\(\int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx\) [57]

Optimal result

Integrand size = 21, antiderivative size = 109 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\frac {3 b e^3 (c+d x) \sqrt {1-(c+d x)^2}}{32 d}+\frac {b e^3 (c+d x)^3 \sqrt {1-(c+d x)^2}}{16 d}-\frac {3 b e^3 \arcsin (c+d x)}{32 d}+\frac {e^3 (c+d x)^4 (a+b \arcsin (c+d x))}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.80 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\frac {e^3 \left (3 b (c+d x) \sqrt {1-(c+d x)^2}+2 b (c+d x)^3 \sqrt {1-(c+d x)^2}-3 b \arcsin (c+d x)+8 (c+d x)^4 (a+b \arcsin (c+d x))\right )}{32 d} \] Input:

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

Output:

(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])))/(3 
2*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5304, 27, 5138, 262, 262, 223}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx\)

\(\Big \downarrow \) 5304

\(\displaystyle \frac {\int e^3 (c+d x)^3 (a+b \arcsin (c+d x))d(c+d x)}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {e^3 \int (c+d x)^3 (a+b \arcsin (c+d x))d(c+d x)}{d}\)

\(\Big \downarrow \) 5138

\(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \arcsin (c+d x))-\frac {1}{4} b \int \frac {(c+d x)^4}{\sqrt {1-(c+d x)^2}}d(c+d x)\right )}{d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \arcsin (c+d x))-\frac {1}{4} b \left (\frac {3}{4} \int \frac {(c+d x)^2}{\sqrt {1-(c+d x)^2}}d(c+d x)-\frac {1}{4} (c+d x)^3 \sqrt {1-(c+d x)^2}\right )\right )}{d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \arcsin (c+d x))-\frac {1}{4} b \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{\sqrt {1-(c+d x)^2}}d(c+d x)-\frac {1}{2} (c+d x) \sqrt {1-(c+d x)^2}\right )-\frac {1}{4} (c+d x)^3 \sqrt {1-(c+d x)^2}\right )\right )}{d}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {e^3 \left (\frac {1}{4} (c+d x)^4 (a+b \arcsin (c+d x))-\frac {1}{4} b \left (\frac {3}{4} \left (\frac {1}{2} \arcsin (c+d x)-\frac {1}{2} (c+d x) \sqrt {1-(c+d x)^2}\right )-\frac {1}{4} (c+d x)^3 \sqrt {1-(c+d x)^2}\right )\right )}{d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 223

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]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 5138

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[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 5304

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m 
_.), x_Symbol] :> Simp[1/d   Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A 
rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
 

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\frac {e^{3} a \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \arcsin \left (d x +c \right )}{4}+\frac {\left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsin \left (d x +c \right )}{32}\right )}{d}\)	\(90\)
default	\(\frac {\frac {e^{3} a \left (d x +c \right )^{4}}{4}+e^{3} b \left (\frac {\left (d x +c \right )^{4} \arcsin \left (d x +c \right )}{4}+\frac {\left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsin \left (d x +c \right )}{32}\right )}{d}\)	\(90\)
parts	\(\frac {e^{3} a \left (d x +c \right )^{4}}{4 d}+\frac {e^{3} b \left (\frac {\left (d x +c \right )^{4} \arcsin \left (d x +c \right )}{4}+\frac {\left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{16}+\frac {3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{32}-\frac {3 \arcsin \left (d x +c \right )}{32}\right )}{d}\)	\(92\)
orering	\(\frac {\left (14 d^{4} x^{4}+56 c \,d^{3} x^{3}+84 c^{2} d^{2} x^{2}+56 c^{3} d x +14 c^{4}+3 d^{2} x^{2}+6 c d x +3 c^{2}-12\right ) \left (d e x +c e \right )^{3} \left (a +b \arcsin \left (d x +c \right )\right )}{32 \left (d x +c \right )^{3} d}-\frac {\left (d x +c -1\right ) \left (d x +c +1\right ) \left (2 d^{2} x^{2}+4 c d x +2 c^{2}+3\right ) \left (3 \left (d e x +c e \right )^{2} \left (a +b \arcsin \left (d x +c \right )\right ) d e +\frac {\left (d e x +c e \right )^{3} b d}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{32 d^{2} \left (d x +c \right )^{2}}\)	\(189\)

Input:

int((d*e*x+c*e)^3*(a+b*arcsin(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(1/4*e^3*a*(d*x+c)^4+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)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (97) = 194\).

