\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx\) [73]

Optimal result

Integrand size = 36, antiderivative size = 110 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {B c^2 x}{a^3}-\frac {a^2 (A-B) c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac {2 B c^2 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac {2 B c^2 \cos (e+f x)}{f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:

B*c^2*x/a^3-1/5*a^2*(A-B)*c^2*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^5-2/3*B*c^2* 
cos(f*x+e)^3/f/(a+a*sin(f*x+e))^3+2*B*c^2*cos(f*x+e)/f/(a^3+a^3*sin(f*x+e) 
)
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(272\) vs. \(2(110)=220\).

Time = 11.50 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.47 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (24 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )-12 (A-B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-8 (3 A-8 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+4 (3 A-8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+2 (3 A-43 B) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+15 B (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5\right ) (c-c \sin (e+f x))^2}{15 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (1+\sin (e+f x))^3} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(24*(A - B)*Sin[(e + f*x)/2] - 12*( 
A - B)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 8*(3*A - 8*B)*Sin[(e + f*x) 
/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 4*(3*A - 8*B)*(Cos[(e + f*x) 
/2] + Sin[(e + f*x)/2])^3 + 2*(3*A - 43*B)*Sin[(e + f*x)/2]*(Cos[(e + f*x) 
/2] + Sin[(e + f*x)/2])^4 + 15*B*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f* 
x)/2])^5)*(c - c*Sin[e + f*x])^2)/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f 
*x)/2])^4*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3446, 3042, 3338, 3042, 3159, 3042, 3159, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle a^2 c^2 \int \frac {\cos ^4(e+f x) (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^5}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \int \frac {\cos (e+f x)^4 (A+B \sin (e+f x))}{(\sin (e+f x) a+a)^5}dx\)

\(\Big \downarrow \) 3338

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3159

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

\(\Big \downarrow \) 24

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

Input:

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

Output:

a^2*c^2*(-1/5*((A - B)*Cos[e + f*x]^5)/(f*(a + a*Sin[e + f*x])^5) + (B*((- 
2*Cos[e + f*x]^3)/(3*a*f*(a + a*Sin[e + f*x])^3) - (-(x/a^2) - (2*Cos[e + 
f*x])/(f*(a^2 + a^2*Sin[e + f*x])))/a^2))/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 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 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 = 0.71 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {2 c^{2} \left (B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {-32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {24 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{3}}\)	\(127\)
default	\(\frac {2 c^{2} \left (B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {-32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {A -B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {16 A -16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {24 A -16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}\right )}{f \,a^{3}}\)	\(127\)
parallelrisch	\(-\frac {2 \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} x f B}{2}+\left (-\frac {5}{2} f x B +A -B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (-5 f x -4\right ) B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+\left (-5 f x B +2 A -\frac {34}{3} B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+B \left (-\frac {5 f x}{2}-\frac {20}{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {f x B}{2}+\frac {A}{5}-\frac {23 B}{15}\right ) c^{2}}{f \,a^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(135\)
risch	\(\frac {B \,c^{2} x}{a^{3}}-\frac {2 \left (-30 A \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+15 A \,c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+250 B \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-180 i B \,c^{2} {\mathrm e}^{3 i \left (f x +e \right )}+140 i B \,c^{2} {\mathrm e}^{i \left (f x +e \right )}-75 B \,c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+3 A \,c^{2}-43 B \,c^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5}}\)	\(138\)
norman	\(\frac {\frac {B \,c^{2} x}{a}+\frac {8 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a f}+\frac {48 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a f}+\frac {B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{a}-\frac {6 A \,c^{2}-46 B \,c^{2}}{15 a f}+\frac {40 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 a f}+\frac {64 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}+\frac {112 B \,c^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{3 a f}-\frac {\left (2 A \,c^{2}-2 B \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a f}-\frac {\left (30 A \,c^{2}-86 B \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{3 a f}-\frac {2 \left (38 A \,c^{2}-198 B \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{5 a f}-\frac {\left (78 A \,c^{2}-478 B \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15 a f}-\frac {2 \left (138 A \,c^{2}-578 B \,c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{15 a f}+\frac {5 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a}+\frac {13 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}+\frac {25 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{a}+\frac {38 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}+\frac {46 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a}+\frac {46 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}+\frac {38 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a}+\frac {25 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{a}+\frac {13 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{a}+\frac {5 B \,c^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}}{a}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{3} a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\)	\(565\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x,method=_RETUR 
NVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (106) = 212\).

Time = 0.09 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.54 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=-\frac {60 \, B c^{2} f x - {\left (15 \, B c^{2} f x - {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A - B\right )} c^{2} - {\left (45 \, B c^{2} f x - {\left (9 \, A + 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x - {\left (A - 11 \, B\right )} c^{2}\right )} \cos \left (f x + e\right ) + {\left (60 \, B c^{2} f x + 12 \, {\left (A - B\right )} c^{2} - {\left (15 \, B c^{2} f x + {\left (3 \, A - 43 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B c^{2} f x + {\left (A + 9 \, B\right )} c^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori 
thm="fricas")
 

Output:

-1/15*(60*B*c^2*f*x - (15*B*c^2*f*x - (3*A - 43*B)*c^2)*cos(f*x + e)^3 - 1 
2*(A - B)*c^2 - (45*B*c^2*f*x - (9*A + 11*B)*c^2)*cos(f*x + e)^2 + 6*(5*B* 
c^2*f*x - (A - 11*B)*c^2)*cos(f*x + e) + (60*B*c^2*f*x + 12*(A - B)*c^2 - 
(15*B*c^2*f*x + (3*A - 43*B)*c^2)*cos(f*x + e)^2 + 6*(5*B*c^2*f*x + (A + 9 
*B)*c^2)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(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. 1647 vs. \(2 (102) = 204\).

