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

Optimal. Leaf size=165 \[ -\frac{3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac{3 b e (c+d x) \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac{e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac{e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b^3 e (c+d x) \sqrt{1-(c+d x)^2}}{8 d}+\frac{3 b^3 e \sin ^{-1}(c+d x)}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.20941, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4805, 12, 4627, 4707, 4641, 321, 216} \[ -\frac{3 b^2 e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )}{4 d}+\frac{3 b e (c+d x) \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{4 d}+\frac{e (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3}{2 d}-\frac{e \left (a+b \sin ^{-1}(c+d x)\right )^3}{4 d}-\frac{3 b^3 e (c+d x) \sqrt{1-(c+d x)^2}}{8 d}+\frac{3 b^3 e \sin ^{-1}(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.216465, size = 137, normalized size = 0.83 \[ \frac{e \left (-3 b^2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )+3 b (c+d x) \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2+2 (c+d x)^2 \left (a+b \sin ^{-1}(c+d x)\right )^3-\left (a+b \sin ^{-1}(c+d x)\right )^3+\frac{3}{2} b^3 \left (\sin ^{-1}(c+d x)-(c+d x) \sqrt{1-(c+d x)^2}\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.036, size = 266, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({\frac{e \left ( dx+c \right ) ^{2}{a}^{3}}{2}}+e{b}^{3} \left ({\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{3} \left ( \left ( dx+c \right ) ^{2}-1 \right ) }{2}}+{\frac{3\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}}{4} \left ( \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}+\arcsin \left ( dx+c \right ) \right ) }-{\frac{3\,\arcsin \left ( dx+c \right ) \left ( \left ( dx+c \right ) ^{2}-1 \right ) }{4}}-{\frac{3\,dx+3\,c}{8}\sqrt{1- \left ( dx+c \right ) ^{2}}}-{\frac{3\,\arcsin \left ( dx+c \right ) }{8}}-{\frac{ \left ( \arcsin \left ( dx+c \right ) \right ) ^{3}}{2}} \right ) +3\,ea{b}^{2} \left ( 1/2\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2} \left ( \left ( dx+c \right ) ^{2}-1 \right ) +1/2\,\arcsin \left ( dx+c \right ) \left ( \left ( dx+c \right ) \sqrt{1- \left ( dx+c \right ) ^{2}}+\arcsin \left ( dx+c \right ) \right ) -1/4\, \left ( \arcsin \left ( dx+c \right ) \right ) ^{2}-1/4\, \left ( dx+c \right ) ^{2} \right ) +3\,e{a}^{2}b \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))^3,x)

[Out]

1/d*(1/2*e*(d*x+c)^2*a^3+e*b^3*(1/2*arcsin(d*x+c)^3*((d*x+c)^2-1)+3/4*arcsin(d*x+c)^2*((d*x+c)*(1-(d*x+c)^2)^(
1/2)+arcsin(d*x+c))-3/4*arcsin(d*x+c)*((d*x+c)^2-1)-3/8*(d*x+c)*(1-(d*x+c)^2)^(1/2)-3/8*arcsin(d*x+c)-1/2*arcs
in(d*x+c)^3)+3*e*a*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)+arcsi
n(d*x+c))-1/4*arcsin(d*x+c)^2-1/4*(d*x+c)^2)+3*e*a^2*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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.56369, size = 716, normalized size = 4.34 \begin{align*} \frac{2 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} d^{2} e x^{2} + 4 \,{\left (2 \, a^{3} - 3 \, a b^{2}\right )} c d e x + 2 \,{\left (2 \, b^{3} d^{2} e x^{2} + 4 \, b^{3} c d e x +{\left (2 \, b^{3} c^{2} - b^{3}\right )} e\right )} \arcsin \left (d x + c\right )^{3} + 6 \,{\left (2 \, a b^{2} d^{2} e x^{2} + 4 \, a b^{2} c d e x +{\left (2 \, a b^{2} c^{2} - a b^{2}\right )} e\right )} \arcsin \left (d x + c\right )^{2} + 3 \,{\left (2 \,{\left (2 \, a^{2} b - b^{3}\right )} d^{2} e x^{2} + 4 \,{\left (2 \, a^{2} b - b^{3}\right )} c d e x -{\left (2 \, a^{2} b - b^{3} - 2 \,{\left (2 \, a^{2} b - b^{3}\right )} c^{2}\right )} e\right )} \arcsin \left (d x + c\right ) + 3 \,{\left ({\left (2 \, a^{2} b - b^{3}\right )} d e x +{\left (2 \, a^{2} b - b^{3}\right )} c e + 2 \,{\left (b^{3} d e x + b^{3} c e\right )} \arcsin \left (d x + c\right )^{2} + 4 \,{\left (a b^{2} d e x + a b^{2} c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt{-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{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))^3,x, algorithm="fricas")

[Out]

