3.188 \(\int (c e+d e x)^4 (a+b \sin ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=203 \[ \frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}+\frac{2 b e^4 \sqrt{1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{8 b e^4 \sqrt{1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{16 b e^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}-\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{8 b^2 e^4 (c+d x)^3}{225 d}-\frac{16}{75} b^2 e^4 x \]

[Out]

(-16*b^2*e^4*x)/75 - (8*b^2*e^4*(c + d*x)^3)/(225*d) - (2*b^2*e^4*(c + d*x)^5)/(125*d) + (16*b*e^4*Sqrt[1 - (c
+ d*x)^2]*(a + b*ArcSin[c + d*x]))/(75*d) + (8*b*e^4*(c + d*x)^2*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]
))/(75*d) + (2*b*e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a +
b*ArcSin[c + d*x])^2)/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.304755, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4805, 12, 4627, 4707, 4677, 8, 30} \[ \frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}+\frac{2 b e^4 \sqrt{1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{8 b e^4 \sqrt{1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{16 b e^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}-\frac{2 b^2 e^4 (c+d x)^5}{125 d}-\frac{8 b^2 e^4 (c+d x)^3}{225 d}-\frac{16}{75} b^2 e^4 x \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(-16*b^2*e^4*x)/75 - (8*b^2*e^4*(c + d*x)^3)/(225*d) - (2*b^2*e^4*(c + d*x)^5)/(125*d) + (16*b*e^4*Sqrt[1 - (c
+ d*x)^2]*(a + b*ArcSin[c + d*x]))/(75*d) + (8*b*e^4*(c + d*x)^2*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]
))/(75*d) + (2*b*e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/(25*d) + (e^4*(c + d*x)^5*(a +
b*ArcSin[c + d*x])^2)/(5*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 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 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)^4 \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int e^4 x^4 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 \operatorname{Subst}\left (\int x^4 \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (2 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^5 \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{5 d}\\ &=\frac{2 b e^4 (c+d x)^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (8 b e^4\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{25 d}-\frac{\left (2 b^2 e^4\right ) \operatorname{Subst}\left (\int x^4 \, dx,x,c+d x\right )}{25 d}\\ &=-\frac{2 b^2 e^4 (c+d x)^5}{125 d}+\frac{8 b e^4 (c+d x)^2 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{2 b e^4 (c+d x)^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (16 b e^4\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sin ^{-1}(x)\right )}{\sqrt{1-x^2}} \, dx,x,c+d x\right )}{75 d}-\frac{\left (8 b^2 e^4\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,c+d x\right )}{75 d}\\ &=-\frac{8 b^2 e^4 (c+d x)^3}{225 d}-\frac{2 b^2 e^4 (c+d x)^5}{125 d}+\frac{16 b e^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{8 b e^4 (c+d x)^2 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{2 b e^4 (c+d x)^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}-\frac{\left (16 b^2 e^4\right ) \operatorname{Subst}(\int 1 \, dx,x,c+d x)}{75 d}\\ &=-\frac{16}{75} b^2 e^4 x-\frac{8 b^2 e^4 (c+d x)^3}{225 d}-\frac{2 b^2 e^4 (c+d x)^5}{125 d}+\frac{16 b e^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{8 b e^4 (c+d x)^2 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{75 d}+\frac{2 b e^4 (c+d x)^4 \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )}{25 d}+\frac{e^4 (c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.376562, size = 164, normalized size = 0.81 \[ \frac{e^4 \left ((c+d x)^5 \left (a+b \sin ^{-1}(c+d x)\right )^2-\frac{2}{25} b \left (-5 \sqrt{1-(c+d x)^2} (c+d x)^4 \left (a+b \sin ^{-1}(c+d x)\right )-\frac{20}{3} \sqrt{1-(c+d x)^2} (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )+\frac{40}{3} \left (b d x-\sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )\right )+b (c+d x)^5+\frac{20}{9} b (c+d x)^3\right )\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4*(a + b*ArcSin[c + d*x])^2,x]

[Out]

(e^4*((c + d*x)^5*(a + b*ArcSin[c + d*x])^2 - (2*b*((20*b*(c + d*x)^3)/9 + b*(c + d*x)^5 - (20*(c + d*x)^2*Sqr
t[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]))/3 - 5*(c + d*x)^4*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x]) +
(40*(b*d*x - Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x])))/3))/25))/(5*d)

