\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx\) [277]

Optimal result

Integrand size = 35, antiderivative size = 275 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^2 (c-d)^3 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{3 a^2 (c-d)^3 (c+d) f (c+d \sin (e+f x))}-\frac {(A c+2 B c-6 A d+3 B d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x)) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \] Output:

2*d*(A*d*(3*c+2*d)-B*(2*c^2+2*c*d+d^2))*arctan((d+c*tan(1/2*f*x+1/2*e))/(c 
^2-d^2)^(1/2))/a^2/(c-d)^3/(c+d)/(c^2-d^2)^(1/2)/f-1/3*d*(A*(c^2-6*c*d-10* 
d^2)+B*(2*c^2+9*c*d+4*d^2))*cos(f*x+e)/a^2/(c-d)^3/(c+d)/f/(c+d*sin(f*x+e) 
)-1/3*(A*c-6*A*d+2*B*c+3*B*d)*cos(f*x+e)/a^2/(c-d)^2/f/(1+sin(f*x+e))/(c+d 
*sin(f*x+e))-1/3*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+ 
e))
 

Mathematica [A] (verified)

Time = 9.43 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.14 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (A-B) (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+(-A+B) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (A (c-7 d)+2 B (c+2 d)) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-\frac {6 d \left (-A d (3 c+2 d)+B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{(c+d) \sqrt {c^2-d^2}}+\frac {3 d^2 (-B c+A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}{(c+d) (c+d \sin (e+f x))}\right )}{3 a^2 (c-d)^3 f (1+\sin (e+f x))^2} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x] 
)^2),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 - 7*d) 
 + 2*B*(c + 2*d))*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 
 - (6*d*(-(A*d*(3*c + 2*d)) + B*(2*c^2 + 2*c*d + d^2))*ArcTan[(d + c*Tan[( 
e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/((c 
 + d)*Sqrt[c^2 - d^2]) + (3*d^2*(-(B*c) + A*d)*Cos[e + f*x]*(Cos[(e + f*x) 
/2] + Sin[(e + f*x)/2])^3)/((c + d)*(c + d*Sin[e + f*x]))))/(3*a^2*(c - d) 
^3*f*(1 + Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3457, 25, 3042, 3457, 3042, 3233, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {-\frac {\int \frac {2 a^2 d (3 B c-5 A d+2 B d)-a^2 d (A c+2 B c-6 A d+3 B d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {2 a^2 d (3 B c-5 A d+2 B d)-a^2 d (A c+2 B c-6 A d+3 B d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {-\frac {\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {\int \frac {3 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 d c+d^2\right )\right )}{c+d \sin (e+f x)}dx}{c^2-d^2}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {3 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {3 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {-\frac {\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {6 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {-\frac {\frac {12 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}+\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {-\frac {\frac {a^2 d \left (A \left (c^2-6 c d-10 d^2\right )+B \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}-\frac {6 a^2 d \left (A d (3 c+2 d)-B \left (2 c^2+2 c d+d^2\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}}{a^2 (c-d)}-\frac {(A c-6 A d+2 B c+3 B d) \cos (e+f x)}{f (c-d) (\sin (e+f x)+1) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}\)

Input:

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

Output:

-1/3*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e 
 + f*x])) + (-(((A*c + 2*B*c - 6*A*d + 3*B*d)*Cos[e + f*x])/((c - d)*f*(1 
+ Sin[e + f*x])*(c + d*Sin[e + f*x]))) - ((-6*a^2*d*(A*d*(3*c + 2*d) - B*( 
2*c^2 + 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^2*d*(A*(c^2 - 6*c*d - 10*d^2) + B*(2*c^2 
+ 9*c*d + 4*d^2))*Cos[e + f*x])/((c^2 - d^2)*f*(c + d*Sin[e + f*x])))/(a^2 
*(c - d)))/(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 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 = 1.51 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 d \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (3 A c d +2 A \,d^{2}-2 B \,c^{2}-2 B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}-\frac {-2 A +2 B}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 A -2 B \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A c -3 A d +2 B d \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f}\)	\(263\)
default	\(\frac {\frac {2 d \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (3 A c d +2 A \,d^{2}-2 B \,c^{2}-2 B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}-\frac {-2 A +2 B}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 A -2 B \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A c -3 A d +2 B d \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{2} f}\)	\(263\)
risch	\(\text {Expression too large to display}\)	\(1411\)

Input:

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

Output:

2/f/a^2*(1/(c-d)^3*d*((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)+(3*A*c*d+2*A*d^ 
2-2*B*c^2-2*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)))-1/2*(-2*A+2*B)/(c-d)^2/(tan(1/2*f*x+1/2*e)+ 
1)^2-1/3*(2*A-2*B)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^3-(A*c-3*A*d+2*B*d)/(c-d 
)^3/(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. 1518 vs. \(2 (264) = 528\).

