\(\int \frac {(d+e x)^3}{a+b \arcsin (c x)} \, dx\) [16]

Optimal result

Integrand size = 18, antiderivative size = 393 \[ \int \frac {(d+e x)^3}{a+b \arcsin (c x)} \, dx=\frac {d^3 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{b c}+\frac {3 d e^2 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{4 b c^3}-\frac {3 d e^2 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 a}{b}+3 \arcsin (c x)\right )}{4 b c^3}-\frac {3 d^2 e \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right ) \sin \left (\frac {2 a}{b}\right )}{2 b c^2}-\frac {e^3 \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right ) \sin \left (\frac {2 a}{b}\right )}{4 b c^4}+\frac {e^3 \operatorname {CosIntegral}\left (\frac {4 a}{b}+4 \arcsin (c x)\right ) \sin \left (\frac {4 a}{b}\right )}{8 b c^4}+\frac {d^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{b c}+\frac {3 d e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{4 b c^3}+\frac {3 d^2 e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b c^2}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{4 b c^4}-\frac {3 d e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \arcsin (c x)\right )}{4 b c^3}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \arcsin (c x)\right )}{8 b c^4} \] Output:

d^3*cos(a/b)*Ci(a/b+arcsin(c*x))/b/c+3/4*d*e^2*cos(a/b)*Ci(a/b+arcsin(c*x) 
)/b/c^3-3/4*d*e^2*cos(3*a/b)*Ci(3*a/b+3*arcsin(c*x))/b/c^3-3/2*d^2*e*Ci(2* 
a/b+2*arcsin(c*x))*sin(2*a/b)/b/c^2-1/4*e^3*Ci(2*a/b+2*arcsin(c*x))*sin(2* 
a/b)/b/c^4+1/8*e^3*Ci(4*a/b+4*arcsin(c*x))*sin(4*a/b)/b/c^4+d^3*sin(a/b)*S 
i(a/b+arcsin(c*x))/b/c+3/4*d*e^2*sin(a/b)*Si(a/b+arcsin(c*x))/b/c^3+3/2*d^ 
2*e*cos(2*a/b)*Si(2*a/b+2*arcsin(c*x))/b/c^2+1/4*e^3*cos(2*a/b)*Si(2*a/b+2 
*arcsin(c*x))/b/c^4-3/4*d*e^2*sin(3*a/b)*Si(3*a/b+3*arcsin(c*x))/b/c^3-1/8 
*e^3*cos(4*a/b)*Si(4*a/b+4*arcsin(c*x))/b/c^4
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.77 \[ \int \frac {(d+e x)^3}{a+b \arcsin (c x)} \, dx=\frac {d^3 \left (\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )\right )}{b c}+\frac {3 d e^2 \left (\cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )-\cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\arcsin (c x)\right )\right )\right )}{4 b c^3}+\frac {e^3 \left (-2 \operatorname {CosIntegral}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {2 a}{b}\right )+\operatorname {CosIntegral}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right ) \sin \left (\frac {4 a}{b}\right )+2 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (2 \left (\frac {a}{b}+\arcsin (c x)\right )\right )-\cos \left (\frac {4 a}{b}\right ) \text {Si}\left (4 \left (\frac {a}{b}+\arcsin (c x)\right )\right )\right )}{8 b c^4}+\frac {3 d^2 e \left (-\operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right ) \sin \left (\frac {2 a}{b}\right )+\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )\right )}{2 b c^2} \] Input:

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

Output:

(d^3*(Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]] + Sin[a/b]*SinIntegral[a/b + 
 ArcSin[c*x]]))/(b*c) + (3*d*e^2*(Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]] 
- Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcSin[c*x])] + Sin[a/b]*SinIntegral[a 
/b + ArcSin[c*x]] - Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c*x])]))/(4*b 
*c^3) + (e^3*(-2*CosIntegral[2*(a/b + ArcSin[c*x])]*Sin[(2*a)/b] + CosInte 
gral[4*(a/b + ArcSin[c*x])]*Sin[(4*a)/b] + 2*Cos[(2*a)/b]*SinIntegral[2*(a 
/b + ArcSin[c*x])] - Cos[(4*a)/b]*SinIntegral[4*(a/b + ArcSin[c*x])]))/(8* 
b*c^4) + (3*d^2*e*(-(CosIntegral[(2*a)/b + 2*ArcSin[c*x]]*Sin[(2*a)/b]) + 
Cos[(2*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]]))/(2*b*c^2)
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 377, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5246, 7293, 2009}

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.

