\(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\) [264]

Optimal result

Integrand size = 35, antiderivative size = 305 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=-\frac {a^3 (3 B c-A d-3 B d) x}{d^4}-\frac {a^3 (c-d) \left (A d \left (2 c^2+6 c d+7 d^2\right )-3 B \left (2 c^3+4 c^2 d+c d^2-2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^4 (c+d)^2 \sqrt {c^2-d^2} f}-\frac {a^3 (3 B c (2 c+3 d)-A d (2 c+5 d)) \cos (e+f x)}{2 d^3 (c+d)^2 f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {\left (A d (c+4 d)-B \left (3 c^2+4 c d-2 d^2\right )\right ) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d)^2 f (c+d \sin (e+f x))} \] Output:

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(830\) vs. \(2(305)=610\).

Time = 9.03 (sec) , antiderivative size = 830, normalized size of antiderivative = 2.72 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\frac {a^3 (1+\sin (e+f x))^3 \left (\frac {4 (c-d) \left (-A d \left (2 c^2+6 c d+7 d^2\right )+3 B \left (2 c^3+4 c^2 d+c d^2-2 d^3\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+\frac {-12 B c^5 e+4 A c^4 d e-12 B c^4 d e+8 A c^3 d^2 e+6 B c^3 d^2 e+6 A c^2 d^3 e+6 B c^2 d^3 e+4 A c d^4 e+6 B c d^4 e+2 A d^5 e+6 B d^5 e-12 B c^5 f x+4 A c^4 d f x-12 B c^4 d f x+8 A c^3 d^2 f x+6 B c^3 d^2 f x+6 A c^2 d^3 f x+6 B c^2 d^3 f x+4 A c d^4 f x+6 B c d^4 f x+2 A d^5 f x+6 B d^5 f x-d \left (2 A d \left (-2 c^3-4 c^2 d+5 c d^2+d^3\right )+B \left (12 c^4+12 c^3 d-9 c^2 d^2+4 c d^3+d^4\right )\right ) \cos (e+f x)-2 d^2 (c+d)^2 (-3 B c+A d+3 B d) (e+f x) \cos (2 (e+f x))+B c^2 d^3 \cos (3 (e+f x))+2 B c d^4 \cos (3 (e+f x))+B d^5 \cos (3 (e+f x))-24 B c^4 d e \sin (e+f x)+8 A c^3 d^2 e \sin (e+f x)-24 B c^3 d^2 e \sin (e+f x)+16 A c^2 d^3 e \sin (e+f x)+24 B c^2 d^3 e \sin (e+f x)+8 A c d^4 e \sin (e+f x)+24 B c d^4 e \sin (e+f x)-24 B c^4 d f x \sin (e+f x)+8 A c^3 d^2 f x \sin (e+f x)-24 B c^3 d^2 f x \sin (e+f x)+16 A c^2 d^3 f x \sin (e+f x)+24 B c^2 d^3 f x \sin (e+f x)+8 A c d^4 f x \sin (e+f x)+24 B c d^4 f x \sin (e+f x)-9 B c^3 d^2 \sin (2 (e+f x))+3 A c^2 d^3 \sin (2 (e+f x))-9 B c^2 d^3 \sin (2 (e+f x))+3 A c d^4 \sin (2 (e+f x))+4 B c d^4 \sin (2 (e+f x))-6 A d^5 \sin (2 (e+f x))-2 B d^5 \sin (2 (e+f x))}{(c+d \sin (e+f x))^2}\right )}{4 d^4 (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6} \] Input:

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

Output:

(a^3*(1 + Sin[e + f*x])^3*((4*(c - d)*(-(A*d*(2*c^2 + 6*c*d + 7*d^2)) + 3* 
B*(2*c^3 + 4*c^2*d + c*d^2 - 2*d^3))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[ 
c^2 - d^2]])/Sqrt[c^2 - d^2] + (-12*B*c^5*e + 4*A*c^4*d*e - 12*B*c^4*d*e + 
 8*A*c^3*d^2*e + 6*B*c^3*d^2*e + 6*A*c^2*d^3*e + 6*B*c^2*d^3*e + 4*A*c*d^4 
*e + 6*B*c*d^4*e + 2*A*d^5*e + 6*B*d^5*e - 12*B*c^5*f*x + 4*A*c^4*d*f*x - 
12*B*c^4*d*f*x + 8*A*c^3*d^2*f*x + 6*B*c^3*d^2*f*x + 6*A*c^2*d^3*f*x + 6*B 
*c^2*d^3*f*x + 4*A*c*d^4*f*x + 6*B*c*d^4*f*x + 2*A*d^5*f*x + 6*B*d^5*f*x - 
 d*(2*A*d*(-2*c^3 - 4*c^2*d + 5*c*d^2 + d^3) + B*(12*c^4 + 12*c^3*d - 9*c^ 
2*d^2 + 4*c*d^3 + d^4))*Cos[e + f*x] - 2*d^2*(c + d)^2*(-3*B*c + A*d + 3*B 
*d)*(e + f*x)*Cos[2*(e + f*x)] + B*c^2*d^3*Cos[3*(e + f*x)] + 2*B*c*d^4*Co 
s[3*(e + f*x)] + B*d^5*Cos[3*(e + f*x)] - 24*B*c^4*d*e*Sin[e + f*x] + 8*A* 
c^3*d^2*e*Sin[e + f*x] - 24*B*c^3*d^2*e*Sin[e + f*x] + 16*A*c^2*d^3*e*Sin[ 
e + f*x] + 24*B*c^2*d^3*e*Sin[e + f*x] + 8*A*c*d^4*e*Sin[e + f*x] + 24*B*c 
*d^4*e*Sin[e + f*x] - 24*B*c^4*d*f*x*Sin[e + f*x] + 8*A*c^3*d^2*f*x*Sin[e 
+ f*x] - 24*B*c^3*d^2*f*x*Sin[e + f*x] + 16*A*c^2*d^3*f*x*Sin[e + f*x] + 2 
4*B*c^2*d^3*f*x*Sin[e + f*x] + 8*A*c*d^4*f*x*Sin[e + f*x] + 24*B*c*d^4*f*x 
*Sin[e + f*x] - 9*B*c^3*d^2*Sin[2*(e + f*x)] + 3*A*c^2*d^3*Sin[2*(e + f*x) 
] - 9*B*c^2*d^3*Sin[2*(e + f*x)] + 3*A*c*d^4*Sin[2*(e + f*x)] + 4*B*c*d^4* 
Sin[2*(e + f*x)] - 6*A*d^5*Sin[2*(e + f*x)] - 2*B*d^5*Sin[2*(e + f*x)])/(c 
 + d*Sin[e + f*x])^2))/(4*d^4*(c + d)^2*f*(Cos[(e + f*x)/2] + Sin[(e + ...
 