________________________________________________________________________________________

Maple [A]  time = 0.035, size = 194, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{e}^{4} \left ( dx+c \right ) ^{5}{a}^{2}}{5}}+{e}^{4}{b}^{2} \left ({\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{2} \left ( dx+c \right ) ^{5}}{5}}+{\frac{2\,\arcsin \left ( dx+c \right ) \left ( 3\, \left ( dx+c \right ) ^{4}+4\, \left ( dx+c \right ) ^{2}+8 \right ) }{75}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{2\, \left ( dx+c \right ) ^{5}}{125}}-{\frac{8\, \left ( dx+c \right ) ^{3}}{225}}-{\frac{16\,dx}{75}}-{\frac{16\,c}{75}} \right ) +2\,{e}^{4}ab \left ( 1/5\, \left ( dx+c \right ) ^{5}\arcsin \left ( dx+c \right ) +1/25\, \left ( dx+c \right ) ^{4}\sqrt{1- \left ( dx+c \right ) ^{2}}+{\frac{4\, \left ( dx+c \right ) ^{2}\sqrt{1- \left ( dx+c \right ) ^{2}}}{75}}+{\frac{8\,\sqrt{1- \left ( dx+c \right ) ^{2}}}{75}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x)

[Out]

1/d*(1/5*e^4*(d*x+c)^5*a^2+e^4*b^2*(1/5*arcsin(d*x+c)^2*(d*x+c)^5+2/75*arcsin(d*x+c)*(3*(d*x+c)^4+4*(d*x+c)^2+
8)*(1-(d*x+c)^2)^(1/2)-2/125*(d*x+c)^5-8/225*(d*x+c)^3-16/75*d*x-16/75*c)+2*e^4*a*b*(1/5*(d*x+c)^5*arcsin(d*x+
c)+1/25*(d*x+c)^4*(1-(d*x+c)^2)^(1/2)+4/75*(d*x+c)^2*(1-(d*x+c)^2)^(1/2)+8/75*(1-(d*x+c)^2)^(1/2)))

