
*(a + b*ArcSin[c + d*x])^2)/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSin[c + d*x])^2)/(4*d)

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Rubi [A]  time = 0.258837, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4805, 12, 4627, 4707, 4641, 30} \[ \frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac{b e^3 \sqrt{1-(c+d x)^2} (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{8 d}+\frac{3 b e^3 \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )}{16 d}-\frac{3 e^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{32 d}-\frac{b^2 e^3 (c+d x)^4}{32 d}-\frac{3 b^2 e^3 (c+d x)^2}{32 d} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(-3*b^2*e^3*(c + d*x)^2)/(32*d) - (b^2*e^3*(c + d*x)^4)/(32*d) + (3*b*e^3*(c + d*x)*Sqrt[1 - (c + d*x)^2]*(a +
b*ArcSin[c + d*x]))/(16*d) + (b*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(8*d) - (3*e^3
*(a + b*ArcSin[c + d*x])^2)/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSin[c + d*x])^2)/(4*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)^3 \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^3 x^3 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int x^3 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac{\left (b e^3\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=\frac{b e^3 (c+d x)^3 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{8 d}+\frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac{\left (3 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{8 d}-\frac{\left (b^2 e^3\right ) \operatorname{Subst}\left (\int x^3 \, dx,x,c+d x\right )}{8 d}\\ &=-\frac{b^2 e^3 (c+d x)^4}{32 d}+\frac{3 b e^3 (c+d x) \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{16 d}+\frac{b e^3 (c+d x)^3 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{8 d}+\frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}-\frac{\left (3 b e^3\right ) \operatorname{Subst}\left (\int \frac{a+b \sin ^{-1}(x)}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{16 d}-\frac{\left (3 b^2 e^3\right ) \operatorname{Subst}(\int x \, dx,x,c+d x)}{16 d}\\ &=-\frac{3 b^2 e^3 (c+d x)^2}{32 d}-\frac{b^2 e^3 (c+d x)^4}{32 d}+\frac{3 b e^3 (c+d x) \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{16 d}+\frac{b e^3 (c+d x)^3 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{8 d}-\frac{3 e^3 \left (a+b \sin ^{-1}(c+d x)\right )^2}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.201898, size = 142, normalized size = 0.81 \[ \frac{e^3 \left (\frac{1}{8} \left (-3 \left (-2 b \sqrt{1-(c+d x)^2} (c+d x) \left (a+b \sin ^{-1}(c+d x)\right )+\left (a+b \sin ^{-1}(c+d x)\right )^2+b^2 (c+d x)^2\right )+4 b \sqrt{1-(c+d x)^2} (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )-b^2 (c+d x)^4\right )+(c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^3*(a + b*ArcSin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.036, size = 206, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{3} \left ( dx+c \right ) ^{4}{a}^{2}}{4}}+{e}^{3}{b}^{2} \left ({\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) ^{4}}{4}}-{\frac{\arcsin \left ( dx+c \right ) }{16} \left ( -2\, \left ( dx+c \right ) ^{3}\sqrt{1- \left ( dx+c \right ) ^{2}}-3\, \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}+3\,\arcsin \left ( dx+c \right ) \right ) }+{\frac{3\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{32}}-{\frac{ \left ( dx+c \right ) ^{4}}{32}}-{\frac{3\, \left ( dx+c \right ) ^{2}}{32}} \right ) +2\,{e}^{3}ab \left ( 1/4\, \left ( dx+c \right ) ^{4}\arcsin \left ( dx+c \right ) +1/16\, \left ( dx+c \right ) ^{3}\sqrt{1- \left ( dx+c \right ) ^{2}}+{\frac{ \left ( 3\,dx+3\,c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}}{32}}-{\frac{3\,\arcsin \left ( dx+c \right ) }{32}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3*(a+b*arcsin(d*x+c))^2,x)

[Out]

