\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\) [276]

Optimal result

Integrand size = 35, antiderivative size = 152 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=-\frac {2 d (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^2 (c-d)^2 \sqrt {c^2-d^2} f}-\frac {(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2} \] Output:

-2*d*(-A*d+B*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a^2/(c-d) 
^2/(c^2-d^2)^(1/2)/f-1/3*(A*(c-4*d)+B*(2*c+d))*cos(f*x+e)/a^2/(c-d)^2/f/(1 
+sin(f*x+e))-1/3*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^2
 

Mathematica [A] (verified)

Time = 4.17 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.51 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (A-B) (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+(-A+B) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (A (c-4 d)+B (2 c+d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+\frac {6 d (-B c+A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{\sqrt {c^2-d^2}}\right )}{3 a^2 (c-d)^2 f (1+\sin (e+f x))^2} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*(c - d)*Sin[(e + f*x)/2] 
 + (-A + B)*(c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(A*(c - 4*d) 
 + B*(2*c + d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 
 (6*d*(-(B*c) + A*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos 
[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/Sqrt[c^2 - d^2]))/(3*a^2*(c - d)^2*f* 
(1 + Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3457, 25, 3042, 3457, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

\(\displaystyle \frac {-\frac {6 d (B c-A d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (c-d)}-\frac {(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {12 d (B c-A d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f (c-d)}-\frac {(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {6 d (B c-A d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f (c-d) \sqrt {c^2-d^2}}-\frac {(A (c-4 d)+B (2 c+d)) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1)}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}\)

Input:

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

Output:

-1/3*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^2) + ((-6*d*(B 
*c - A*d)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((c - 
d)*Sqrt[c^2 - d^2]*f) - ((A*(c - 4*d) + B*(2*c + d))*Cos[e + f*x])/((c - d 
)*f*(1 + Sin[e + f*x])))/(3*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 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 = 159, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {-\frac {-2 A +2 B}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 A -2 B \right )}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A c -2 A d +B d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 d \left (A d -B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}}{a^{2} f}\)	\(159\)
default	\(\frac {-\frac {-2 A +2 B}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 A -2 B \right )}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A c -2 A d +B d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {2 d \left (A d -B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}}{a^{2} f}\)	\(159\)
risch	\(\frac {\frac {2 A c}{3}-\frac {8 A d}{3}-2 i A c \,{\mathrm e}^{i \left (f x +e \right )}+6 i A d \,{\mathrm e}^{i \left (f x +e \right )}+2 A d \,{\mathrm e}^{2 i \left (f x +e \right )}+\frac {4 B c}{3}+\frac {2 B d}{3}-2 i B c \,{\mathrm e}^{i \left (f x +e \right )}-2 i B d \,{\mathrm e}^{i \left (f x +e \right )}-2 B c \,{\mathrm e}^{2 i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} \left (c -d \right )^{2} f \,a^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}\)	\(444\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 599 vs. \(2 (143) = 286\).

Time = 0.11 (sec) , antiderivative size = 1285, normalized size of antiderivative = 8.45 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),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 [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.64 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (B c d - A d^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} + \frac {3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 9 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A c + B c - 5 \, A d + 2 \, B d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \] Input:

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

Output:

-2/3*(3*(B*c*d - A*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^2*c^2 - 2*a^2*c*d + a^2 
*d^2)*sqrt(c^2 - d^2)) + (3*A*c*tan(1/2*f*x + 1/2*e)^2 - 6*A*d*tan(1/2*f*x 
 + 1/2*e)^2 + 3*B*d*tan(1/2*f*x + 1/2*e)^2 + 3*A*c*tan(1/2*f*x + 1/2*e) + 
3*B*c*tan(1/2*f*x + 1/2*e) - 9*A*d*tan(1/2*f*x + 1/2*e) + 3*B*d*tan(1/2*f* 
x + 1/2*e) + 2*A*c + B*c - 5*A*d + 2*B*d)/((a^2*c^2 - 2*a^2*c*d + a^2*d^2) 
*(tan(1/2*f*x + 1/2*e) + 1)^3))/f
 