Time = 0.18 (sec) , antiderivative size = 3123, normalized size of antiderivative = 11.36 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algori 
thm="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 = 0.21 (sec) , antiderivative size = 411, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {3 \, {\left (2 \, B c^{2} d - 3 \, A c d^{2} + 2 \, B c d^{2} - 2 \, A d^{3} + B d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {3 \, {\left (B c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B c^{2} d^{2} - A c d^{3}\right )}}{{\left (a^{2} c^{5} - 2 \, a^{2} c^{4} d + 2 \, a^{2} c^{2} d^{3} - a^{2} c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} + \frac {3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 9 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, A c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, B c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 15 \, A d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 9 \, B d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A c + B c - 8 \, A d + 5 \, B d}{{\left (a^{2} c^{3} - 3 \, a^{2} c^{2} d + 3 \, a^{2} c d^{2} - a^{2} d^{3}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \] Input:

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

Output:

-2/3*(3*(2*B*c^2*d - 3*A*c*d^2 + 2*B*c*d^2 - 2*A*d^3 + B*d^3)*(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^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*sqrt(c^2 - d^2 
)) + 3*(B*c*d^3*tan(1/2*f*x + 1/2*e) - A*d^4*tan(1/2*f*x + 1/2*e) + B*c^2* 
d^2 - A*c*d^3)/((a^2*c^5 - 2*a^2*c^4*d + 2*a^2*c^2*d^3 - a^2*c*d^4)*(c*tan 
(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) + (3*A*c*tan(1/2*f*x 
+ 1/2*e)^2 - 9*A*d*tan(1/2*f*x + 1/2*e)^2 + 6*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) - 15*A*d*tan(1/2* 
f*x + 1/2*e) + 9*B*d*tan(1/2*f*x + 1/2*e) + 2*A*c + B*c - 8*A*d + 5*B*d)/( 
(a^2*c^3 - 3*a^2*c^2*d + 3*a^2*c*d^2 - a^2*d^3)*(tan(1/2*f*x + 1/2*e) + 1) 
^3))/f
 

Mupad [B] (verification not implemented)

Time = 40.45 (sec) , antiderivative size = 844, normalized size of antiderivative = 3.07 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*d*atan(((d*(2*a^2*d^5 - 4*a^2*c*d^4 - 2*a^2*c^4*d + 4*a^2*c^3*d^2)*(2*B 
*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2*B*c*d))/(a^2*(c + d)^(3/2)*(c - d)^(7 
/2)) - (2*c*d*tan(e/2 + (f*x)/2)*(a^2*c^4 - a^2*d^4 + 2*a^2*c*d^3 - 2*a^2* 
c^3*d)*(2*B*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2*B*c*d))/(a^2*(c + d)^(3/2) 
*(c - d)^(7/2)))/(2*B*d^3 - 4*A*d^3 - 6*A*c*d^2 + 4*B*c*d^2 + 4*B*c^2*d))* 
(2*B*c^2 - 2*A*d^2 + B*d^2 - 3*A*c*d + 2*B*c*d))/(a^2*f*(c + d)^(3/2)*(c - 
 d)^(7/2)) - ((2*(2*A*c^3 - 3*A*d^3 + B*c^3 - 8*A*c*d^2 - 6*A*c^2*d + 8*B* 
c*d^2 + 6*B*c^2*d))/(3*(c + d)*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + 
 (f*x)/2)^2*(5*A*c^3 - 9*A*d^3 + B*c^3 - 30*A*c*d^2 - 11*A*c^2*d + 27*B*c* 
d^2 + 17*B*c^2*d))/(3*c*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/ 
2)^3*(A*c^4 - 3*A*d^4 + B*c^4 - 9*A*c^2*d^2 + 8*B*c^2*d^2 - 7*A*c*d^3 - 2* 
A*c^3*d + 7*B*c*d^3 + 4*B*c^3*d))/(c*(c + d)*(c - d)*(c^2 - 2*c*d + d^2)) 
+ (2*tan(e/2 + (f*x)/2)*(3*A*c^4 - 3*A*d^4 + 3*B*c^4 - 27*A*c^2*d^2 + 30*B 
*c^2*d^2 - 25*A*c*d^3 - 8*A*c^3*d + 13*B*c*d^3 + 14*B*c^3*d))/(3*c*(c + d) 
*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^4*(A*c^4 - A*d^4 - 3 
*A*c^2*d^2 + 2*B*c^2*d^2 - 2*A*c^3*d + B*c*d^3 + 2*B*c^3*d))/(c*(c + d)*(c 
 - d)*(c^2 - 2*c*d + d^2)))/(f*(a^2*c + tan(e/2 + (f*x)/2)*(3*a^2*c + 2*a^ 
2*d) + tan(e/2 + (f*x)/2)^4*(3*a^2*c + 2*a^2*d) + tan(e/2 + (f*x)/2)^2*(4* 
a^2*c + 6*a^2*d) + tan(e/2 + (f*x)/2)^3*(4*a^2*c + 6*a^2*d) + a^2*c*tan(e/ 
2 + (f*x)/2)^5))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 4579, normalized size of antiderivative = 16.65 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*(27*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))* 
tan((e + f*x)/2)**5*a*c**3*d**2 + 36*sqrt(c**2 - d**2)*atan((tan((e + f*x) 
/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*a*c**2*d**3 + 12*sqrt(c* 
*2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/ 
2)**5*a*c*d**4 - 18*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c 
**2 - d**2))*tan((e + f*x)/2)**5*b*c**4*d - 30*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*b*c**3*d**2 - 
21*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan( 
(e + f*x)/2)**5*b*c**2*d**3 - 6*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c 
 + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*b*c*d**4 + 81*sqrt(c**2 - d** 
2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**4*a* 
c**3*d**2 + 162*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 
- d**2))*tan((e + f*x)/2)**4*a*c**2*d**3 + 108*sqrt(c**2 - d**2)*atan((tan 
((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**4*a*c*d**4 + 24* 
sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e 
+ f*x)/2)**4*a*d**5 - 54*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/s 
qrt(c**2 - d**2))*tan((e + f*x)/2)**4*b*c**4*d - 126*sqrt(c**2 - d**2)*ata 
n((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**4*b*c**3*d 
**2 - 123*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2 
))*tan((e + f*x)/2)**4*b*c**2*d**3 - 60*sqrt(c**2 - d**2)*atan((tan((e ...