Time = 0.09 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.92 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\frac {8 \, a d^{4} e^{3} x^{4} + 32 \, a c d^{3} e^{3} x^{3} + 48 \, a c^{2} d^{2} e^{3} x^{2} + 32 \, a c^{3} d e^{3} x + {\left (8 \, b d^{4} e^{3} x^{4} + 32 \, b c d^{3} e^{3} x^{3} + 48 \, b c^{2} d^{2} e^{3} x^{2} + 32 \, b c^{3} d e^{3} x + {\left (8 \, b c^{4} - 3 \, b\right )} e^{3}\right )} \arcsin \left (d x + c\right ) + {\left (2 \, b d^{3} e^{3} x^{3} + 6 \, b c d^{2} e^{3} x^{2} + 3 \, {\left (2 \, b c^{2} + b\right )} d e^{3} x + {\left (2 \, b c^{3} + 3 \, b c\right )} e^{3}\right )} \sqrt {-d^{2} x^{2} - 2 \, c d x - c^{2} + 1}}{32 \, d} \] Input:

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

Output:

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
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (94) = 188\).

Time = 0.28 (sec) , antiderivative size = 394, normalized size of antiderivative = 3.61 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\begin {cases} a c^{3} e^{3} x + \frac {3 a c^{2} d e^{3} x^{2}}{2} + a c d^{2} e^{3} x^{3} + \frac {a d^{3} e^{3} x^{4}}{4} + \frac {b c^{4} e^{3} \operatorname {asin}{\left (c + d x \right )}}{4 d} + b c^{3} e^{3} x \operatorname {asin}{\left (c + d x \right )} + \frac {b c^{3} e^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16 d} + \frac {3 b c^{2} d e^{3} x^{2} \operatorname {asin}{\left (c + d x \right )}}{2} + \frac {3 b c^{2} e^{3} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + b c d^{2} e^{3} x^{3} \operatorname {asin}{\left (c + d x \right )} + \frac {3 b c d e^{3} x^{2} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac {3 b c e^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32 d} + \frac {b d^{3} e^{3} x^{4} \operatorname {asin}{\left (c + d x \right )}}{4} + \frac {b d^{2} e^{3} x^{3} \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{16} + \frac {3 b e^{3} x \sqrt {- c^{2} - 2 c d x - d^{2} x^{2} + 1}}{32} - \frac {3 b e^{3} \operatorname {asin}{\left (c + d x \right )}}{32 d} & \text {for}\: d \neq 0 \\c^{3} e^{3} x \left (a + b \operatorname {asin}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

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*sq 
rt(-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))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 816 vs. \(2 (97) = 194\).

