\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx\) [283]

Optimal result

Integrand size = 35, antiderivative size = 229 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {2 d^2 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^3 \sqrt {c^2-d^2} f}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {(2 A c+3 B c-7 A d+2 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (B \left (3 c^2-16 c d-2 d^2\right )+A \left (2 c^2-9 c d+22 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )} \] Output:

2*d^2*(-A*d+B*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/a^3/(c-d 
)^3/(c^2-d^2)^(1/2)/f-1/5*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^3-1/15 
*(2*A*c-7*A*d+3*B*c+2*B*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin(f*x+e))^2-1/15* 
(B*(3*c^2-16*c*d-2*d^2)+A*(2*c^2-9*c*d+22*d^2))*cos(f*x+e)/(c-d)^3/f/(a^3+ 
a^3*sin(f*x+e))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(502\) vs. \(2(229)=458\).

Time = 7.62 (sec) , antiderivative size = 502, normalized size of antiderivative = 2.19 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (15 B c^2 \cos \left (\frac {1}{2} (e+f x)\right )-15 A c d \cos \left (\frac {1}{2} (e+f x)\right )-75 B c d \cos \left (\frac {1}{2} (e+f x)\right )+75 A d^2 \cos \left (\frac {1}{2} (e+f x)\right )-10 A c^2 \cos \left (\frac {3}{2} (e+f x)\right )-15 B c^2 \cos \left (\frac {3}{2} (e+f x)\right )+45 A c d \cos \left (\frac {3}{2} (e+f x)\right )+65 B c d \cos \left (\frac {3}{2} (e+f x)\right )-95 A d^2 \cos \left (\frac {3}{2} (e+f x)\right )+10 B d^2 \cos \left (\frac {3}{2} (e+f x)\right )+20 A c^2 \sin \left (\frac {1}{2} (e+f x)\right )+15 B c^2 \sin \left (\frac {1}{2} (e+f x)\right )-75 A c d \sin \left (\frac {1}{2} (e+f x)\right )-85 B c d \sin \left (\frac {1}{2} (e+f x)\right )+145 A d^2 \sin \left (\frac {1}{2} (e+f x)\right )-20 B d^2 \sin \left (\frac {1}{2} (e+f x)\right )-\frac {60 d^2 (-B c+A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}{\sqrt {c^2-d^2}}-15 B c d \sin \left (\frac {3}{2} (e+f x)\right )+15 A d^2 \sin \left (\frac {3}{2} (e+f x)\right )-2 A c^2 \sin \left (\frac {5}{2} (e+f x)\right )-3 B c^2 \sin \left (\frac {5}{2} (e+f x)\right )+9 A c d \sin \left (\frac {5}{2} (e+f x)\right )+16 B c d \sin \left (\frac {5}{2} (e+f x)\right )-22 A d^2 \sin \left (\frac {5}{2} (e+f x)\right )+2 B d^2 \sin \left (\frac {5}{2} (e+f x)\right )\right )}{30 a^3 (c-d)^3 f (1+\sin (e+f x))^3} \] Input:

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

Output:

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(15*B*c^2*Cos[(e + f*x)/2] - 15*A*c 
*d*Cos[(e + f*x)/2] - 75*B*c*d*Cos[(e + f*x)/2] + 75*A*d^2*Cos[(e + f*x)/2 
] - 10*A*c^2*Cos[(3*(e + f*x))/2] - 15*B*c^2*Cos[(3*(e + f*x))/2] + 45*A*c 
*d*Cos[(3*(e + f*x))/2] + 65*B*c*d*Cos[(3*(e + f*x))/2] - 95*A*d^2*Cos[(3* 
(e + f*x))/2] + 10*B*d^2*Cos[(3*(e + f*x))/2] + 20*A*c^2*Sin[(e + f*x)/2] 
+ 15*B*c^2*Sin[(e + f*x)/2] - 75*A*c*d*Sin[(e + f*x)/2] - 85*B*c*d*Sin[(e 
+ f*x)/2] + 145*A*d^2*Sin[(e + f*x)/2] - 20*B*d^2*Sin[(e + f*x)/2] - (60*d 
^2*(-(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])^5)/Sqrt[c^2 - d^2] - 15*B*c*d*Sin[(3*(e + f 
*x))/2] + 15*A*d^2*Sin[(3*(e + f*x))/2] - 2*A*c^2*Sin[(5*(e + f*x))/2] - 3 
*B*c^2*Sin[(5*(e + f*x))/2] + 9*A*c*d*Sin[(5*(e + f*x))/2] + 16*B*c*d*Sin[ 
(5*(e + f*x))/2] - 22*A*d^2*Sin[(5*(e + f*x))/2] + 2*B*d^2*Sin[(5*(e + f*x 
))/2]))/(30*a^3*(c - d)^3*f*(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 1.23 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.13, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {3042, 3457, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {-\frac {\int -\frac {\left (B c (3 c-13 d)+A \left (2 c^2-7 d c+15 d^2\right )\right ) a^2+d (2 A c+3 B c-7 A d+2 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (B c (3 c-13 d)+A \left (2 c^2-7 d c+15 d^2\right )\right ) a^2+d (2 A c+3 B c-7 A d+2 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (B c (3 c-13 d)+A \left (2 c^2-7 d c+15 d^2\right )\right ) a^2+d (2 A c+3 B c-7 A d+2 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))}dx}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {-\frac {\int -\frac {15 a^3 d^2 (B c-A d)}{c+d \sin (e+f x)}dx}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {15 a d^2 (B c-A d) \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {15 a d^2 (B c-A d) \int \frac {1}{c+d \sin (e+f x)}dx}{c-d}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {\frac {30 a d^2 (B c-A d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f (c-d)}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {-\frac {60 a d^2 (B c-A d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f (c-d)}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {30 a d^2 (B c-A d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f (c-d) \sqrt {c^2-d^2}}-\frac {a^2 \left (A \left (2 c^2-9 c d+22 d^2\right )+B \left (3 c^2-16 c d-2 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a)}}{3 a^2 (c-d)}-\frac {a (2 A c-7 A d+3 B c+2 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3}\)

