\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx\) [271]

Optimal result

Integrand size = 35, antiderivative size = 283 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=-\frac {\left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d) \left (c^2-d^2\right )^{5/2} f}-\frac {d (2 A c-3 B c+3 A d-2 B d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f (c+d \sin (e+f x))^2}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))^2}-\frac {d \left (2 A c^2-5 B c^2+9 A c d-6 B c d+4 A d^2-4 B d^2\right ) \cos (e+f x)}{2 a (c-d)^3 (c+d)^2 f (c+d \sin (e+f x))} \] Output:

-(3*A*d*(2*c^2+2*c*d+d^2)-B*(2*c^3+4*c^2*d+7*c*d^2+2*d^3))*arctan((d+c*tan 
(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a/(c-d)/(c^2-d^2)^(5/2)/f-1/2*d*(2*A*c+3 
*A*d-3*B*c-2*B*d)*cos(f*x+e)/a/(c-d)^2/(c+d)/f/(c+d*sin(f*x+e))^2-(A-B)*co 
s(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^2-1/2*d*(2*A*c^2+9*A*c* 
d+4*A*d^2-5*B*c^2-6*B*c*d-4*B*d^2)*cos(f*x+e)/a/(c-d)^3/(c+d)^2/f/(c+d*sin 
(f*x+e))
 

Mathematica [A] (verified)

Time = 5.51 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.11 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )+\frac {2 \left (-3 A d \left (2 c^2+2 c d+d^2\right )+B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^2 \sqrt {c^2-d^2}}+\frac {(c-d) d (B c-A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))^2}+\frac {d \left (-A d (5 c+2 d)+B \left (3 c^2+2 c d+2 d^2\right )\right ) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d)^2 (c+d \sin (e+f x))}\right )}{2 a (c-d)^3 f (1+\sin (e+f x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3457, 3042, 3233, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\frac {a \left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a d \left (B \left (5 c^2+6 c d+4 d^2\right )-A \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}+\frac {a d (2 A c+3 A d-3 B c-2 B d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {a \left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a d \left (B \left (5 c^2+6 c d+4 d^2\right )-A \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}+\frac {a d (2 A c+3 A d-3 B c-2 B d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 3139

\(\displaystyle -\frac {\frac {\frac {2 a \left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {a d \left (B \left (5 c^2+6 c d+4 d^2\right )-A \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}+\frac {a d (2 A c+3 A d-3 B c-2 B d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle -\frac {\frac {-\frac {4 a \left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a d \left (B \left (5 c^2+6 c d+4 d^2\right )-A \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}+\frac {a d (2 A c+3 A d-3 B c-2 B d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {\frac {\frac {2 a \left (3 A d \left (2 c^2+2 c d+d^2\right )-B \left (2 c^3+4 c^2 d+7 c d^2+2 d^3\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a d \left (B \left (5 c^2+6 c d+4 d^2\right )-A \left (2 c^2+9 c d+4 d^2\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}+\frac {a d (2 A c+3 A d-3 B c-2 B d) \cos (e+f x)}{2 f \left (c^2-d^2\right ) (c+d \sin (e+f x))^2}}{a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))^2}\)

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])*(c + d*Sin[e + f*x])^3),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*d*(2*A*c - 3*B*c + 3*A*d - 2*B*d)*Cos[e + f*x])/(2*(c^2 - d^ 
2)*f*(c + d*Sin[e + f*x])^2) + ((2*a*(3*A*d*(2*c^2 + 2*c*d + d^2) - B*(2*c 
^3 + 4*c^2*d + 7*c*d^2 + 2*d^3))*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sq 
rt[c^2 - d^2])])/((c^2 - d^2)^(3/2)*f) - (a*d*(B*(5*c^2 + 6*c*d + 4*d^2) - 
 A*(2*c^2 + 9*c*d + 4*d^2))*Cos[e + f*x])/((c^2 - d^2)*f*(c + d*Sin[e + f* 
x])))/(2*(c^2 - d^2)))/(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 = 1.82 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.70

