\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx\) [71]

Optimal result

Integrand size = 36, antiderivative size = 201 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {7 (2 A-7 B) c^4 x}{2 a^3}-\frac {7 (2 A-7 B) c^4 \cos (e+f x)}{2 a^3 f}-\frac {a^4 (A-B) c^4 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac {2 a^2 (2 A-7 B) c^4 \cos ^7(e+f x)}{15 f (a+a \sin (e+f x))^5}-\frac {14 (2 A-7 B) c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^3}-\frac {7 (2 A-7 B) c^4 \cos ^3(e+f x)}{6 f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:

-7/2*(2*A-7*B)*c^4*x/a^3-7/2*(2*A-7*B)*c^4*cos(f*x+e)/a^3/f-1/5*a^4*(A-B)* 
c^4*cos(f*x+e)^9/f/(a+a*sin(f*x+e))^7+2/15*a^2*(2*A-7*B)*c^4*cos(f*x+e)^7/ 
f/(a+a*sin(f*x+e))^5-14/15*(2*A-7*B)*c^4*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^3 
-7/6*(2*A-7*B)*c^4*cos(f*x+e)^3/f/(a^3+a^3*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 12.17 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.73 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c-c \sin (e+f x))^4 \left (384 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )-192 (A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-128 (8 A-13 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+64 (8 A-13 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+64 (29 A-79 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-210 (2 A-7 B) (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-60 (A-7 B) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5-15 B \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5 \sin (2 (e+f x))\right )}{60 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8 (1+\sin (e+f x))^3} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4)/(a + a*Sin[e + f*x 
])^3,x]
 

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^4*(384*(A - B) 
*Sin[(e + f*x)/2] - 192*(A - B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 12 
8*(8*A - 13*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 
64*(8*A - 13*B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 64*(29*A - 79*B) 
*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 210*(2*A - 7*B 
)*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - 60*(A - 7*B)*Cos[e + 
 f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - 15*B*(Cos[(e + f*x)/2] + S 
in[(e + f*x)/2])^5*Sin[2*(e + f*x)]))/(60*a^3*f*(Cos[(e + f*x)/2] - Sin[(e 
 + f*x)/2])^8*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.92, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 3042, 3159, 3042, 3158, 3042, 3161, 24}

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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^4 (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^4 (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3}dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a^4 c^4 \int \frac {\cos ^8(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^7}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 c^4 \int \frac {\cos (e+f x)^8 (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^7}dx\)

\(\Big \downarrow \) 3338

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \int \frac {\cos ^8(e+f x)}{(\sin (e+f x) a+a)^6}dx}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \int \frac {\cos (e+f x)^8}{(\sin (e+f x) a+a)^6}dx}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \int \frac {\cos ^6(e+f x)}{(\sin (e+f x) a+a)^4}dx}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \int \frac {\cos (e+f x)^6}{(\sin (e+f x) a+a)^4}dx}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3159

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \int \frac {\cos ^4(e+f x)}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \int \frac {\cos (e+f x)^4}{(\sin (e+f x) a+a)^2}dx}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3158

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \left (\frac {3 \int \frac {\cos (e+f x)^2}{\sin (e+f x) a+a}dx}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 3161

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \left (\frac {3 \left (\frac {\int 1dx}{a}+\frac {\cos (e+f x)}{a f}\right )}{2 a}+\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^4 c^4 \left (-\frac {(2 A-7 B) \left (-\frac {7 \left (-\frac {5 \left (\frac {\cos ^3(e+f x)}{2 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac {3 \left (\frac {\cos (e+f x)}{a f}+\frac {x}{a}\right )}{2 a}\right )}{a^2}-\frac {2 \cos ^5(e+f x)}{a f (a \sin (e+f x)+a)^3}\right )}{3 a^2}-\frac {2 \cos ^7(e+f x)}{3 a f (a \sin (e+f x)+a)^5}\right )}{5 a}-\frac {(A-B) \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}\right )\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4)/(a + a*Sin[e + f*x])^3,x 
]
 