Input:

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

Output:

-1/5*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^3) + (-1/3*(a* 
(2*A*c + 3*B*c - 7*A*d + 2*B*d)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f* 
x])^2) + ((30*a*d^2*(B*c - A*d)*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqr 
t[c^2 - d^2])])/((c - d)*Sqrt[c^2 - d^2]*f) - (a^2*(B*(3*c^2 - 16*c*d - 2* 
d^2) + A*(2*c^2 - 9*c*d + 22*d^2))*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + 
 f*x])))/(3*a^2*(c - d)))/(5*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 = 1.20 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {-\frac {2 d^{2} \left (A d -B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}-\frac {-8 A +8 B}{2 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +6 A d +2 B c -4 B d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -10 A d -6 B c +8 B d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-3 A c d +3 A \,d^{2}-B \,d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} f}\)	\(252\)
default	\(\frac {-\frac {2 d^{2} \left (A d -B c \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{3} \sqrt {c^{2}-d^{2}}}-\frac {-8 A +8 B}{2 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +6 A d +2 B c -4 B d}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -10 A d -6 B c +8 B d \right )}{3 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-3 A c d +3 A \,d^{2}-B \,d^{2}\right )}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} f}\)	\(252\)
risch	\(\frac {-\frac {4 A \,c^{2}}{15}-\frac {2 B \,c^{2}}{5}-\frac {44 A \,d^{2}}{15}+\frac {6 A c d}{5}+\frac {4 B \,d^{2}}{15}+\frac {32 B c d}{15}-10 A c d \,{\mathrm e}^{2 i \left (f x +e \right )}-\frac {34 B c d \,{\mathrm e}^{2 i \left (f x +e \right )}}{3}+2 B c d \,{\mathrm e}^{4 i \left (f x +e \right )}+2 i B \,c^{2} {\mathrm e}^{i \left (f x +e \right )}-2 i B \,c^{2} {\mathrm e}^{3 i \left (f x +e \right )}-\frac {26 i B c d \,{\mathrm e}^{i \left (f x +e \right )}}{3}+\frac {38 i A \,d^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}+\frac {4 i A \,c^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}+2 i A c d \,{\mathrm e}^{3 i \left (f x +e \right )}+\frac {58 A \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}+2 B \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}-\frac {8 B \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}-2 A \,d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {8 A \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}-6 i A c d \,{\mathrm e}^{i \left (f x +e \right )}-10 i A \,d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+10 i B c d \,{\mathrm e}^{3 i \left (f x +e \right )}-\frac {4 i B \,d^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{5} \left (c -d \right )^{3} f \,a^{3}}-\frac {d^{3} \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 )^{3} f \,a^{3}}+\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 ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}+\frac {d^{3} \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 )^{3} f \,a^{3}}-\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 ) B c}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{3} f \,a^{3}}\)	\(659\)

Input:

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

Output:

2/f/a^3*(-d^2*(A*d-B*c)/(c-d)^3/(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))-1/4*(-8*A+8*B)/(c-d)/(tan(1/2*f*x+1/2*e)+1) 
^4-1/5*(4*A-4*B)/(c-d)/(tan(1/2*f*x+1/2*e)+1)^5-1/2*(-4*A*c+6*A*d+2*B*c-4* 
B*d)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(8*A*c-10*A*d-6*B*c+8*B*d)/(c-d) 
^2/(tan(1/2*f*x+1/2*e)+1)^3-(A*c^2-3*A*c*d+3*A*d^2-B*d^2)/(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. 1104 vs. \(2 (218) = 436\).