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 \left (\frac {\frac {d^{2} \left (7 A \,c^{2} d +2 A c \,d^{2}-2 A \,d^{3}-5 B \,c^{3}-2 B \,c^{2} d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 A \,c^{4} d +2 A \,c^{3} d^{2}+11 A \,c^{2} d^{3}+4 A c \,d^{4}-2 A \,d^{5}-4 B \,c^{5}-2 B \,c^{4} d -9 B \,c^{3} d^{2}-4 B \,c^{2} d^{3}-2 B c \,d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}+\frac {d^{2} \left (17 A \,c^{2} d +6 A c \,d^{2}-2 A \,d^{3}-11 B \,c^{3}-6 B \,c^{2} d -4 B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 A \,c^{2} d +2 A c \,d^{2}-A \,d^{3}-4 B \,c^{3}-2 B \,c^{2} d -B c \,d^{2}\right )}{2 c^{2}+4 c d +2 d^{2}}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{2}}+\frac {\left (6 A \,c^{2} d +6 A c \,d^{2}+3 A \,d^{3}-2 B \,c^{3}-4 B \,c^{2} d -7 B c \,d^{2}-2 B \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}}{a f}\)	\(482\)
default	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 \left (\frac {\frac {d^{2} \left (7 A \,c^{2} d +2 A c \,d^{2}-2 A \,d^{3}-5 B \,c^{3}-2 B \,c^{2} d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 A \,c^{4} d +2 A \,c^{3} d^{2}+11 A \,c^{2} d^{3}+4 A c \,d^{4}-2 A \,d^{5}-4 B \,c^{5}-2 B \,c^{4} d -9 B \,c^{3} d^{2}-4 B \,c^{2} d^{3}-2 B c \,d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}+\frac {d^{2} \left (17 A \,c^{2} d +6 A c \,d^{2}-2 A \,d^{3}-11 B \,c^{3}-6 B \,c^{2} d -4 B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}+\frac {d \left (6 A \,c^{2} d +2 A c \,d^{2}-A \,d^{3}-4 B \,c^{3}-2 B \,c^{2} d -B c \,d^{2}\right )}{2 c^{2}+4 c d +2 d^{2}}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{2}}+\frac {\left (6 A \,c^{2} d +6 A c \,d^{2}+3 A \,d^{3}-2 B \,c^{3}-4 B \,c^{2} d -7 B c \,d^{2}-2 B \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3}}}{a f}\)	\(482\)
risch	\(\text {Expression too large to display}\)	\(1881\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

2/f/a*(-(A-B)/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)-1/(c-d)^3*((1/2*d^2*(7*A*c^2* 
d+2*A*c*d^2-2*A*d^3-5*B*c^3-2*B*c^2*d)/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e 
)^3+1/2*d*(6*A*c^4*d+2*A*c^3*d^2+11*A*c^2*d^3+4*A*c*d^4-2*A*d^5-4*B*c^5-2* 
B*c^4*d-9*B*c^3*d^2-4*B*c^2*d^3-2*B*c*d^4)/(c^2+2*c*d+d^2)/c^2*tan(1/2*f*x 
+1/2*e)^2+1/2*d^2*(17*A*c^2*d+6*A*c*d^2-2*A*d^3-11*B*c^3-6*B*c^2*d-4*B*c*d 
^2)/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)+1/2*d*(6*A*c^2*d+2*A*c*d^2-A*d^3- 
4*B*c^3-2*B*c^2*d-B*c*d^2)/(c^2+2*c*d+d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*ta 
n(1/2*f*x+1/2*e)+c)^2+1/2*(6*A*c^2*d+6*A*c*d^2+3*A*d^3-2*B*c^3-4*B*c^2*d-7 
*B*c*d^2-2*B*d^3)/(c^2+2*c*d+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. 1608 vs. \(2 (274) = 548\).

Time = 0.21 (sec) , antiderivative size = 3303, normalized size of antiderivative = 11.67 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorith 
m="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,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. 727 vs. \(2 (274) = 548\).

Time = 0.26 (sec) , antiderivative size = 727, normalized size of antiderivative = 2.57 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorith 
m="giac")
 

Output:

((2*B*c^3 - 6*A*c^2*d + 4*B*c^2*d - 6*A*c*d^2 + 7*B*c*d^2 - 3*A*d^3 + 2*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*c^5 - a*c^4*d - 2*a*c^3*d^2 + 2*a*c^2*d^3 
+ a*c*d^4 - a*d^5)*sqrt(c^2 - d^2)) - 2*(A - B)/((a*c^3 - 3*a*c^2*d + 3*a* 
c*d^2 - a*d^3)*(tan(1/2*f*x + 1/2*e) + 1)) + (5*B*c^4*d^2*tan(1/2*f*x + 1/ 
2*e)^3 - 7*A*c^3*d^3*tan(1/2*f*x + 1/2*e)^3 + 2*B*c^3*d^3*tan(1/2*f*x + 1/ 
2*e)^3 - 2*A*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*A*c*d^5*tan(1/2*f*x + 1/2* 
e)^3 + 4*B*c^5*d*tan(1/2*f*x + 1/2*e)^2 - 6*A*c^4*d^2*tan(1/2*f*x + 1/2*e) 
^2 + 2*B*c^4*d^2*tan(1/2*f*x + 1/2*e)^2 - 2*A*c^3*d^3*tan(1/2*f*x + 1/2*e) 
^2 + 9*B*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 - 11*A*c^2*d^4*tan(1/2*f*x + 1/2*e 
)^2 + 4*B*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 - 4*A*c*d^5*tan(1/2*f*x + 1/2*e)^ 
2 + 2*B*c*d^5*tan(1/2*f*x + 1/2*e)^2 + 2*A*d^6*tan(1/2*f*x + 1/2*e)^2 + 11 
*B*c^4*d^2*tan(1/2*f*x + 1/2*e) - 17*A*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*B* 
c^3*d^3*tan(1/2*f*x + 1/2*e) - 6*A*c^2*d^4*tan(1/2*f*x + 1/2*e) + 4*B*c^2* 
d^4*tan(1/2*f*x + 1/2*e) + 2*A*c*d^5*tan(1/2*f*x + 1/2*e) + 4*B*c^5*d - 6* 
A*c^4*d^2 + 2*B*c^4*d^2 - 2*A*c^3*d^3 + B*c^3*d^3 + A*c^2*d^4)/((a*c^7 - a 
*c^6*d - 2*a*c^5*d^2 + 2*a*c^4*d^3 + a*c^3*d^4 - a*c^2*d^5)*(c*tan(1/2*f*x 
 + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f
 