Output:

a^4*c^4*(-1/5*((A - B)*Cos[e + f*x]^9)/(f*(a + a*Sin[e + f*x])^7) - ((2*A 
- 7*B)*((-2*Cos[e + f*x]^7)/(3*a*f*(a + a*Sin[e + f*x])^5) - (7*((-2*Cos[e 
 + f*x]^5)/(a*f*(a + a*Sin[e + f*x])^3) - (5*((3*(x/a + Cos[e + f*x]/(a*f) 
))/(2*a) + Cos[e + f*x]^3/(2*f*(a^2 + a^2*Sin[e + f*x]))))/a^2))/(3*a^2))) 
/(5*a))
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3158

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x 
])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(a*(m + p)))   Int[(g*Cos 
[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, 
f, g}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || 
 EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p, 0] && In 
tegersQ[2*m, 2*p]
 

rule 3159

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[2*g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f 
*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Simp[g^2*((p - 1)/(b^2*(2*m + p + 1 
)))   Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; 
FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] & 
& NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]
 

rule 3161

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[g*((g*Cos[e + f*x])^(p - 1)/(b*f*(p - 1))), x] + Si 
mp[g^2/a   Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g}, x 
] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]
 

rule 3338

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - 
 a*d)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1) 
)), x] + Simp[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1))   Int[(g*Cos[e 
+ f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[m + p], 0 
]) && NeQ[2*m + p + 1, 0]
 

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin 
[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* 
d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 11.06 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {2 c^{4} \left (-\frac {-128 A +128 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8 A -24 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {64 A -64 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {64 A -32 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {16 B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A -7 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {7 \left (2 A -7 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,a^{3}}\)	\(201\)
default	\(\frac {2 c^{4} \left (-\frac {-128 A +128 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {8 A -24 B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {64 A -64 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {64 A -32 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {16 B}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {-\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A -7 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A -7 B}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}-\frac {7 \left (2 A -7 B \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}\right )}{f \,a^{3}}\)	\(201\)
parallelrisch	\(-\frac {7 \left (\left (-\frac {281}{14} A +\frac {471}{7} B +35 f x B -10 f x A \right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (-\frac {761}{84} B +\frac {131}{42} A +5 f x A -\frac {35}{2} f x B \right ) \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (-\frac {1937}{140} B +\frac {291}{70} A -\frac {7}{2} f x B +f x A \right ) \cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\left (\frac {895}{21} B -\frac {493}{42} A +35 f x B -10 f x A \right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )+\left (\frac {1285}{28} B -\frac {179}{14} A -5 f x A +\frac {35}{2} f x B \right ) \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\left (-\frac {41}{42} A +\frac {239}{84} B +f x A -\frac {7}{2} f x B \right ) \sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\frac {\left (A -\frac {23 B}{4}\right ) \cos \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )}{14}+\frac {\left (A -\frac {23 B}{4}\right ) \sin \left (\frac {7 f x}{2}+\frac {7 e}{2}\right )}{14}-\frac {B \left (\cos \left (\frac {9 f x}{2}+\frac {9 e}{2}\right )-\sin \left (\frac {9 f x}{2}+\frac {9 e}{2}\right )\right )}{56}\right ) c^{4}}{f \,a^{3} \left (-5 \sin \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )-10 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-10 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+5 \cos \left (\frac {3 f x}{2}+\frac {3 e}{2}\right )+\cos \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )+\sin \left (\frac {5 f x}{2}+\frac {5 e}{2}\right )\right )}\)	\(294\)
risch	\(-\frac {7 c^{4} x A}{a^{3}}+\frac {49 c^{4} x B}{2 a^{3}}+\frac {i B \,c^{4} {\mathrm e}^{2 i \left (f x +e \right )}}{8 a^{3} f}-\frac {c^{4} {\mathrm e}^{i \left (f x +e \right )} A}{2 a^{3} f}+\frac {7 c^{4} {\mathrm e}^{i \left (f x +e \right )} B}{2 a^{3} f}-\frac {c^{4} {\mathrm e}^{-i \left (f x +e \right )} A}{2 a^{3} f}+\frac {7 c^{4} {\mathrm e}^{-i \left (f x +e \right )} B}{2 a^{3} f}-\frac {i B \,c^{4} {\mathrm e}^{-2 i \left (f x +e \right )}}{8 a^{3} f}-\frac {16 \left (-170 A \,c^{4} {\mathrm e}^{2 i \left (f x +e \right )}+120 i A \,c^{4} {\mathrm e}^{3 i \left (f x +e \right )}-100 i A \,c^{4} {\mathrm e}^{i \left (f x +e \right )}+45 A \,c^{4} {\mathrm e}^{4 i \left (f x +e \right )}+460 B \,c^{4} {\mathrm e}^{2 i \left (f x +e \right )}-330 i B \,c^{4} {\mathrm e}^{3 i \left (f x +e \right )}+290 i B \,c^{4} {\mathrm e}^{i \left (f x +e \right )}-105 B \,c^{4} {\mathrm e}^{4 i \left (f x +e \right )}+29 A \,c^{4}-79 B \,c^{4}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\)	\(309\)
norman	\(\text {Expression too large to display}\)	\(920\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