1/8*(2*(2*a^3 - 3*a*b^2)*d^2*e*x^2 + 4*(2*a^3 - 3*a*b^2)*c*d*e*x + 2*(2*b^3*d^2*e*x^2 + 4*b^3*c*d*e*x + (2*b^3
*c^2 - b^3)*e)*arcsin(d*x + c)^3 + 6*(2*a*b^2*d^2*e*x^2 + 4*a*b^2*c*d*e*x + (2*a*b^2*c^2 - a*b^2)*e)*arcsin(d*
x + c)^2 + 3*(2*(2*a^2*b - b^3)*d^2*e*x^2 + 4*(2*a^2*b - b^3)*c*d*e*x - (2*a^2*b - b^3 - 2*(2*a^2*b - b^3)*c^2
)*e)*arcsin(d*x + c) + 3*((2*a^2*b - b^3)*d*e*x + (2*a^2*b - b^3)*c*e + 2*(b^3*d*e*x + b^3*c*e)*arcsin(d*x + c
)^2 + 4*(a*b^2*d*e*x + a*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 = 2.99633, size = 685, normalized size = 4.15 \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)*(a+b*asin(d*x+c))**3,x)

[Out]

Piecewise((a**3*c*e*x + a**3*d*e*x**2/2 + 3*a**2*b*c**2*e*asin(c + d*x)/(2*d) + 3*a**2*b*c*e*x*asin(c + d*x) +
3*a**2*b*c*e*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/(4*d) + 3*a**2*b*d*e*x**2*asin(c + d*x)/2 + 3*a**2*b*e*x*s
qrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/4 - 3*a**2*b*e*asin(c + d*x)/(4*d) + 3*a*b**2*c**2*e*asin(c + d*x)**2/(2*
d) + 3*a*b**2*c*e*x*asin(c + d*x)**2 - 3*a*b**2*c*e*x/2 + 3*a*b**2*c*e*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*a
sin(c + d*x)/(2*d) + 3*a*b**2*d*e*x**2*asin(c + d*x)**2/2 - 3*a*b**2*d*e*x**2/4 + 3*a*b**2*e*x*sqrt(-c**2 - 2*
c*d*x - d**2*x**2 + 1)*asin(c + d*x)/2 - 3*a*b**2*e*asin(c + d*x)**2/(4*d) + b**3*c**2*e*asin(c + d*x)**3/(2*d
) - 3*b**3*c**2*e*asin(c + d*x)/(4*d) + b**3*c*e*x*asin(c + d*x)**3 - 3*b**3*c*e*x*asin(c + d*x)/2 + 3*b**3*c*
e*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)*asin(c + d*x)**2/(4*d) - 3*b**3*c*e*sqrt(-c**2 - 2*c*d*x - d**2*x**2 +
1)/(8*d) + b**3*d*e*x**2*asin(c + d*x)**3/2 - 3*b**3*d*e*x**2*asin(c + d*x)/4 + 3*b**3*e*x*sqrt(-c**2 - 2*c*d
*x - d**2*x**2 + 1)*asin(c + d*x)**2/4 - 3*b**3*e*x*sqrt(-c**2 - 2*c*d*x - d**2*x**2 + 1)/8 - b**3*e*asin(c +
d*x)**3/(4*d) + 3*b**3*e*asin(c + d*x)/(8*d), Ne(d, 0)), (c*e*x*(a + b*asin(c))**3, True))

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Giac [B]  time = 1.26782, size = 479, normalized size = 2.9 \begin{align*} \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{3} \arcsin \left (d x + c\right )^{3} e}{2 \, d} + \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b^{3} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac{3 \,{\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{2} \arcsin \left (d x + c\right )^{2} e}{2 \, d} + \frac{b^{3} \arcsin \left (d x + c\right )^{3} e}{4 \, d} + \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} a b^{2} \arcsin \left (d x + c\right ) e}{2 \, d} + \frac{3 \,{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{2} b \arcsin \left (d x + c\right ) e}{2 \, d} - \frac{3 \,{\left ({\left (d x + c\right )}^{2} - 1\right )} b^{3} \arcsin \left (d x + c\right ) e}{4 \, d} + \frac{3 \, a b^{2} \arcsin \left (d x + c\right )^{2} e}{4 \, d} + \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} a^{2} b e}{4 \, d} - \frac{3 \, \sqrt{-{\left (d x + c\right )}^{2} + 1}{\left (d x + c\right )} b^{3} e}{8 \, d} + \frac{{\left ({\left (d x + c\right )}^{2} - 1\right )} a^{3} e}{2 \, d} - \frac{3 \,{\left ({\left (d x + c\right )}^{2} - 1\right )} a b^{2} e}{4 \, d} + \frac{3 \, a^{2} b \arcsin \left (d x + c\right ) e}{4 \, d} - \frac{3 \, b^{3} \arcsin \left (d x + c\right ) e}{8 \, d} - \frac{3 \, a 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))^3,x, algorithm="giac")

[Out]

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