1/d*(1/4*e^3*(d*x+c)^4*a^2+e^3*b^2*(1/4*arcsin(d*x+c)^2*(d*x+c)^4-1/16*arcsin(d*x+c)*(-2*(d*x+c)^3*(1-(d*x+c)^
2)^(1/2)-3*(d*x+c)*(1-(d*x+c)^2)^(1/2)+3*arcsin(d*x+c))+3/32*arcsin(d*x+c)^2-1/32*(d*x+c)^4-3/32*(d*x+c)^2)+2*
e^3*a*b*(1/4*(d*x+c)^4*arcsin(d*x+c)+1/16*(d*x+c)^3*(1-(d*x+c)^2)^(1/2)+3/32*(d*x+c)*(1-(d*x+c)^2)^(1/2)-3/32*
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)^3*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.61765, size = 922, normalized size = 5.24 \begin{align*} \frac{{\left (8 \, a^{2} - b^{2}\right )} d^{4} e^{3} x^{4} + 4 \,{\left (8 \, a^{2} - b^{2}\right )} c d^{3} e^{3} x^{3} + 3 \,{\left (2 \,{\left (8 \, a^{2} - b^{2}\right )} c^{2} - b^{2}\right )} d^{2} e^{3} x^{2} + 2 \,{\left (2 \,{\left (8 \, a^{2} - b^{2}\right )} c^{3} - 3 \, b^{2} c\right )} d e^{3} x +{\left (8 \, b^{2} d^{4} e^{3} x^{4} + 32 \, b^{2} c d^{3} e^{3} x^{3} + 48 \, b^{2} c^{2} d^{2} e^{3} x^{2} + 32 \, b^{2} c^{3} d e^{3} x +{\left (8 \, b^{2} c^{4} - 3 \, b^{2}\right )} e^{3}\right )} \arcsin \left (d x + c\right )^{2} + 2 \,{\left (8 \, a b d^{4} e^{3} x^{4} + 32 \, a b c d^{3} e^{3} x^{3} + 48 \, a b c^{2} d^{2} e^{3} x^{2} + 32 \, a b c^{3} d e^{3} x +{\left (8 \, a b c^{4} - 3 \, a b\right )} e^{3}\right )} \arcsin \left (d x + c\right ) + 2 \,{\left (2 \, a b d^{3} e^{3} x^{3} + 6 \, a b c d^{2} e^{3} x^{2} + 3 \,{\left (2 \, a b c^{2} + a b\right )} d e^{3} x +{\left (2 \, a b c^{3} + 3 \, a b c\right )} e^{3} +{\left (2 \, b^{2} d^{3} e^{3} x^{3} + 6 \, b^{2} c d^{2} e^{3} x^{2} + 3 \,{\left (2 \, b^{2} c^{2} + b^{2}\right )} d e^{3} x +{\left (2 \, b^{2} c^{3} + 3 \, b^{2} c\right )} e^{3}\right )} \arcsin \left (d x + c\right )\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3*(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/32*((8*a^2 - b^2)*d^4*e^3*x^4 + 4*(8*a^2 - b^2)*c*d^3*e^3*x^3 + 3*(2*(8*a^2 - b^2)*c^2 - b^2)*d^2*e^3*x^2 +
2*(2*(8*a^2 - b^2)*c^3 - 3*b^2*c)*d*e^3*x + (8*b^2*d^4*e^3*x^4 + 32*b^2*c*d^3*e^3*x^3 + 48*b^2*c^2*d^2*e^3*x^2
+ 32*b^2*c^3*d*e^3*x + (8*b^2*c^4 - 3*b^2)*e^3)*arcsin(d*x + c)^2 + 2*(8*a*b*d^4*e^3*x^4 + 32*a*b*c*d^3*e^3*x
^3 + 48*a*b*c^2*d^2*e^3*x^2 + 32*a*b*c^3*d*e^3*x + (8*a*b*c^4 - 3*a*b)*e^3)*arcsin(d*x + c) + 2*(2*a*b*d^3*e^3
*x^3 + 6*a*b*c*d^2*e^3*x^2 + 3*(2*a*b*c^2 + a*b)*d*e^3*x + (2*a*b*c^3 + 3*a*b*c)*e^3 + (2*b^2*d^3*e^3*x^3 + 6*
b^2*c*d^2*e^3*x^2 + 3*(2*b^2*c^2 + b^2)*d*e^3*x + (2*b^2*c^3 + 3*b^2*c)*e^3)*arcsin(d*x + c))*sqrt(-d^2*x^2 -
2*c*d*x - c^2 + 1))/d

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Sympy [A]  time = 6.92324, size = 916, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3*(a+b*asin(d*x+c))**2,x)

[Out]