2/f*c^4/a^3*(-1/4*(-128*A+128*B)/(tan(1/2*f*x+1/2*e)+1)^4-(8*A-24*B)/(tan( 
1/2*f*x+1/2*e)+1)-1/5*(64*A-64*B)/(tan(1/2*f*x+1/2*e)+1)^5-1/3*(64*A-32*B) 
/(tan(1/2*f*x+1/2*e)+1)^3+16*B/(tan(1/2*f*x+1/2*e)+1)^2-(-1/2*B*tan(1/2*f* 
x+1/2*e)^3+(A-7*B)*tan(1/2*f*x+1/2*e)^2+1/2*B*tan(1/2*f*x+1/2*e)+A-7*B)/(1 
+tan(1/2*f*x+1/2*e)^2)^2-7/2*(2*A-7*B)*arctan(tan(1/2*f*x+1/2*e)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (189) = 378\).

Time = 0.10 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.95 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {15 \, B c^{4} \cos \left (f x + e\right )^{5} - 30 \, {\left (A - 6 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} + 420 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x + 96 \, {\left (A - B\right )} c^{4} - {\left (105 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x + {\left (554 \, A - 1819 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )^{3} - {\left (315 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x - 2 \, {\left (134 \, A - 619 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (35 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x + 2 \, {\left (74 \, A - 249 \, B\right )} c^{4}\right )} \cos \left (f x + e\right ) - {\left (15 \, B c^{4} \cos \left (f x + e\right )^{4} + 15 \, {\left (2 \, A - 11 \, B\right )} c^{4} \cos \left (f x + e\right )^{3} - 420 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x + 96 \, {\left (A - B\right )} c^{4} + {\left (105 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x - 2 \, {\left (262 \, A - 827 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )^{2} - 6 \, {\left (35 \, {\left (2 \, A - 7 \, B\right )} c^{4} f x + 2 \, {\left (66 \, A - 241 \, B\right )} c^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algori 
thm="fricas")
 

Output:

1/30*(15*B*c^4*cos(f*x + e)^5 - 30*(A - 6*B)*c^4*cos(f*x + e)^4 + 420*(2*A 
 - 7*B)*c^4*f*x + 96*(A - B)*c^4 - (105*(2*A - 7*B)*c^4*f*x + (554*A - 181 
9*B)*c^4)*cos(f*x + e)^3 - (315*(2*A - 7*B)*c^4*f*x - 2*(134*A - 619*B)*c^ 
4)*cos(f*x + e)^2 + 6*(35*(2*A - 7*B)*c^4*f*x + 2*(74*A - 249*B)*c^4)*cos( 
f*x + e) - (15*B*c^4*cos(f*x + e)^4 + 15*(2*A - 11*B)*c^4*cos(f*x + e)^3 - 
 420*(2*A - 7*B)*c^4*f*x + 96*(A - B)*c^4 + (105*(2*A - 7*B)*c^4*f*x - 2*( 
262*A - 827*B)*c^4)*cos(f*x + e)^2 - 6*(35*(2*A - 7*B)*c^4*f*x + 2*(66*A - 
 241*B)*c^4)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*c 
os(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2 
*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(f*x + e))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7337 vs. \(2 (185) = 370\).