Mupad [B] (verification not implemented)

Time = 40.75 (sec) , antiderivative size = 1076, normalized size of antiderivative = 3.80 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

((A*d^4 - 2*A*c^4 + 2*B*c^4 - 8*A*c^2*d^2 + 4*B*c^2*d^2 - 2*A*c*d^3 - 4*A* 
c^3*d + B*c*d^3 + 8*B*c^3*d)/((c + d)*(c^2 - d^2)*(c^2 - 2*c*d + d^2)) - ( 
tan(e/2 + (f*x)/2)^3*(2*A*d^6 - 13*A*c^2*d^4 - 17*A*c^3*d^3 - 22*A*c^4*d^2 
 + 4*B*c^2*d^4 + 19*B*c^3*d^3 + 23*B*c^4*d^2 - 2*A*c*d^5 - 8*A*c^5*d + 2*B 
*c*d^5 + 12*B*c^5*d))/(c^2*(c^2 - 2*c*d + d^2)*(c*d^2 - c^2*d - c^3 + d^3) 
) + (tan(e/2 + (f*x)/2)^2*(2*A*d^5 - 4*A*c^5 + 4*B*c^5 - 21*A*c^2*d^3 - 14 
*A*c^3*d^2 + 14*B*c^2*d^3 + 17*B*c^3*d^2 - 4*A*c*d^4 - 4*A*c^4*d + 2*B*c*d 
^4 + 8*B*c^4*d))/(c^2*(c^2 - d^2)*(c^2 - 2*c*d + d^2)) + (tan(e/2 + (f*x)/ 
2)^4*(2*A*c^5 - 2*A*d^5 - 2*B*c^5 + 7*A*c^2*d^3 + 2*A*c^3*d^2 - 2*B*c^2*d^ 
3 - 7*B*c^3*d^2 + 2*A*c*d^4 + 4*A*c^4*d - 4*B*c^4*d))/(c*(c^2 - 2*c*d + d^ 
2)*(c*d^2 - c^2*d - c^3 + d^3)) + (tan(e/2 + (f*x)/2)*(2*A*d^5 - 27*A*c^2* 
d^3 - 22*A*c^3*d^2 + 15*B*c^2*d^3 + 29*B*c^3*d^2 - 5*A*c*d^4 - 8*A*c^4*d + 
 4*B*c*d^4 + 12*B*c^4*d))/(c*(c + d)*(c^2 - d^2)*(c^2 - 2*c*d + d^2)))/(f* 
(tan(e/2 + (f*x)/2)^2*(2*a*c^2 + 4*a*d^2 + 4*a*c*d) + tan(e/2 + (f*x)/2)^3 
*(2*a*c^2 + 4*a*d^2 + 4*a*c*d) + a*c^2 + tan(e/2 + (f*x)/2)*(a*c^2 + 4*a*c 
*d) + tan(e/2 + (f*x)/2)^4*(a*c^2 + 4*a*c*d) + a*c^2*tan(e/2 + (f*x)/2)^5) 
) - (atan((((2*a*d^6 - 4*a*c^2*d^4 + 4*a*c^3*d^3 + 2*a*c^4*d^2 - 2*a*c*d^5 
 - 2*a*c^5*d)*(2*B*c^3 - 3*A*d^3 + 2*B*d^3 - 6*A*c*d^2 - 6*A*c^2*d + 7*B*c 
*d^2 + 4*B*c^2*d))/(2*a*(c + d)^(5/2)*(c - d)^(7/2)) - (c*tan(e/2 + (f*x)/ 
2)*(a*c^5 - a*d^5 + 2*a*c^2*d^3 - 2*a*c^3*d^2 + a*c*d^4 - a*c^4*d)*(2*B...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 5820, normalized size of antiderivative = 20.57 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*t 
an((e + f*x)/2)**5*a*c**6*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*a*c**5*d**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**4*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**3*d**4 + 2*sqrt(c**2 - d**2)*atan((ta 
n((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**5*b*c**7 + 12*s 
qrt(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**6*d + 23*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**5*d**2 + 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** 
4*d**3 + 8*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**4 - 6*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**6*d - 54*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**5*d**2 - 147*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sq 
rt(c**2 - d**2))*tan((e + f*x)/2)**4*a*c**4*d**3 - 120*sqrt(c**2 - d**2)*a 
tan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**4*a*c**3 
*d**4 - 48*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**5 + 2*sqrt(c**2 - d**2)*atan((tan((e ...