\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\) [270]

Optimal result

Integrand size = 35, antiderivative size = 181 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=-\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d) \left (c^2-d^2\right )^{3/2} f}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 4.81 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )+\frac {2 \left (-A d (2 c+d)+B \left (c^2+c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) \sqrt {c^2-d^2}}+\frac {d (B c-A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}\right )}{a (c-d)^2 f (1+\sin (e+f x))} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*Sin[(e + f*x)/2] + (2*(- 
(A*d*(2*c + d)) + B*(c^2 + c*d + d^2))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqr 
t[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/((c + d)*Sqrt[c^2 - d 
^2]) + (d*(B*c - A*d)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/ 
((c + d)*(c + d*Sin[e + f*x]))))/(a*(c - d)^2*f*(1 + Sin[e + f*x]))
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3457, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {-\frac {4 a \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\frac {2 a \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}\)

Input:

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

Output:

-(((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])*(c + d*Sin[e + f* 
x]))) - ((2*a*(A*d*(2*c + d) - B*(c^2 + c*d + d^2))*ArcTan[(2*d + 2*c*Tan[ 
(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c^2 - d^2)^(3/2)*f) - (a*d*(B*(2*c + 
 d) - A*(c + 2*d))*Cos[e + f*x])/((c^2 - d^2)*f*(c + d*Sin[e + f*x])))/(a^ 
2*(c - d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3233

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

rule 3457

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( 
n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) 
  Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b 
*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 
2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ 
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] 
 &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
 

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.09

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A c d +A \,d^{2}-B \,c^{2}-B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}-\frac {2 \left (A -B \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a f}\)	\(197\)
default	\(\frac {-\frac {2 \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A c d +A \,d^{2}-B \,c^{2}-B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}-\frac {2 \left (A -B \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a f}\)	\(197\)
risch	\(\text {Expression too large to display}\)	\(1081\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x,method=_RETURNV 
ERBOSE)
 

Output:

2/f/a*(-1/(c-d)^2*((d^2*(A*d-B*c)/(c+d)/c*tan(1/2*f*x+1/2*e)+d*(A*d-B*c)/( 
c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(2*A*c*d+A*d^2-B*c 
^2-B*c*d-B*d^2)/(c+d)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2 
*d)/(c^2-d^2)^(1/2)))-(A-B)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 725 vs. \(2 (176) = 352\).

Time = 0.13 (sec) , antiderivative size = 1538, normalized size of antiderivative = 8.50 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="fricas")
 

Output:

[1/2*(2*(A - B)*c^4 - 4*(A - B)*c^2*d^2 + 2*(A - B)*d^4 + 2*((A - 2*B)*c^3 
*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + 
 (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c^2*d - ( 
2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e)^2 + (B*c^3 - (2*A - B)*c^2*d - 
(A - B)*c*d^2)*cos(f*x + e) + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 
 - (A - B)*d^3 + (B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e))*s 
in(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*si 
n(f*x + e) - c^2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))* 
sqrt(-c^2 + d^2))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 
 2*((A - B)*c^4 + (A - 2*B)*c^3*d + B*c^2*d^2 - (A - 2*B)*c*d^3 - A*d^4)*c 
os(f*x + e) - 2*((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4 - ((A - 2*B 
)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e 
))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d 
^5 - a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^ 
3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d 
^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^ 
5 - a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f) 
*sin(f*x + e)), ((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4 + ((A - 2*B 
)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e 
)^2 + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (176) = 352\).

Time = 0.24 (sec) , antiderivative size = 425, normalized size of antiderivative = 2.35 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (B c^{2} - 2 \, A c d + B c d - A d^{2} + B d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a c^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {c^{2} - d^{2}}} - \frac {A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c^{3} - B c^{3} + A c^{2} d - 2 \, B c^{2} d + A c d^{2}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2,x, algorith 
m="giac")
 

Output:

2*((B*c^2 - 2*A*c*d + B*c*d - A*d^2 + B*d^2)*(pi*floor(1/2*(f*x + e)/pi + 
1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((a*c^ 
3 - a*c^2*d - a*c*d^2 + a*d^3)*sqrt(c^2 - d^2)) - (A*c^3*tan(1/2*f*x + 1/2 
*e)^2 - B*c^3*tan(1/2*f*x + 1/2*e)^2 + A*c^2*d*tan(1/2*f*x + 1/2*e)^2 - B* 
c^2*d*tan(1/2*f*x + 1/2*e)^2 - B*c*d^2*tan(1/2*f*x + 1/2*e)^2 + A*d^3*tan( 
1/2*f*x + 1/2*e)^2 + 2*A*c^2*d*tan(1/2*f*x + 1/2*e) - 3*B*c^2*d*tan(1/2*f* 
x + 1/2*e) + 3*A*c*d^2*tan(1/2*f*x + 1/2*e) - 3*B*c*d^2*tan(1/2*f*x + 1/2* 
e) + A*d^3*tan(1/2*f*x + 1/2*e) + A*c^3 - B*c^3 + A*c^2*d - 2*B*c^2*d + A* 
c*d^2)/((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2*e)^3 
+ c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 
1/2*e) + 2*d*tan(1/2*f*x + 1/2*e) + c)))/f
 

