\(\int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx\) [410]

Optimal result

Integrand size = 30, antiderivative size = 188 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}}-\frac {3 \cos (e+f x)}{8 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{3/2}}+\frac {3 \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:

-1/4*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2)-3/8*cos(f* 
x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(3/2)+3/8*cos(f*x+e)/a^2/ 
f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2)+3/8*arctanh(sin(f*x+e))*co 
s(f*x+e)/a^2/c/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.45 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.44 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (154+48 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )+6 \cos (2 (e+f x)) \left (31+8 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-8 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )-48 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+3 \left (-9+8 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-8 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)+69 \sin (3 (e+f x))+24 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right ) \sin (3 (e+f x))-24 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right ) \sin (3 (e+f x))\right )}{256 c f (-1+\sin (e+f x)) (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2 
])*(154 + 48*Log[1 - Tan[(e + f*x)/2]] + 6*Cos[2*(e + f*x)]*(31 + 8*Log[1 
- Tan[(e + f*x)/2]] - 8*Log[1 + Tan[(e + f*x)/2]]) - 48*Log[1 + Tan[(e + f 
*x)/2]] + 3*(-9 + 8*Log[1 - Tan[(e + f*x)/2]] - 8*Log[1 + Tan[(e + f*x)/2] 
])*Sin[e + f*x] + 69*Sin[3*(e + f*x)] + 24*Log[1 - Tan[(e + f*x)/2]]*Sin[3 
*(e + f*x)] - 24*Log[1 + Tan[(e + f*x)/2]]*Sin[3*(e + f*x)]))/(256*c*f*(-1 
 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3222, 3042, 3222, 3042, 3222, 3042, 3220, 3042, 4257}

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.

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {3 \int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{3/2}}dx}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{3/2}}dx}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3220

\(\displaystyle \frac {3 \left (\frac {\frac {\cos (e+f x) \int \sec (e+f x)dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \left (\frac {\frac {\cos (e+f x) \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {3 \left (\frac {\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{3/2}}\right )}{4 a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{3/2}}\)

Input:

Int[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2)),x]
 

Output:

-1/4*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(3/2) 
) + (3*(-1/2*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x 
])^(3/2)) + (Cos[e + f*x]/(2*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x 
])^(3/2)) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*c*f*Sqrt[a + a*Sin[e + 
 f*x]]*Sqrt[c - c*Sin[e + f*x]]))/a))/(4*a)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3220

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ 
.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x] 
]*Sqrt[c + d*Sin[e + f*x]])   Int[1/Cos[e + f*x], x], x] /; FreeQ[{a, b, c, 
 d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
 

rule 3222

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( 
f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( 
(c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) 
)   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free 
Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && 
 ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || 
!SumSimplerQ[n, 1])
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 2.00 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.61

method	result	size
default	\(-\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (96 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )-96 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right )+96 \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )-3 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}-42 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+69 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-16\right ) \sqrt {4}}{512 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2} c}\)	\(303\)

Input:

int(1/(a+sin(f*x+e)*a)^(5/2)/(c-c*sin(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

-1/512/f*sec(1/4*Pi+1/2*f*x+1/2*e)*csc(1/4*Pi+1/2*f*x+1/2*e)^3*(96*sin(1/4 
*Pi+1/2*f*x+1/2*e)^4*cos(1/4*Pi+1/2*f*x+1/2*e)^2*ln(-cot(1/4*Pi+1/2*f*x+1/ 
2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)-1)-96*sin(1/4*Pi+1/2*f*x+1/2*e)^4*cos(1/4*P 
i+1/2*f*x+1/2*e)^2*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e) 
)+96*sin(1/4*Pi+1/2*f*x+1/2*e)^4*cos(1/4*Pi+1/2*f*x+1/2*e)^2*ln(-cot(1/4*P 
i+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)-3*cos(1/4*Pi+1/2*f*x+1/2*e)^ 
6-42*cos(1/4*Pi+1/2*f*x+1/2*e)^4+69*cos(1/4*Pi+1/2*f*x+1/2*e)^2-16)/(a*sin 
(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c*cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a^2/ 
c*4^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.96 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\left [\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {a c} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) - 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} - 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{16 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}}, -\frac {3 \, {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )}\right ) + {\left (3 \, \cos \left (f x + e\right )^{2} - 3 \, \sin \left (f x + e\right ) - 1\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{8 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a^{3} c^{2} f \cos \left (f x + e\right )^{3}\right )}}\right ] \] Input:

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

[1/16*(3*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(a*c)*log(-(a* 
c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + 
a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/cos(f*x + e)^3) - 2*(3*cos(f*x 
+ e)^2 - 3*sin(f*x + e) - 1)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) 
 + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x + e) + a^3*c^2*f*cos(f*x + e)^3), 
 -1/8*(3*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(-a*c)*arctan( 
sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e) 
/(a*c*cos(f*x + e))) + (3*cos(f*x + e)^2 - 3*sin(f*x + e) - 1)*sqrt(a*sin( 
f*x + e) + a)*sqrt(-c*sin(f*x + e) + c))/(a^3*c^2*f*cos(f*x + e)^3*sin(f*x 
 + e) + a^3*c^2*f*cos(f*x + e)^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

integrate(1/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(3/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{3/2}} \,d x \] Input:

int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2)),x)
 

Output:

int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}+\sin \left (f x +e \right )^{4}-2 \sin \left (f x +e \right )^{3}-2 \sin \left (f x +e \right )^{2}+\sin \left (f x +e \right )+1}d x \right )}{a^{3} c^{2}} \] Input:

int(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(3/2),x)
 

Output:

(sqrt(c)*sqrt(a)*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(s 
in(e + f*x)**5 + sin(e + f*x)**4 - 2*sin(e + f*x)**3 - 2*sin(e + f*x)**2 + 
 sin(e + f*x) + 1),x))/(a**3*c**2)