Time = 0.11 (sec) , antiderivative size = 816, normalized size of antiderivative = 7.49 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/4*a*d^3*e^3*x^4 + a*c*d^2*e^3*x^3 + 3/2*a*c^2*d*e^3*x^2 + 3/4*(2*x^2*arc 
sin(d*x + c) + d*(3*c^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2 
))/d^3 + sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x/d^2 - (c^2 - 1)*arcsin(-(d^2 
*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^3 - 3*sqrt(-d^2*x^2 - 2*c*d*x - 
 c^2 + 1)*c/d^3))*b*c^2*d*e^3 + 1/6*(6*x^3*arcsin(d*x + c) + d*(2*sqrt(-d^ 
2*x^2 - 2*c*d*x - c^2 + 1)*x^2/d^2 - 15*c^3*arcsin(-(d^2*x + c*d)/sqrt(c^2 
*d^2 - (c^2 - 1)*d^2))/d^4 - 5*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c*x/d^3 
+ 9*(c^2 - 1)*c*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^4 + 
 15*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2/d^4 - 4*sqrt(-d^2*x^2 - 2*c*d*x 
 - c^2 + 1)*(c^2 - 1)/d^4))*b*c*d^2*e^3 + 1/96*(24*x^4*arcsin(d*x + c) + ( 
6*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*x^3/d^2 - 14*sqrt(-d^2*x^2 - 2*c*d*x 
- c^2 + 1)*c*x^2/d^3 + 105*c^4*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 
 1)*d^2))/d^5 + 35*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^2*x/d^4 - 90*(c^2 
- 1)*c^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - (c^2 - 1)*d^2))/d^5 - 105*sq 
rt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*c^3/d^5 - 9*sqrt(-d^2*x^2 - 2*c*d*x - c^2 
 + 1)*(c^2 - 1)*x/d^4 + 9*(c^2 - 1)^2*arcsin(-(d^2*x + c*d)/sqrt(c^2*d^2 - 
 (c^2 - 1)*d^2))/d^5 + 55*sqrt(-d^2*x^2 - 2*c*d*x - c^2 + 1)*(c^2 - 1)*c/d 
^5)*d)*b*d^3*e^3 + a*c^3*e^3*x + ((d*x + c)*arcsin(d*x + c) + sqrt(-(d*x + 
 c)^2 + 1))*b*c^3*e^3/d
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.25 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\frac {{\left (d x + c\right )}^{4} a e^{3}}{4 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )}^{2} b e^{3} \arcsin \left (d x + c\right )}{4 \, d} - \frac {{\left (-{\left (d x + c\right )}^{2} + 1\right )}^{\frac {3}{2}} {\left (d x + c\right )} b e^{3}}{16 \, d} + \frac {{\left ({\left (d x + c\right )}^{2} - 1\right )} b e^{3} \arcsin \left (d x + c\right )}{2 \, d} + \frac {5 \, \sqrt {-{\left (d x + c\right )}^{2} + 1} {\left (d x + c\right )} b e^{3}}{32 \, d} + \frac {5 \, b e^{3} \arcsin \left (d x + c\right )}{32 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\int {\left (c\,e+d\,e\,x\right )}^3\,\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right ) \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.84 \[ \int (c e+d e x)^3 (a+b \arcsin (c+d x)) \, dx=\frac {e^{3} \left (24 \mathit {asin} \left (d x +c \right ) b \,c^{4}+96 \mathit {asin} \left (d x +c \right ) b \,c^{3} d x +144 \mathit {asin} \left (d x +c \right ) b \,c^{2} d^{2} x^{2}+96 \mathit {asin} \left (d x +c \right ) b c \,d^{3} x^{3}+24 \mathit {asin} \left (d x +c \right ) b \,d^{4} x^{4}-9 \mathit {asin} \left (d x +c \right ) b +6 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b \,c^{3}+18 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b \,c^{2} d x +18 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b c \,d^{2} x^{2}+9 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b c +6 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b \,d^{3} x^{3}+9 \sqrt {-d^{2} x^{2}-2 c d x -c^{2}+1}\, b d x +96 a \,c^{3} d x +144 a \,c^{2} d^{2} x^{2}+96 a c \,d^{3} x^{3}+24 a \,d^{4} x^{4}+384 b \,c^{3}+64 b c \right )}{96 d} \] Input:

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

Output:

(e**3*(24*asin(c + d*x)*b*c**4 + 96*asin(c + d*x)*b*c**3*d*x + 144*asin(c 
+ d*x)*b*c**2*d**2*x**2 + 96*asin(c + d*x)*b*c*d**3*x**3 + 24*asin(c + d*x 
)*b*d**4*x**4 - 9*asin(c + d*x)*b + 6*sqrt( - c**2 - 2*c*d*x - d**2*x**2 + 
 1)*b*c**3 + 18*sqrt( - c**2 - 2*c*d*x - d**2*x**2 + 1)*b*c**2*d*x + 18*sq 
rt( - c**2 - 2*c*d*x - d**2*x**2 + 1)*b*c*d**2*x**2 + 9*sqrt( - c**2 - 2*c 
*d*x - d**2*x**2 + 1)*b*c + 6*sqrt( - c**2 - 2*c*d*x - d**2*x**2 + 1)*b*d* 
*3*x**3 + 9*sqrt( - c**2 - 2*c*d*x - d**2*x**2 + 1)*b*d*x + 96*a*c**3*d*x 
+ 144*a*c**2*d**2*x**2 + 96*a*c*d**3*x**3 + 24*a*d**4*x**4 + 384*b*c**3 + 
64*b*c))/(96*d)