Piecewise((a**2*c**3*e**3*x + 3*a**2*c**2*d*e**3*x**2/2 + a**2*c*d**2*e**3*x**3 + a**2*d**3*e**3*x**4/4 + a*b*
c**4*e**3*asin(c + d*x)/(2*d) + 2*a*b*c**3*e**3*x*asin(c + d*x) + a*b*c**3*e**3*sqrt(-c**2 - 2*c*d*x - d**2*x*
*2 + 1)/(8*d) + 3*a*b*c**2*d*e**3*x**2*asin(c + d*x) + 3*a*b*c**2*e**3*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)
/8 + 2*a*b*c*d**2*e**3*x**3*asin(c + d*x) + 3*a*b*c*d*e**3*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/8 + 3*a*
b*c*e**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(16*d) + a*b*d**3*e**3*x**4*asin(c + d*x)/2 + a*b*d**2*e**3*x**
3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/8 + 3*a*b*e**3*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/16 - 3*a*b*e**3
*asin(c + d*x)/(16*d) + b**2*c**4*e**3*asin(c + d*x)**2/(4*d) + b**2*c**3*e**3*x*asin(c + d*x)**2 - b**2*c**3*
e**3*x/8 + b**2*c**3*e**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(8*d) + 3*b**2*c**2*d*e**3*x**2*
asin(c + d*x)**2/2 - 3*b**2*c**2*d*e**3*x**2/16 + 3*b**2*c**2*e**3*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asi
n(c + d*x)/8 + b**2*c*d**2*e**3*x**3*asin(c + d*x)**2 - b**2*c*d**2*e**3*x**3/8 + 3*b**2*c*d*e**3*x**2*sqrt(-c
**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/8 - 3*b**2*c*e**3*x/16 + 3*b**2*c*e**3*sqrt(-c**2 - 2*c*d*x - d**
2*x**2 + 1)*asin(c + d*x)/(16*d) + b**2*d**3*e**3*x**4*asin(c + d*x)**2/4 - b**2*d**3*e**3*x**4/32 + b**2*d**2
*e**3*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/8 - 3*b**2*d*e**3*x**2/32 + 3*b**2*e**3*x*sqrt(
-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/16 - 3*b**2*e**3*asin(c + d*x)**2/(32*d), Ne(d, 0)), (c**3*e**3
*x*(a + b*asin(c))**2, True))

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Giac [B]  time = 1.24333, size = 474, normalized size = 2.69 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} b^{2} \arcsin \left (d x + c\right )^{2} e^{3}}{4 \, d} - \frac{{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}}{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right ) e^{3}}{8 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} a b \arcsin \left (d x + c\right ) e^{3}}{2 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{3}}{2 \, d} - \frac{{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}}{\left (d x + c\right )} a b e^{3}}{8 \, d} + \frac{5 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right ) e^{3}}{16 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} a^{2} e^{3}}{4 \, d} - \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} b^{2} e^{3}}{32 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a b \arcsin \left (d x + c\right ) e^{3}}{d} + \frac{5 \, b^{2} \arcsin \left (d x + c\right )^{2} e^{3}}{32 \, d} + \frac{5 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} a b e^{3}}{16 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} e^{3}}{2 \, d} - \frac{5 \,{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{2} e^{3}}{32 \, d} + \frac{5 \, a b \arcsin \left (d x + c\right ) e^{3}}{16 \, d} - \frac{17 \, b^{2} e^{3}}{256 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

1/4*((d*x + c)^2 - 1)^2*b^2*arcsin(d*x + c)^2*e^3/d - 1/8*(-(d*x + c)^2 + 1)^(3/2)*(d*x + c)*b^2*arcsin(d*x +
c)*e^3/d + 1/2*((d*x + c)^2 - 1)^2*a*b*arcsin(d*x + c)*e^3/d + 1/2*((d*x + c)^2 - 1)*b^2*arcsin(d*x + c)^2*e^3
/d - 1/8*(-(d*x + c)^2 + 1)^(3/2)*(d*x + c)*a*b*e^3/d + 5/16*sqrt(-(d*x + c)^2 + 1)*(d*x + c)*b^2*arcsin(d*x +
c)*e^3/d + 1/4*((d*x + c)^2 - 1)^2*a^2*e^3/d - 1/32*((d*x + c)^2 - 1)^2*b^2*e^3/d + ((d*x + c)^2 - 1)*a*b*arc
sin(d*x + c)*e^3/d + 5/32*b^2*arcsin(d*x + c)^2*e^3/d + 5/16*sqrt(-(d*x + c)^2 + 1)*(d*x + c)*a*b*e^3/d + 1/2*
((d*x + c)^2 - 1)*a^2*e^3/d - 5/32*((d*x + c)^2 - 1)*b^2*e^3/d + 5/16*a*b*arcsin(d*x + c)*e^3/d - 17/256*b^2*e
^3/d