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

Optimal. Leaf size=80 \[ \frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac{b e^2 \sqrt{1-(c+d x)^2}}{3 d} \]

[Out]

(b*e^2*Sqrt[1 - (c + d*x)^2])/(3*d) - (b*e^2*(1 - (c + d*x)^2)^(3/2))/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSin[c
+ d*x]))/(3*d)

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Rubi [A]  time = 0.0713548, antiderivative size = 80, 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, 266, 43} \[ \frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac{b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac{b e^2 \sqrt{1-(c+d x)^2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^2*Sqrt[1 - (c + d*x)^2])/(3*d) - (b*e^2*(1 - (c + d*x)^2)^(3/2))/(9*d) + (e^2*(c + d*x)^3*(a + b*ArcSin[c
+ d*x]))/(3*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \sin ^{-1}(c+d x)\right ) \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \sin ^{-1}(x)\right ) \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{3 d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1-x}} \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \left (\frac{1}{\sqrt{1-x}}-\sqrt{1-x}\right ) \, dx,x,(c+d x)^2\right )}{6 d}\\ &=\frac{b e^2 \sqrt{1-(c+d x)^2}}{3 d}-\frac{b e^2 \left (1-(c+d x)^2\right )^{3/2}}{9 d}+\frac{e^2 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0483664, size = 64, normalized size = 0.8 \[ \frac{e^2 \left (3 (c+d x)^3 \left (a+b \sin ^{-1}(c+d x)\right )+b \left (c^2+2 c d x+d^2 x^2+2\right ) \sqrt{1-(c+d x)^2}\right )}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^2*(b*(2 + c^2 + 2*c*d*x + d^2*x^2)*Sqrt[1 - (c + d*x)^2] + 3*(c + d*x)^3*(a + b*ArcSin[c + d*x])))/(9*d)

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Maple [A]  time = 0.004, size = 77, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{2} \left ( dx+c \right ) ^{3}a}{3}}+{e}^{2}b \left ({\frac{ \left ( dx+c \right ) ^{3}\arcsin \left ( dx+c \right ) }{3}}+{\frac{ \left ( dx+c \right ) ^{2}}{9}\sqrt{1- \left ( dx+c \right ) ^{2}}}+{\frac{2}{9}\sqrt{1- \left ( dx+c \right ) ^{2}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*e^2*(d*x+c)^3*a+e^2*b*(1/3*(d*x+c)^3*arcsin(d*x+c)+1/9*(d*x+c)^2*(1-(d*x+c)^2)^(1/2)+2/9*(1-(d*x+c)^2
)^(1/2)))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.20817, size = 321, normalized size = 4.01 \begin{align*} \frac{3 \, a d^{3} e^{2} x^{3} + 9 \, a c d^{2} e^{2} x^{2} + 9 \, a c^{2} d e^{2} x + 3 \,{\left (b d^{3} e^{2} x^{3} + 3 \, b c d^{2} e^{2} x^{2} + 3 \, b c^{2} d e^{2} x + b c^{3} e^{2}\right )} \arcsin \left (d x + c\right ) +{\left (b d^{2} e^{2} x^{2} + 2 \, b c d e^{2} x +{\left (b c^{2} + 2 \, b\right )} e^{2}\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{9 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*(3*a*d^3*e^2*x^3 + 9*a*c*d^2*e^2*x^2 + 9*a*c^2*d*e^2*x + 3*(b*d^3*e^2*x^3 + 3*b*c*d^2*e^2*x^2 + 3*b*c^2*d*
e^2*x + b*c^3*e^2)*arcsin(d*x + c) + (b*d^2*e^2*x^2 + 2*b*c*d*e^2*x + (b*c^2 + 2*b)*e^2)*sqrt(-d^2*x^2 - 2*c*d
*x - c^2 + 1))/d

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Sympy [A]  time = 1.25899, size = 258, normalized size = 3.22 \begin{align*} \begin{cases} a c^{2} e^{2} x + a c d e^{2} x^{2} + \frac{a d^{2} e^{2} x^{3}}{3} + \frac{b c^{3} e^{2} \operatorname{asin}{\left (c + d x \right )}}{3 d} + b c^{2} e^{2} x \operatorname{asin}{\left (c + d x \right )} + \frac{b c^{2} e^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} + b c d e^{2} x^{2} \operatorname{asin}{\left (c + d x \right )} + \frac{2 b c e^{2} x \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac{b d^{2} e^{2} x^{3} \operatorname{asin}{\left (c + d x \right )}}{3} + \frac{b d e^{2} x^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9} + \frac{2 b e^{2} \sqrt{- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{9 d} & \text{for}\: d \neq 0 \\c^{2} e^{2} 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)**2*(a+b*asin(d*x+c)),x)

[Out]

Piecewise((a*c**2*e**2*x + a*c*d*e**2*x**2 + a*d**2*e**2*x**3/3 + b*c**3*e**2*asin(c + d*x)/(3*d) + b*c**2*e**
2*x*asin(c + d*x) + b*c**2*e**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(9*d) + b*c*d*e**2*x**2*asin(c + d*x) +
2*b*c*e**2*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/9 + b*d**2*e**2*x**3*asin(c + d*x)/3 + b*d*e**2*x**2*sqrt(-
c**2 - 2*c*d*x - d**2*x**2 + 1)/9 + 2*b*e**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(9*d), Ne(d, 0)), (c**2*e**
2*x*(a + b*asin(c)), True))

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Giac [A]  time = 1.32084, size = 142, normalized size = 1.78 \begin{align*} \frac{{\left (d x + c\right )}^{3} a e^{2}}{3 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} + \frac{{\left (d x + c\right )} b \arcsin \left (d x + c\right ) e^{2}}{3 \, d} - \frac{{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}} b e^{2}}{9 \, d} + \frac{\sqrt{-{\left (d x + c\right )}^{2} + 1} b e^{2}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(d*x + c)^3*a*e^2/d + 1/3*((d*x + c)^2 - 1)*(d*x + c)*b*arcsin(d*x + c)*e^2/d + 1/3*(d*x + c)*b*arcsin(d*x
+ c)*e^2/d - 1/9*(-(d*x + c)^2 + 1)^(3/2)*b*e^2/d + 1/3*sqrt(-(d*x + c)^2 + 1)*b*e^2/d