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

\(\Big \downarrow \) 5246

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {c^3 d^3 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{b}+\frac {c^3 d^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{b}-\frac {3 c^2 d^2 e \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b}+\frac {3 c^2 d^2 e \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{2 b}+\frac {3 c d e^2 \cos \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right )}{4 b}-\frac {3 c d e^2 \cos \left (\frac {3 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {3 a}{b}+3 \arcsin (c x)\right )}{4 b}-\frac {e^3 \sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{4 b}+\frac {e^3 \sin \left (\frac {4 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {4 a}{b}+4 \arcsin (c x)\right )}{8 b}+\frac {3 c d e^2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{4 b}-\frac {3 c d e^2 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \arcsin (c x)\right )}{4 b}+\frac {e^3 \cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arcsin (c x)\right )}{4 b}-\frac {e^3 \cos \left (\frac {4 a}{b}\right ) \text {Si}\left (\frac {4 a}{b}+4 \arcsin (c x)\right )}{8 b}}{c^4}\)

Input:

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

Output:

((c^3*d^3*Cos[a/b]*CosIntegral[a/b + ArcSin[c*x]])/b + (3*c*d*e^2*Cos[a/b] 
*CosIntegral[a/b + ArcSin[c*x]])/(4*b) - (3*c*d*e^2*Cos[(3*a)/b]*CosIntegr 
al[(3*a)/b + 3*ArcSin[c*x]])/(4*b) - (3*c^2*d^2*e*CosIntegral[(2*a)/b + 2* 
ArcSin[c*x]]*Sin[(2*a)/b])/(2*b) - (e^3*CosIntegral[(2*a)/b + 2*ArcSin[c*x 
]]*Sin[(2*a)/b])/(4*b) + (e^3*CosIntegral[(4*a)/b + 4*ArcSin[c*x]]*Sin[(4* 
a)/b])/(8*b) + (c^3*d^3*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]])/b + (3*c* 
d*e^2*Sin[a/b]*SinIntegral[a/b + ArcSin[c*x]])/(4*b) + (3*c^2*d^2*e*Cos[(2 
*a)/b]*SinIntegral[(2*a)/b + 2*ArcSin[c*x]])/(2*b) + (e^3*Cos[(2*a)/b]*Sin 
Integral[(2*a)/b + 2*ArcSin[c*x]])/(4*b) - (3*c*d*e^2*Sin[(3*a)/b]*SinInte 
gral[(3*a)/b + 3*ArcSin[c*x]])/(4*b) - (e^3*Cos[(4*a)/b]*SinIntegral[(4*a) 
/b + 4*ArcSin[c*x]])/(8*b))/c^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5246

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((d_.) + (e_.)*(x_))^(m_.), x_S 
ymbol] :> Simp[1/c^(m + 1)   Subst[Int[(a + b*x)^n*Cos[x]*(c*d + e*Sin[x])^ 
m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[m, 0]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 327, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {8 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) c^{3} d^{3}+8 \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) c^{3} d^{3}-12 \sin \left (\frac {2 a}{b}\right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) c^{2} d^{2} e +12 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) c^{2} d^{2} e +6 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) c d \,e^{2}-6 \sin \left (\frac {3 a}{b}\right ) \operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) c d \,e^{2}-6 \cos \left (\frac {3 a}{b}\right ) \operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) c d \,e^{2}+6 \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) c d \,e^{2}-2 \sin \left (\frac {2 a}{b}\right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) e^{3}+2 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) e^{3}-\cos \left (\frac {4 a}{b}\right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) e^{3}+\sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) e^{3}}{8 c^{4} b}\)	\(327\)
default	\(\frac {8 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) c^{3} d^{3}+8 \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) c^{3} d^{3}-12 \sin \left (\frac {2 a}{b}\right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) c^{2} d^{2} e +12 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) c^{2} d^{2} e +6 \,\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) c d \,e^{2}-6 \sin \left (\frac {3 a}{b}\right ) \operatorname {Si}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) c d \,e^{2}-6 \cos \left (\frac {3 a}{b}\right ) \operatorname {Ci}\left (3 \arcsin \left (c x \right )+\frac {3 a}{b}\right ) c d \,e^{2}+6 \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) c d \,e^{2}-2 \sin \left (\frac {2 a}{b}\right ) \operatorname {Ci}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) e^{3}+2 \,\operatorname {Si}\left (2 \arcsin \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right ) e^{3}-\cos \left (\frac {4 a}{b}\right ) \operatorname {Si}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) e^{3}+\sin \left (\frac {4 a}{b}\right ) \operatorname {Ci}\left (4 \arcsin \left (c x \right )+\frac {4 a}{b}\right ) e^{3}}{8 c^{4} b}\)	\(327\)