Time = 8.08 (sec) , antiderivative size = 1647, normalized size of antiderivative = 14.97 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((-30*A*c**2*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 
 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3 
*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 60*A*c* 
*2*tan(e/2 + f*x/2)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 
+ f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2) 
**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 6*A*c**2/(15*a**3*f*tan(e/ 
2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2 
)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a* 
*3*f) + 15*B*c**2*f*x*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**5 + 
 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3 
*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 75*B*c* 
*2*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan( 
e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f* 
x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 150*B*c**2*f*x*tan(e/2 
 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)** 
4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a 
**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 150*B*c**2*f*x*tan(e/2 + f*x/2)**2/( 
15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f 
*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 
+ f*x/2) + 15*a**3*f) + 75*B*c**2*f*x*tan(e/2 + f*x/2)/(15*a**3*f*tan(e...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1134 vs. \(2 (106) = 212\).

Time = 0.14 (sec) , antiderivative size = 1134, normalized size of antiderivative = 10.31 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori 
thm="maxima")
 

Output:

2/15*(B*c^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos 
(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e 
)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1 
) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co 
s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f* 
x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1) 
)/a^3) - A*c^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(co 
s(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + 
e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1 
) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(co 
s(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f* 
x + e)^5/(cos(f*x + e) + 1)^5) - 2*A*c^2*(5*sin(f*x + e)/(cos(f*x + e) + 1 
) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/ 
(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*s 
in(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 
1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 4*B*c^2*(5*sin(f*x + e)/ 
(cos(f*x + e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5* 
a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) 
+ 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4 
/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*A*...
 

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.37 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {\frac {15 \, {\left (f x + e\right )} B c^{2}}{a^{3}} - \frac {2 \, {\left (15 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 15 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 170 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c^{2} - 23 \, B c^{2}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2/(a+a*sin(f*x+e))^3,x, algori 
thm="giac")
 

Output:

1/15*(15*(f*x + e)*B*c^2/a^3 - 2*(15*A*c^2*tan(1/2*f*x + 1/2*e)^4 - 15*B*c 
^2*tan(1/2*f*x + 1/2*e)^4 - 60*B*c^2*tan(1/2*f*x + 1/2*e)^3 + 30*A*c^2*tan 
(1/2*f*x + 1/2*e)^2 - 170*B*c^2*tan(1/2*f*x + 1/2*e)^2 - 100*B*c^2*tan(1/2 
*f*x + 1/2*e) + 3*A*c^2 - 23*B*c^2)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 39.75 (sec) , antiderivative size = 230, normalized size of antiderivative = 2.09 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {c^2\,\left (120\,B+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\frac {c^2\,\left (46\,B-6\,A+15\,B\,\left (e+f\,x\right )\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {c^2\,\left (30\,B-30\,A+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {c^2\,\left (340\,B-60\,A+150\,B\,\left (e+f\,x\right )\right )}{15}-10\,B\,c^2\,\left (e+f\,x\right )\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {c^2\,\left (200\,B+75\,B\,\left (e+f\,x\right )\right )}{15}-5\,B\,c^2\,\left (e+f\,x\right )\right )-B\,c^2\,\left (e+f\,x\right )}{a^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5}+\frac {B\,c^2\,x}{a^3} \] Input:

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

Output:

(tan(e/2 + (f*x)/2)^3*((c^2*(120*B + 150*B*(e + f*x)))/15 - 10*B*c^2*(e + 
f*x)) + (c^2*(46*B - 6*A + 15*B*(e + f*x)))/15 + tan(e/2 + (f*x)/2)^4*((c^ 
2*(30*B - 30*A + 75*B*(e + f*x)))/15 - 5*B*c^2*(e + f*x)) + tan(e/2 + (f*x 
)/2)^2*((c^2*(340*B - 60*A + 150*B*(e + f*x)))/15 - 10*B*c^2*(e + f*x)) + 
tan(e/2 + (f*x)/2)*((c^2*(200*B + 75*B*(e + f*x)))/15 - 5*B*c^2*(e + f*x)) 
 - B*c^2*(e + f*x))/(a^3*f*(tan(e/2 + (f*x)/2) + 1)^5) + (B*c^2*x)/a^3
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.34 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx=\frac {c^{2} \left (6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} a +15 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b f x -6 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} b +75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} b f x +60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} a +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b f x +60 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} b +150 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b f x +280 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} b +30 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +75 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b f x +170 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) b +15 b f x +40 b \right )}{15 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}+10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \] Input:

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

Output:

(c**2*(6*tan((e + f*x)/2)**5*a + 15*tan((e + f*x)/2)**5*b*f*x - 6*tan((e + 
 f*x)/2)**5*b + 75*tan((e + f*x)/2)**4*b*f*x + 60*tan((e + f*x)/2)**3*a + 
150*tan((e + f*x)/2)**3*b*f*x + 60*tan((e + f*x)/2)**3*b + 150*tan((e + f* 
x)/2)**2*b*f*x + 280*tan((e + f*x)/2)**2*b + 30*tan((e + f*x)/2)*a + 75*ta 
n((e + f*x)/2)*b*f*x + 170*tan((e + f*x)/2)*b + 15*b*f*x + 40*b))/(15*a**3 
*f*(tan((e + f*x)/2)**5 + 5*tan((e + f*x)/2)**4 + 10*tan((e + f*x)/2)**3 + 
 10*tan((e + f*x)/2)**2 + 5*tan((e + f*x)/2) + 1))