________________________________________________________________________________________

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)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.396, size = 1207, normalized size = 5.95 \begin{align*} \frac{9 \,{\left (25 \, a^{2} - 2 \, b^{2}\right )} d^{5} e^{4} x^{5} + 45 \,{\left (25 \, a^{2} - 2 \, b^{2}\right )} c d^{4} e^{4} x^{4} + 10 \,{\left (9 \,{\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{2} - 4 \, b^{2}\right )} d^{3} e^{4} x^{3} + 30 \,{\left (3 \,{\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{3} - 4 \, b^{2} c\right )} d^{2} e^{4} x^{2} + 15 \,{\left (3 \,{\left (25 \, a^{2} - 2 \, b^{2}\right )} c^{4} - 8 \, b^{2} c^{2} - 16 \, b^{2}\right )} d e^{4} x + 225 \,{\left (b^{2} d^{5} e^{4} x^{5} + 5 \, b^{2} c d^{4} e^{4} x^{4} + 10 \, b^{2} c^{2} d^{3} e^{4} x^{3} + 10 \, b^{2} c^{3} d^{2} e^{4} x^{2} + 5 \, b^{2} c^{4} d e^{4} x + b^{2} c^{5} e^{4}\right )} \arcsin \left (d x + c\right )^{2} + 450 \,{\left (a b d^{5} e^{4} x^{5} + 5 \, a b c d^{4} e^{4} x^{4} + 10 \, a b c^{2} d^{3} e^{4} x^{3} + 10 \, a b c^{3} d^{2} e^{4} x^{2} + 5 \, a b c^{4} d e^{4} x + a b c^{5} e^{4}\right )} \arcsin \left (d x + c\right ) + 30 \,{\left (3 \, a b d^{4} e^{4} x^{4} + 12 \, a b c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, a b c^{2} + 2 \, a b\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, a b c^{3} + 2 \, a b c\right )} d e^{4} x +{\left (3 \, a b c^{4} + 4 \, a b c^{2} + 8 \, a b\right )} e^{4} +{\left (3 \, b^{2} d^{4} e^{4} x^{4} + 12 \, b^{2} c d^{3} e^{4} x^{3} + 2 \,{\left (9 \, b^{2} c^{2} + 2 \, b^{2}\right )} d^{2} e^{4} x^{2} + 4 \,{\left (3 \, b^{2} c^{3} + 2 \, b^{2} c\right )} d e^{4} x +{\left (3 \, b^{2} c^{4} + 4 \, b^{2} c^{2} + 8 \, b^{2}\right )} e^{4}\right )} \arcsin \left (d x + c\right )\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{1125 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/1125*(9*(25*a^2 - 2*b^2)*d^5*e^4*x^5 + 45*(25*a^2 - 2*b^2)*c*d^4*e^4*x^4 + 10*(9*(25*a^2 - 2*b^2)*c^2 - 4*b^
2)*d^3*e^4*x^3 + 30*(3*(25*a^2 - 2*b^2)*c^3 - 4*b^2*c)*d^2*e^4*x^2 + 15*(3*(25*a^2 - 2*b^2)*c^4 - 8*b^2*c^2 -
16*b^2)*d*e^4*x + 225*(b^2*d^5*e^4*x^5 + 5*b^2*c*d^4*e^4*x^4 + 10*b^2*c^2*d^3*e^4*x^3 + 10*b^2*c^3*d^2*e^4*x^2
+ 5*b^2*c^4*d*e^4*x + b^2*c^5*e^4)*arcsin(d*x + c)^2 + 450*(a*b*d^5*e^4*x^5 + 5*a*b*c*d^4*e^4*x^4 + 10*a*b*c^
2*d^3*e^4*x^3 + 10*a*b*c^3*d^2*e^4*x^2 + 5*a*b*c^4*d*e^4*x + a*b*c^5*e^4)*arcsin(d*x + c) + 30*(3*a*b*d^4*e^4*
x^4 + 12*a*b*c*d^3*e^4*x^3 + 2*(9*a*b*c^2 + 2*a*b)*d^2*e^4*x^2 + 4*(3*a*b*c^3 + 2*a*b*c)*d*e^4*x + (3*a*b*c^4
+ 4*a*b*c^2 + 8*a*b)*e^4 + (3*b^2*d^4*e^4*x^4 + 12*b^2*c*d^3*e^4*x^3 + 2*(9*b^2*c^2 + 2*b^2)*d^2*e^4*x^2 + 4*(
3*b^2*c^3 + 2*b^2*c)*d*e^4*x + (3*b^2*c^4 + 4*b^2*c^2 + 8*b^2)*e^4)*arcsin(d*x + c))*sqrt(-d^2*x^2 - 2*c*d*x -
c^2 + 1))/d

________________________________________________________________________________________

Sympy [A]  time = 13.3175, size = 1268, normalized size = 6.25 \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)**4*(a+b*asin(d*x+c))**2,x)

[Out]

Piecewise((a**2*c**4*e**4*x + 2*a**2*c**3*d*e**4*x**2 + 2*a**2*c**2*d**2*e**4*x**3 + a**2*c*d**3*e**4*x**4 + a
**2*d**4*e**4*x**5/5 + 2*a*b*c**5*e**4*asin(c + d*x)/(5*d) + 2*a*b*c**4*e**4*x*asin(c + d*x) + 2*a*b*c**4*e**4
*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(25*d) + 4*a*b*c**3*d*e**4*x**2*asin(c + d*x) + 8*a*b*c**3*e**4*x*sqrt(
-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 4*a*b*c**2*d**2*e**4*x**3*asin(c + d*x) + 12*a*b*c**2*d*e**4*x**2*sqrt(-
c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*a*b*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + 2*a*b*c*d*
*3*e**4*x**4*asin(c + d*x) + 8*a*b*c*d**2*e**4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 16*a*b*c*e**4*x
*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 2*a*b*d**4*e**4*x**5*asin(c + d*x)/5 + 2*a*b*d**3*e**4*x**4*sqrt(-
c**2 - 2*c*d*x - d**2*x**2 + 1)/25 + 8*a*b*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/75 + 16*a*b*e**4*
sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(75*d) + b**2*c**5*e**4*asin(c + d*x)**2/(5*d) + b**2*c**4*e**4*x*asin(c
+ d*x)**2 - 2*b**2*c**4*e**4*x/25 + 2*b**2*c**4*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(25*
d) + 2*b**2*c**3*d*e**4*x**2*asin(c + d*x)**2 - 4*b**2*c**3*d*e**4*x**2/25 + 8*b**2*c**3*e**4*x*sqrt(-c**2 - 2
*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 + 2*b**2*c**2*d**2*e**4*x**3*asin(c + d*x)**2 - 4*b**2*c**2*d**2*e**4
*x**3/25 + 12*b**2*c**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 - 8*b**2*c**2*e**4*
x/75 + 8*b**2*c**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/(75*d) + b**2*c*d**3*e**4*x**4*asi
n(c + d*x)**2 - 2*b**2*c*d**3*e**4*x**4/25 + 8*b**2*c*d**2*e**4*x**3*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asi
n(c + d*x)/25 - 8*b**2*c*d*e**4*x**2/75 + 16*b**2*c*e**4*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)
/75 + b**2*d**4*e**4*x**5*asin(c + d*x)**2/5 - 2*b**2*d**4*e**4*x**5/125 + 2*b**2*d**3*e**4*x**4*sqrt(-c**2 -
2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)/25 - 8*b**2*d**2*e**4*x**3/225 + 8*b**2*d*e**4*x**2*sqrt(-c**2 - 2*c*d*
x - d**2*x**2 + 1)*asin(c + d*x)/75 - 16*b**2*e**4*x/75 + 16*b**2*e**4*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*a
sin(c + d*x)/(75*d), Ne(d, 0)), (c**4*e**4*x*(a + b*asin(c))**2, True))