Input:

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

Output:

1/8/c^4*(8*Ci(arcsin(c*x)+a/b)*cos(a/b)*c^3*d^3+8*sin(a/b)*Si(arcsin(c*x)+ 
a/b)*c^3*d^3-12*sin(2*a/b)*Ci(2*arcsin(c*x)+2*a/b)*c^2*d^2*e+12*Si(2*arcsi 
n(c*x)+2*a/b)*cos(2*a/b)*c^2*d^2*e+6*Ci(arcsin(c*x)+a/b)*cos(a/b)*c*d*e^2- 
6*sin(3*a/b)*Si(3*arcsin(c*x)+3*a/b)*c*d*e^2-6*cos(3*a/b)*Ci(3*arcsin(c*x) 
+3*a/b)*c*d*e^2+6*sin(a/b)*Si(arcsin(c*x)+a/b)*c*d*e^2-2*sin(2*a/b)*Ci(2*a 
rcsin(c*x)+2*a/b)*e^3+2*Si(2*arcsin(c*x)+2*a/b)*cos(2*a/b)*e^3-cos(4*a/b)* 
Si(4*arcsin(c*x)+4*a/b)*e^3+sin(4*a/b)*Ci(4*arcsin(c*x)+4*a/b)*e^3)/b
 

Fricas [F]

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

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

Output:

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)/(b*arcsin(c*x) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate((e*x + d)^3/(b*arcsin(c*x) + a), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 609, normalized size of antiderivative = 1.55 \[ \int \frac {(d+e x)^3}{a+b \arcsin (c x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-3*d*e^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) + d^3*cos( 
a/b)*cos_integral(a/b + arcsin(c*x))/(b*c) + e^3*cos(a/b)^3*cos_integral(4 
*a/b + 4*arcsin(c*x))*sin(a/b)/(b*c^4) - 3*d^2*e*cos(a/b)*cos_integral(2*a 
/b + 2*arcsin(c*x))*sin(a/b)/(b*c^2) - e^3*cos(a/b)^4*sin_integral(4*a/b + 
 4*arcsin(c*x))/(b*c^4) - 3*d*e^2*cos(a/b)^2*sin(a/b)*sin_integral(3*a/b + 
 3*arcsin(c*x))/(b*c^3) + 3*d^2*e*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin 
(c*x))/(b*c^2) + d^3*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c) + 9/4* 
d*e^2*cos(a/b)*cos_integral(3*a/b + 3*arcsin(c*x))/(b*c^3) + 3/4*d*e^2*cos 
(a/b)*cos_integral(a/b + arcsin(c*x))/(b*c^3) - 1/2*e^3*cos(a/b)*cos_integ 
ral(4*a/b + 4*arcsin(c*x))*sin(a/b)/(b*c^4) - 1/2*e^3*cos(a/b)*cos_integra 
l(2*a/b + 2*arcsin(c*x))*sin(a/b)/(b*c^4) + e^3*cos(a/b)^2*sin_integral(4* 
a/b + 4*arcsin(c*x))/(b*c^4) + 3/4*d*e^2*sin(a/b)*sin_integral(3*a/b + 3*a 
rcsin(c*x))/(b*c^3) - 3/2*d^2*e*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^2 
) + 1/2*e^3*cos(a/b)^2*sin_integral(2*a/b + 2*arcsin(c*x))/(b*c^4) + 3/4*d 
*e^2*sin(a/b)*sin_integral(a/b + arcsin(c*x))/(b*c^3) - 1/8*e^3*sin_integr 
al(4*a/b + 4*arcsin(c*x))/(b*c^4) - 1/4*e^3*sin_integral(2*a/b + 2*arcsin( 
c*x))/(b*c^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(d+e x)^3}{a+b \arcsin (c x)} \, dx=\left (\int \frac {x^{3}}{\mathit {asin} \left (c x \right ) b +a}d x \right ) e^{3}+3 \left (\int \frac {x^{2}}{\mathit {asin} \left (c x \right ) b +a}d x \right ) d \,e^{2}+3 \left (\int \frac {x}{\mathit {asin} \left (c x \right ) b +a}d x \right ) d^{2} e +\left (\int \frac {1}{\mathit {asin} \left (c x \right ) b +a}d x \right ) d^{3} \] Input:

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

Output:

int(x**3/(asin(c*x)*b + a),x)*e**3 + 3*int(x**2/(asin(c*x)*b + a),x)*d*e** 
2 + 3*int(x/(asin(c*x)*b + a),x)*d**2*e + int(1/(asin(c*x)*b + a),x)*d**3