Time = 24.79 (sec) , antiderivative size = 7337, normalized size of antiderivative = 36.50 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**4/(a+a*sin(f*x+e))**3,x)
 

Output:

Piecewise((-210*A*c**4*f*x*tan(e/2 + f*x/2)**9/(30*a**3*f*tan(e/2 + f*x/2) 
**9 + 150*a**3*f*tan(e/2 + f*x/2)**8 + 360*a**3*f*tan(e/2 + f*x/2)**7 + 60 
0*a**3*f*tan(e/2 + f*x/2)**6 + 780*a**3*f*tan(e/2 + f*x/2)**5 + 780*a**3*f 
*tan(e/2 + f*x/2)**4 + 600*a**3*f*tan(e/2 + f*x/2)**3 + 360*a**3*f*tan(e/2 
 + f*x/2)**2 + 150*a**3*f*tan(e/2 + f*x/2) + 30*a**3*f) - 1050*A*c**4*f*x* 
tan(e/2 + f*x/2)**8/(30*a**3*f*tan(e/2 + f*x/2)**9 + 150*a**3*f*tan(e/2 + 
f*x/2)**8 + 360*a**3*f*tan(e/2 + f*x/2)**7 + 600*a**3*f*tan(e/2 + f*x/2)** 
6 + 780*a**3*f*tan(e/2 + f*x/2)**5 + 780*a**3*f*tan(e/2 + f*x/2)**4 + 600* 
a**3*f*tan(e/2 + f*x/2)**3 + 360*a**3*f*tan(e/2 + f*x/2)**2 + 150*a**3*f*t 
an(e/2 + f*x/2) + 30*a**3*f) - 2520*A*c**4*f*x*tan(e/2 + f*x/2)**7/(30*a** 
3*f*tan(e/2 + f*x/2)**9 + 150*a**3*f*tan(e/2 + f*x/2)**8 + 360*a**3*f*tan( 
e/2 + f*x/2)**7 + 600*a**3*f*tan(e/2 + f*x/2)**6 + 780*a**3*f*tan(e/2 + f* 
x/2)**5 + 780*a**3*f*tan(e/2 + f*x/2)**4 + 600*a**3*f*tan(e/2 + f*x/2)**3 
+ 360*a**3*f*tan(e/2 + f*x/2)**2 + 150*a**3*f*tan(e/2 + f*x/2) + 30*a**3*f 
) - 4200*A*c**4*f*x*tan(e/2 + f*x/2)**6/(30*a**3*f*tan(e/2 + f*x/2)**9 + 1 
50*a**3*f*tan(e/2 + f*x/2)**8 + 360*a**3*f*tan(e/2 + f*x/2)**7 + 600*a**3* 
f*tan(e/2 + f*x/2)**6 + 780*a**3*f*tan(e/2 + f*x/2)**5 + 780*a**3*f*tan(e/ 
2 + f*x/2)**4 + 600*a**3*f*tan(e/2 + f*x/2)**3 + 360*a**3*f*tan(e/2 + f*x/ 
2)**2 + 150*a**3*f*tan(e/2 + f*x/2) + 30*a**3*f) - 5460*A*c**4*f*x*tan(e/2 
 + f*x/2)**5/(30*a**3*f*tan(e/2 + f*x/2)**9 + 150*a**3*f*tan(e/2 + f*x/...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2394 vs. \(2 (189) = 378\).

Time = 0.19 (sec) , antiderivative size = 2394, normalized size of antiderivative = 11.91 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algori 
thm="maxima")
 

Output:

1/15*(B*c^4*((1325*sin(f*x + e)/(cos(f*x + e) + 1) + 2673*sin(f*x + e)^2/( 
cos(f*x + e) + 1)^2 + 3805*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 4329*sin( 
f*x + e)^4/(cos(f*x + e) + 1)^4 + 3575*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 
 + 2275*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 975*sin(f*x + e)^7/(cos(f*x 
+ e) + 1)^7 + 195*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 304)/(a^3 + 5*a^3* 
sin(f*x + e)/(cos(f*x + e) + 1) + 12*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1) 
^2 + 20*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 26*a^3*sin(f*x + e)^4/(c 
os(f*x + e) + 1)^4 + 26*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 20*a^3*s 
in(f*x + e)^6/(cos(f*x + e) + 1)^6 + 12*a^3*sin(f*x + e)^7/(cos(f*x + e) + 
 1)^7 + 5*a^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + a^3*sin(f*x + e)^9/(co 
s(f*x + e) + 1)^9) + 195*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) - 6* 
A*c^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(cos(f*x 
+ e) + 1)^2 + 200*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4 
/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f* 
x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e 
) + 1) + 11*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f*x + e)^ 
3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 11*a 
^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(cos(f*x + e 
) + 1)^6 + a^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin(f*x + 
e)/(cos(f*x + e) + 1))/a^3) + 24*B*c^4*((105*sin(f*x + e)/(cos(f*x + e)...
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.44 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {\frac {105 \, {\left (2 \, A c^{4} - 7 \, B c^{4}\right )} {\left (f x + e\right )}}{a^{3}} - \frac {30 \, {\left (B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 14 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, A c^{4} + 14 \, B c^{4}\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} a^{3}} + \frac {32 \, {\left (15 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 45 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 60 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 210 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 130 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 380 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 80 \, A c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 250 \, B c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 19 \, A c^{4} - 59 \, B c^{4}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{30 \, f} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algori 
thm="giac")
 