Mupad [B] (verification not implemented)

Time = 37.78 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.41 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx=\frac {2\,\mathrm {atan}\left (\frac {\frac {\left (2\,a\,c^3\,d-2\,a\,c^2\,d^2-2\,a\,c\,d^3+2\,a\,d^4\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c^3-a\,c^2\,d-a\,c\,d^2+a\,d^3\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}}{2\,B\,c^2-2\,A\,d^2+2\,B\,d^2-4\,A\,c\,d+2\,B\,c\,d}\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,c^2+A\,d^2-B\,c^2+A\,c\,d-2\,B\,c\,d\right )}{\left (c+d\right )\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d^2+2\,A\,c\,d-3\,B\,c\,d\right )}{c\,{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (A\,c^3+A\,d^3-B\,c^3+A\,c^2\,d-B\,c\,d^2-B\,c^2\,d\right )}{c\,\left (c+d\right )\,{\left (c-d\right )}^2}}{f\,\left (a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (a\,c+2\,a\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,c\right )} \] Input:

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

Output:

(2*atan((((2*a*d^4 - 2*a*c^2*d^2 - 2*a*c*d^3 + 2*a*c^3*d)*(B*c^2 - A*d^2 + 
 B*d^2 - 2*A*c*d + B*c*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)) + (2*c*tan(e/2 
+ (f*x)/2)*(a*c^3 + a*d^3 - a*c*d^2 - a*c^2*d)*(B*c^2 - A*d^2 + B*d^2 - 2* 
A*c*d + B*c*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)))/(2*B*c^2 - 2*A*d^2 + 2*B* 
d^2 - 4*A*c*d + 2*B*c*d))*(B*c^2 - A*d^2 + B*d^2 - 2*A*c*d + B*c*d))/(a*f* 
(c + d)^(3/2)*(c - d)^(5/2)) - ((2*(A*c^2 + A*d^2 - B*c^2 + A*c*d - 2*B*c* 
d))/((c + d)*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(A*d^2 + 2*A*c*d - 3*B*c*d 
))/(c*(c - d)^2) + (2*tan(e/2 + (f*x)/2)^2*(A*c^3 + A*d^3 - B*c^3 + A*c^2* 
d - B*c*d^2 - B*c^2*d))/(c*(c + d)*(c - d)^2))/(f*(a*c + tan(e/2 + (f*x)/2 
)^2*(a*c + 2*a*d) + tan(e/2 + (f*x)/2)*(a*c + 2*a*d) + a*c*tan(e/2 + (f*x) 
/2)^3))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 2468, normalized size of antiderivative = 13.64 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*( - 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2) 
)*tan((e + f*x)/2)**3*a*c**3*d - 5*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2 
)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**3*a*c**2*d**2 - 2*sqrt(c**2 
- d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)* 
*3*a*c*d**3 + sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - 
d**2))*tan((e + f*x)/2)**3*b*c**4 + 3*sqrt(c**2 - d**2)*atan((tan((e + f*x 
)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**3*b*c**3*d + 3*sqrt(c**2 
- d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)* 
*3*b*c**2*d**2 + 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c* 
*2 - d**2))*tan((e + f*x)/2)**3*b*c*d**3 - 2*sqrt(c**2 - d**2)*atan((tan(( 
e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**2*a*c**3*d - 9*sqr 
t(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f 
*x)/2)**2*a*c**2*d**2 - 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d) 
/sqrt(c**2 - d**2))*tan((e + f*x)/2)**2*a*c*d**3 - 4*sqrt(c**2 - d**2)*ata 
n((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**2*a*d**4 + 
 sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e 
 + f*x)/2)**2*b*c**4 + 5*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/s 
qrt(c**2 - d**2))*tan((e + f*x)/2)**2*b*c**3*d + 9*sqrt(c**2 - d**2)*atan( 
(tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**2*b*c**2*d** 
2 + 8*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2)...