________________________________________________________________________________________

Giac [B]  time = 1.2839, size = 576, normalized size = 2.84 \begin{align*} \frac{{\left (d x + c\right )}^{5} a^{2} e^{4}}{5 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2}{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2}{\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt{-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right ) e^{4}}{25 \, d} - \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2}{\left (d x + c\right )} b^{2} e^{4}}{125 \, d} + \frac{4 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}{\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} + \frac{{\left (d x + c\right )} b^{2} \arcsin \left (d x + c\right )^{2} e^{4}}{5 \, d} + \frac{2 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} \sqrt{-{\left (d x + c\right )}^{2} + 1} a b e^{4}}{25 \, d} - \frac{4 \,{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}} b^{2} \arcsin \left (d x + c\right ) e^{4}}{15 \, d} - \frac{76 \,{\left ({\left (d x + c\right )}^{2} - 1\right )}{\left (d x + c\right )} b^{2} e^{4}}{1125 \, d} + \frac{2 \,{\left (d x + c\right )} a b \arcsin \left (d x + c\right ) e^{4}}{5 \, d} - \frac{4 \,{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac{3}{2}} a b e^{4}}{15 \, d} + \frac{2 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} b^{2} \arcsin \left (d x + c\right ) e^{4}}{5 \, d} - \frac{298 \,{\left (d x + c\right )} b^{2} e^{4}}{1125 \, d} + \frac{2 \, \sqrt{-{\left (d x + c\right )}^{2} + 1} a b e^{4}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4*(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

1/5*(d*x + c)^5*a^2*e^4/d + 1/5*((d*x + c)^2 - 1)^2*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/d + 2/5*((d*x + c)^2 -
1)^2*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d + 2/5*((d*x + c)^2 - 1)*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/d + 2/25
*((d*x + c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*b^2*arcsin(d*x + c)*e^4/d - 2/125*((d*x + c)^2 - 1)^2*(d*x + c)*b^
2*e^4/d + 4/5*((d*x + c)^2 - 1)*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d + 1/5*(d*x + c)*b^2*arcsin(d*x + c)^2*e^4/
d + 2/25*((d*x + c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*a*b*e^4/d - 4/15*(-(d*x + c)^2 + 1)^(3/2)*b^2*arcsin(d*x +
c)*e^4/d - 76/1125*((d*x + c)^2 - 1)*(d*x + c)*b^2*e^4/d + 2/5*(d*x + c)*a*b*arcsin(d*x + c)*e^4/d - 4/15*(-(
d*x + c)^2 + 1)^(3/2)*a*b*e^4/d + 2/5*sqrt(-(d*x + c)^2 + 1)*b^2*arcsin(d*x + c)*e^4/d - 298/1125*(d*x + c)*b^
2*e^4/d + 2/5*sqrt(-(d*x + c)^2 + 1)*a*b*e^4/d