Output:

-1/30*(105*(2*A*c^4 - 7*B*c^4)*(f*x + e)/a^3 - 30*(B*c^4*tan(1/2*f*x + 1/2 
*e)^3 - 2*A*c^4*tan(1/2*f*x + 1/2*e)^2 + 14*B*c^4*tan(1/2*f*x + 1/2*e)^2 - 
 B*c^4*tan(1/2*f*x + 1/2*e) - 2*A*c^4 + 14*B*c^4)/((tan(1/2*f*x + 1/2*e)^2 
 + 1)^2*a^3) + 32*(15*A*c^4*tan(1/2*f*x + 1/2*e)^4 - 45*B*c^4*tan(1/2*f*x 
+ 1/2*e)^4 + 60*A*c^4*tan(1/2*f*x + 1/2*e)^3 - 210*B*c^4*tan(1/2*f*x + 1/2 
*e)^3 + 130*A*c^4*tan(1/2*f*x + 1/2*e)^2 - 380*B*c^4*tan(1/2*f*x + 1/2*e)^ 
2 + 80*A*c^4*tan(1/2*f*x + 1/2*e) - 250*B*c^4*tan(1/2*f*x + 1/2*e) + 19*A* 
c^4 - 59*B*c^4)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 40.06 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.08 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=-\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {286\,A\,c^4}{3}-\frac {1007\,B\,c^4}{3}\right )+\frac {334\,A\,c^4}{15}-\frac {1154\,B\,c^4}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (16\,A\,c^4-49\,B\,c^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (66\,A\,c^4-243\,B\,c^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {542\,A\,c^4}{3}-\frac {1741\,B\,c^4}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (\frac {706\,A\,c^4}{3}-\frac {2621\,B\,c^4}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {794\,A\,c^4}{3}-\frac {2875\,B\,c^4}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {1006\,A\,c^4}{5}-\frac {3401\,B\,c^4}{5}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {1718\,A\,c^4}{5}-\frac {5633\,B\,c^4}{5}\right )}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+12\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7+20\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+26\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+26\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+20\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+12\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )}-\frac {7\,c^4\,\mathrm {atan}\left (\frac {7\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A-7\,B\right )}{14\,A\,c^4-49\,B\,c^4}\right )\,\left (2\,A-7\,B\right )}{a^3\,f} \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^4)/(a + a*sin(e + f*x))^3,x 
)
 

Output:

- (tan(e/2 + (f*x)/2)*((286*A*c^4)/3 - (1007*B*c^4)/3) + (334*A*c^4)/15 - 
(1154*B*c^4)/15 + tan(e/2 + (f*x)/2)^8*(16*A*c^4 - 49*B*c^4) + tan(e/2 + ( 
f*x)/2)^7*(66*A*c^4 - 243*B*c^4) + tan(e/2 + (f*x)/2)^6*((542*A*c^4)/3 - ( 
1741*B*c^4)/3) + tan(e/2 + (f*x)/2)^5*((706*A*c^4)/3 - (2621*B*c^4)/3) + t 
an(e/2 + (f*x)/2)^3*((794*A*c^4)/3 - (2875*B*c^4)/3) + tan(e/2 + (f*x)/2)^ 
2*((1006*A*c^4)/5 - (3401*B*c^4)/5) + tan(e/2 + (f*x)/2)^4*((1718*A*c^4)/5 
 - (5633*B*c^4)/5))/(f*(12*a^3*tan(e/2 + (f*x)/2)^2 + 20*a^3*tan(e/2 + (f* 
x)/2)^3 + 26*a^3*tan(e/2 + (f*x)/2)^4 + 26*a^3*tan(e/2 + (f*x)/2)^5 + 20*a 
^3*tan(e/2 + (f*x)/2)^6 + 12*a^3*tan(e/2 + (f*x)/2)^7 + 5*a^3*tan(e/2 + (f 
*x)/2)^8 + a^3*tan(e/2 + (f*x)/2)^9 + a^3 + 5*a^3*tan(e/2 + (f*x)/2))) - ( 
7*c^4*atan((7*c^4*tan(e/2 + (f*x)/2)*(2*A - 7*B))/(14*A*c^4 - 49*B*c^4))*( 
2*A - 7*B))/(a^3*f)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.44 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx=\frac {c^{4} \left (-96 a +294 b -420 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a f x +1470 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b f x -294 \cos \left (f x +e \right ) b -1007 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +210 a f x -735 b f x -210 \cos \left (f x +e \right ) a f x +735 \cos \left (f x +e \right ) b f x +630 \sin \left (f x +e \right ) a f x -2205 \sin \left (f x +e \right ) b f x -989 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +286 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +286 a \sin \left (f x +e \right )-165 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +316 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +15 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -180 \sin \left (f x +e \right )^{4} b +762 \sin \left (f x +e \right )^{3} a +15 \sin \left (f x +e \right )^{5} b +30 \sin \left (f x +e \right )^{4} a +96 \cos \left (f x +e \right ) a -2544 \sin \left (f x +e \right )^{3} b +922 \sin \left (f x +e \right )^{2} a -3458 \sin \left (f x +e \right )^{2} b -210 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a f x +735 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b f x +210 \sin \left (f x +e \right )^{3} a f x -735 \sin \left (f x +e \right )^{3} b f x +630 \sin \left (f x +e \right )^{2} a f x -2205 \sin \left (f x +e \right )^{2} b f x -1007 \sin \left (f x +e \right ) b \right )}{30 a^{3} f \left (\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )+\cos \left (f x +e \right )-\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}-3 \sin \left (f x +e \right )-1\right )} \] Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x)
 

Output:

(c**4*(15*cos(e + f*x)*sin(e + f*x)**4*b + 30*cos(e + f*x)*sin(e + f*x)**3 
*a - 165*cos(e + f*x)*sin(e + f*x)**3*b - 210*cos(e + f*x)*sin(e + f*x)**2 
*a*f*x + 316*cos(e + f*x)*sin(e + f*x)**2*a + 735*cos(e + f*x)*sin(e + f*x 
)**2*b*f*x - 989*cos(e + f*x)*sin(e + f*x)**2*b - 420*cos(e + f*x)*sin(e + 
 f*x)*a*f*x + 286*cos(e + f*x)*sin(e + f*x)*a + 1470*cos(e + f*x)*sin(e + 
f*x)*b*f*x - 1007*cos(e + f*x)*sin(e + f*x)*b - 210*cos(e + f*x)*a*f*x + 9 
6*cos(e + f*x)*a + 735*cos(e + f*x)*b*f*x - 294*cos(e + f*x)*b + 15*sin(e 
+ f*x)**5*b + 30*sin(e + f*x)**4*a - 180*sin(e + f*x)**4*b + 210*sin(e + f 
*x)**3*a*f*x + 762*sin(e + f*x)**3*a - 735*sin(e + f*x)**3*b*f*x - 2544*si 
n(e + f*x)**3*b + 630*sin(e + f*x)**2*a*f*x + 922*sin(e + f*x)**2*a - 2205 
*sin(e + f*x)**2*b*f*x - 3458*sin(e + f*x)**2*b + 630*sin(e + f*x)*a*f*x + 
 286*sin(e + f*x)*a - 2205*sin(e + f*x)*b*f*x - 1007*sin(e + f*x)*b + 210* 
a*f*x - 96*a - 735*b*f*x + 294*b))/(30*a**3*f*(cos(e + f*x)*sin(e + f*x)** 
2 + 2*cos(e + f*x)*sin(e + f*x) + cos(e + f*x) - sin(e + f*x)**3 - 3*sin(e 
 + f*x)**2 - 3*sin(e + f*x) - 1))