Mupad [B] (verification not implemented)

Time = 37.97 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.99 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2\,d\,\mathrm {atan}\left (\frac {\frac {d\,\left (A\,d-B\,c\right )\,\left (2\,a^2\,c^2\,d-4\,a^2\,c\,d^2+2\,a^2\,d^3\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d-B\,c\right )\,\left (a^2\,c^2-2\,a^2\,c\,d+a^2\,d^2\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}}{2\,A\,d^2-2\,B\,c\,d}\right )\,\left (A\,d-B\,c\right )}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (2\,A\,c-5\,A\,d+B\,c+2\,B\,d\right )}{3\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,c-3\,A\,d+B\,c+B\,d\right )}{{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (A\,c-2\,A\,d+B\,d\right )}{{\left (c-d\right )}^2}}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )} \] Input:

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

Output:

(2*d*atan(((d*(A*d - B*c)*(2*a^2*d^3 - 4*a^2*c*d^2 + 2*a^2*c^2*d))/(a^2*(c 
 + d)^(1/2)*(c - d)^(5/2)) + (2*c*d*tan(e/2 + (f*x)/2)*(A*d - B*c)*(a^2*c^ 
2 + a^2*d^2 - 2*a^2*c*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)))/(2*A*d^2 - 2* 
B*c*d))*(A*d - B*c))/(a^2*f*(c + d)^(1/2)*(c - d)^(5/2)) - ((2*(2*A*c - 5* 
A*d + B*c + 2*B*d))/(3*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(A*c - 3*A*d + B 
*c + B*d))/(c - d)^2 + (2*tan(e/2 + (f*x)/2)^2*(A*c - 2*A*d + B*d))/(c - d 
)^2)/(f*(3*a^2*tan(e/2 + (f*x)/2)^2 + a^2*tan(e/2 + (f*x)/2)^3 + a^2 + 3*a 
^2*tan(e/2 + (f*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 837, normalized size of antiderivative = 5.51 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(3*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*t 
an((e + f*x)/2)**3*a*d**2 - 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*d + 9*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**2 - 
 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*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)*a*d**2 - 9*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)*b*c*d + 3*sqrt(c* 
*2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*d**2 - 3*sqr 
t(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c*d + ta 
n((e + f*x)/2)**3*a*c**3 - 2*tan((e + f*x)/2)**3*a*c**2*d - tan((e + f*x)/ 
2)**3*a*c*d**2 + 2*tan((e + f*x)/2)**3*a*d**3 + tan((e + f*x)/2)**3*b*c**2 
*d - tan((e + f*x)/2)**3*b*d**3 + 3*tan((e + f*x)/2)*a*c**2*d - 3*tan((e + 
 f*x)/2)*a*d**3 - 3*tan((e + f*x)/2)*b*c**3 + 3*tan((e + f*x)/2)*b*c*d**2 
- a*c**3 + 3*a*c**2*d + a*c*d**2 - 3*a*d**3 - b*c**3 - b*c**2*d + b*c*d**2 
 + b*d**3))/(3*a**2*f*(tan((e + f*x)/2)**3*c**4 - 2*tan((e + f*x)/2)**3*c* 
*3*d + 2*tan((e + f*x)/2)**3*c*d**3 - tan((e + f*x)/2)**3*d**4 + 3*tan((e 
+ f*x)/2)**2*c**4 - 6*tan((e + f*x)/2)**2*c**3*d + 6*tan((e + f*x)/2)**2*c 
*d**3 - 3*tan((e + f*x)/2)**2*d**4 + 3*tan((e + f*x)/2)*c**4 - 6*tan((e + 
f*x)/2)*c**3*d + 6*tan((e + f*x)/2)*c*d**3 - 3*tan((e + f*x)/2)*d**4 + ...