Time = 0.15 (sec) , antiderivative size = 2292, normalized size of antiderivative = 10.01 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/30*(6*(A - B)*c^4 - 12*(A - B)*c^3*d + 12*(A - B)*c*d^3 - 6*(A - B)*d^4 
 - 2*((2*A + 3*B)*c^4 - (9*A + 16*B)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 
16*B)*c*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^3 + 2*(2*(2*A + 3*B)*c^4 - (1 
8*A + 17*B)*c^3*d + 5*(5*A - 2*B)*c^2*d^2 + (18*A + 17*B)*c*d^3 - (29*A - 
4*B)*d^4)*cos(f*x + e)^2 + 15*(4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos 
(f*x + e)^3 - 3*(B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B*c*d^2 - A*d^3)*cos 
(f*x + e) + (4*B*c*d^2 - 4*A*d^3 - (B*c*d^2 - A*d^3)*cos(f*x + e)^2 + 2*(B 
*c*d^2 - A*d^3)*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)) + 6*((3*A + 2*B)*c^4 - (11*A + 9*B)*c^3*d + 
5*(3*A - B)*c^2*d^2 + (11*A + 9*B)*c*d^3 - 3*(6*A - B)*d^4)*cos(f*x + e) - 
 2*(3*(A - B)*c^4 - 6*(A - B)*c^3*d + 6*(A - B)*c*d^3 - 3*(A - B)*d^4 - (( 
2*A + 3*B)*c^4 - (9*A + 16*B)*c^3*d + 5*(4*A - B)*c^2*d^2 + (9*A + 16*B)*c 
*d^3 - 2*(11*A - B)*d^4)*cos(f*x + e)^2 - 3*((2*A + 3*B)*c^4 - (9*A + 11*B 
)*c^3*d + 5*(3*A - B)*c^2*d^2 + (9*A + 11*B)*c*d^3 - (17*A - 2*B)*d^4)*cos 
(f*x + e))*sin(f*x + e))/((a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c 
^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^3 + 3*(a^3*c^5 - 3*a^3*c^4* 
d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*d^4 + a^3*d^5)*f*cos(f*x + e)^ 
2 - 2*(a^3*c^5 - 3*a^3*c^4*d + 2*a^3*c^3*d^2 + 2*a^3*c^2*d^3 - 3*a^3*c*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (218) = 436\).

Time = 0.28 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.41 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

2/15*(15*(B*c*d^2 - A*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arct 
an((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((a^3*c^3 - 3*a^3*c^2*d 
+ 3*a^3*c*d^2 - a^3*d^3)*sqrt(c^2 - d^2)) - (15*A*c^2*tan(1/2*f*x + 1/2*e) 
^4 - 45*A*c*d*tan(1/2*f*x + 1/2*e)^4 + 45*A*d^2*tan(1/2*f*x + 1/2*e)^4 - 1 
5*B*d^2*tan(1/2*f*x + 1/2*e)^4 + 30*A*c^2*tan(1/2*f*x + 1/2*e)^3 + 15*B*c^ 
2*tan(1/2*f*x + 1/2*e)^3 - 105*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 45*B*c*d*tan 
(1/2*f*x + 1/2*e)^3 + 135*A*d^2*tan(1/2*f*x + 1/2*e)^3 - 30*B*d^2*tan(1/2* 
f*x + 1/2*e)^3 + 40*A*c^2*tan(1/2*f*x + 1/2*e)^2 + 15*B*c^2*tan(1/2*f*x + 
1/2*e)^2 - 135*A*c*d*tan(1/2*f*x + 1/2*e)^2 - 65*B*c*d*tan(1/2*f*x + 1/2*e 
)^2 + 185*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 40*B*d^2*tan(1/2*f*x + 1/2*e)^2 + 
 20*A*c^2*tan(1/2*f*x + 1/2*e) + 15*B*c^2*tan(1/2*f*x + 1/2*e) - 75*A*c*d* 
tan(1/2*f*x + 1/2*e) - 55*B*c*d*tan(1/2*f*x + 1/2*e) + 115*A*d^2*tan(1/2*f 
*x + 1/2*e) - 20*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 - 24*A*c*d 
 - 11*B*c*d + 32*A*d^2 - 7*B*d^2)/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - 
a^3*d^3)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 39.95 (sec) , antiderivative size = 591, normalized size of antiderivative = 2.58 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx=\frac {2\,d^2\,\mathrm {atan}\left (\frac {\frac {d^2\,\left (A\,d-B\,c\right )\,\left (-2\,a^3\,c^3\,d+6\,a^3\,c^2\,d^2-6\,a^3\,c\,d^3+2\,a^3\,d^4\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {2\,c\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d-B\,c\right )\,\left (a^3\,c^3-3\,a^3\,c^2\,d+3\,a^3\,c\,d^2-a^3\,d^3\right )}{a^3\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}}{2\,A\,d^3-2\,B\,c\,d^2}\right )\,\left (A\,d-B\,c\right )}{a^3\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{7/2}}-\frac {\frac {2\,\left (7\,A\,c^2+32\,A\,d^2+3\,B\,c^2-7\,B\,d^2-24\,A\,c\,d-11\,B\,c\,d\right )}{15\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A\,c^2+23\,A\,d^2+3\,B\,c^2-4\,B\,d^2-15\,A\,c\,d-11\,B\,c\,d\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,A\,c^2+9\,A\,d^2+B\,c^2-2\,B\,d^2-7\,A\,c\,d-3\,B\,c\,d\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (8\,A\,c^2+37\,A\,d^2+3\,B\,c^2-8\,B\,d^2-27\,A\,c\,d-13\,B\,c\,d\right )}{3\,\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (A\,c^2+3\,A\,d^2-B\,d^2-3\,A\,c\,d\right )}{\left (c-d\right )\,\left (c^2-2\,c\,d+d^2\right )}}{f\,\left (a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5+5\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+10\,a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^3\right )} \] Input:

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

Output:

(2*d^2*atan(((d^2*(A*d - B*c)*(2*a^3*d^4 - 6*a^3*c*d^3 - 2*a^3*c^3*d + 6*a 
^3*c^2*d^2))/(a^3*(c + d)^(1/2)*(c - d)^(7/2)) - (2*c*d^2*tan(e/2 + (f*x)/ 
2)*(A*d - B*c)*(a^3*c^3 - a^3*d^3 + 3*a^3*c*d^2 - 3*a^3*c^2*d))/(a^3*(c + 
d)^(1/2)*(c - d)^(7/2)))/(2*A*d^3 - 2*B*c*d^2))*(A*d - B*c))/(a^3*f*(c + d 
)^(1/2)*(c - d)^(7/2)) - ((2*(7*A*c^2 + 32*A*d^2 + 3*B*c^2 - 7*B*d^2 - 24* 
A*c*d - 11*B*c*d))/(15*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2 
)*(4*A*c^2 + 23*A*d^2 + 3*B*c^2 - 4*B*d^2 - 15*A*c*d - 11*B*c*d))/(3*(c - 
d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^3*(2*A*c^2 + 9*A*d^2 + B*c 
^2 - 2*B*d^2 - 7*A*c*d - 3*B*c*d))/((c - d)*(c^2 - 2*c*d + d^2)) + (2*tan( 
e/2 + (f*x)/2)^2*(8*A*c^2 + 37*A*d^2 + 3*B*c^2 - 8*B*d^2 - 27*A*c*d - 13*B 
*c*d))/(3*(c - d)*(c^2 - 2*c*d + d^2)) + (2*tan(e/2 + (f*x)/2)^4*(A*c^2 + 
3*A*d^2 - B*d^2 - 3*A*c*d))/((c - d)*(c^2 - 2*c*d + d^2)))/(f*(10*a^3*tan( 
e/2 + (f*x)/2)^2 + 10*a^3*tan(e/2 + (f*x)/2)^3 + 5*a^3*tan(e/2 + (f*x)/2)^ 
4 + a^3*tan(e/2 + (f*x)/2)^5 + a^3 + 5*a^3*tan(e/2 + (f*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1863, normalized size of antiderivative = 8.14 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(2*( - 15*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*d**3 + 15*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**2 - 75*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**3 + 75*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*d**2 - 150*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*d**3 + 150*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**2 - 150*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*d**3 + 150*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**2 - 75* 
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**3 + 75*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**2 - 15*sqrt(c**2 - d**2)*atan((tan( 
(e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*d**3 + 15*sqrt(c**2 - d**2)*atan( 
(tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c*d**2 + 3*tan((e + f*x)/2)* 
*5*a*c**4 - 9*tan((e + f*x)/2)**5*a*c**3*d + 6*tan((e + f*x)/2)**5*a*c**2* 
d**2 + 9*tan((e + f*x)/2)**5*a*c*d**3 - 9*tan((e + f*x)/2)**5*a*d**4 - 3*t 
an((e + f*x)/2)**5*b*c**2*d**2 + 3*tan((e + f*x)/2)**5*b*d**4 + 15*tan((e 
+ f*x)/2)**3*a*c**3*d - 45*tan((e + f*x)/2)**3*a*c**2*d**2 - 15*tan((e ...