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

Optimal. Leaf size=109 \[ \frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac{b e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{16 d}+\frac{3 b e^3 \sqrt{1-(c+d x)^2} (c+d x)}{32 d}-\frac{3 b e^3 \sin ^{-1}(c+d x)}{32 d} \]

[Out]

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

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Rubi [A]  time = 0.0756311, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {4805, 12, 4627, 321, 216} \[ \frac{e^3 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac{b e^3 \sqrt{1-(c+d x)^2} (c+d x)^3}{16 d}+\frac{3 b e^3 \sqrt{1-(c+d x)^2} (c+d x)}{32 d}-\frac{3 b e^3 \sin ^{-1}(c+d x)}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*e^3*(c + d*x)*Sqrt[1 - (c + d*x)^2])/(32*d) + (b*e^3*(c + d*x)^3*Sqrt[1 - (c + d*x)^2])/(16*d) - (3*b*e^3
*ArcSin[c + d*x])/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSin[c + d*x]))/(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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 216

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

Rubi steps

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

Mathematica [A]  time = 0.078866, size = 87, normalized size = 0.8 \[ \frac{e^3 \left (8 (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )+2 b \sqrt{1-(c+d x)^2} (c+d x)^3+3 b \sqrt{1-(c+d x)^2} (c+d x)-3 b \sin ^{-1}(c+d x)\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.006, size = 90, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{3} \left ( dx+c \right ) ^{4}a}{4}}+{e}^{3}b \left ({\frac{ \left ( dx+c \right ) ^{4}\arcsin \left ( dx+c \right ) }{4}}+{\frac{ \left ( dx+c \right ) ^{3}}{16}\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{3\,dx+3\,c}{32}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\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)),x)

[Out]

1/d*(1/4*e^3*(d*x+c)^4*a+e^3*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)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.39475, size = 452, normalized size = 4.15 \begin{align*} \frac{8 \, a d^{4} e^{3} x^{4} + 32 \, a c d^{3} e^{3} x^{3} + 48 \, a c^{2} d^{2} e^{3} x^{2} + 32 \, a c^{3} d e^{3} x +{\left (8 \, b d^{4} e^{3} x^{4} + 32 \, b c d^{3} e^{3} x^{3} + 48 \, b c^{2} d^{2} e^{3} x^{2} + 32 \, b c^{3} d e^{3} x +{\left (8 \, b c^{4} - 3 \, b\right )} e^{3}\right )} \arcsin \left (d x + c\right ) +{\left (2 \, b d^{3} e^{3} x^{3} + 6 \, b c d^{2} e^{3} x^{2} + 3 \,{\left (2 \, b c^{2} + b\right )} d e^{3} x +{\left (2 \, b c^{3} + 3 \, b c\right )} e^{3}\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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.81556, size = 394, normalized size = 3.61 \begin{align*} \begin{cases} a c^{3} e^{3} x + \frac{3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac{a d^{3} e^{3} x^{4}}{4} + \frac{b c^{4} e^{3} \operatorname{asin}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname{asin}{\left (c + d x \right )} + \frac{b c^{3} e^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16 d} + \frac{3 b c^{2} d e^{3} x^{2} \operatorname{asin}{\left (c + d x \right )}}{2} + \frac{3 b c^{2} e^{3} x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + b c d^{2} e^{3} x^{3} \operatorname{asin}{\left (c + d x \right )} + \frac{3 b c d e^{3} x^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac{3 b c e^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32 d} + \frac{b d^{3} e^{3} x^{4} \operatorname{asin}{\left (c + d x \right )}}{4} + \frac{b d^{2} e^{3} x^{3} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac{3 b e^{3} x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32} - \frac{3 b e^{3} \operatorname{asin}{\left (c + d x \right )}}{32 d} & \text{for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname{asin}{\left (c \right )}\right ) & \text{otherwise} \end{cases} \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)),x)

[Out]

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

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Giac [A]  time = 1.39028, size = 204, normalized size = 1.87 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} b \arcsin \left (d x + c\right ) e^{3}}{4 \, d} - \frac{{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}}{\left (d x + c\right )} b e^{3}}{16 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} a e^{3}}{4 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} b \arcsin \left (d x + c\right ) e^{3}}{2 \, d} + \frac{5 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b e^{3}}{32 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a e^{3}}{2 \, d} + \frac{5 \, b \arcsin \left (d x + c\right ) e^{3}}{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)),x, algorithm="giac")

[Out]

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