Rubi [A] (verified)

Time = 1.92 (sec) , antiderivative size = 330, normalized size of antiderivative = 1.08, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3454, 25, 3042, 3454, 25, 3042, 3447, 3042, 3502, 3042, 3214, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(2*d*(c + d)*f*(c + d* 
Sin[e + f*x])^2) - (-((((-2*a^3*(c + d)^2*(3*B*c - A*d - 3*B*d)*x)/d - (2* 
a^3*(c - d)*(A*d*(2*c^2 + 6*c*d + 7*d^2) - 3*B*(2*c^3 + 4*c^2*d + c*d^2 - 
2*d^3))*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/(d*Sqrt[ 
c^2 - d^2]*f))/d - (a^3*(3*B*c*(2*c + 3*d) - A*d*(2*c + 5*d))*Cos[e + f*x] 
)/(d*f))/(d*(c + d))) + ((A*d*(c + 4*d) - B*(3*c^2 + 4*c*d - 2*d^2))*Cos[e 
 + f*x]*(a^3 + a^3*Sin[e + f*x]))/(d*(c + d)*f*(c + d*Sin[e + f*x])))/(2*d 
*(c + d))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3214

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

rule 3447

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

rule 3454

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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ 
e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + 
 a*d))   Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp 
[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B 
*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f 
, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
&& GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 
])
 

rule 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 586, normalized size of antiderivative = 1.92

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (\frac {-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d -3 B c +3 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{4}}-\frac {\frac {-\frac {d^{2} \left (A \,c^{3} d +5 A \,c^{2} d^{2}-4 A c \,d^{3}-2 A \,d^{4}-3 B \,c^{4}-3 B \,c^{3} d +6 B \,c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {d \left (2 A \,c^{5} d +4 A \,c^{4} d^{2}-A \,c^{3} d^{3}+7 A \,c^{2} d^{4}-10 A c \,d^{5}-2 A \,d^{6}-4 B \,c^{6}-2 B \,c^{5} d -B \,c^{4} d^{2}-5 B \,c^{3} d^{3}+14 B \,c^{2} d^{4}-2 B c \,d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}-\frac {d^{2} \left (7 A \,c^{3} d +11 A \,c^{2} d^{2}-16 A c \,d^{3}-2 A \,d^{4}-13 B \,c^{4}-5 B \,c^{3} d +22 B \,c^{2} d^{2}-4 B c \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}-\frac {d \left (2 A \,c^{3} d +4 A \,c^{2} d^{2}-5 A c \,d^{3}-A \,d^{4}-4 B \,c^{4}-2 B \,c^{3} d +7 B \,c^{2} d^{2}-B c \,d^{3}\right )}{2 \left (c^{2}+2 c d +d^{2}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{2}}+\frac {\left (2 A \,c^{3} d +4 A \,c^{2} d^{2}+A c \,d^{3}-7 A \,d^{4}-6 B \,c^{4}-6 B \,c^{3} d +9 B \,c^{2} d^{2}+9 B c \,d^{3}-6 B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}\right )}{f}\)	\(586\)
default	\(\frac {2 a^{3} \left (\frac {-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d -3 B c +3 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{4}}-\frac {\frac {-\frac {d^{2} \left (A \,c^{3} d +5 A \,c^{2} d^{2}-4 A c \,d^{3}-2 A \,d^{4}-3 B \,c^{4}-3 B \,c^{3} d +6 B \,c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {d \left (2 A \,c^{5} d +4 A \,c^{4} d^{2}-A \,c^{3} d^{3}+7 A \,c^{2} d^{4}-10 A c \,d^{5}-2 A \,d^{6}-4 B \,c^{6}-2 B \,c^{5} d -B \,c^{4} d^{2}-5 B \,c^{3} d^{3}+14 B \,c^{2} d^{4}-2 B c \,d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2 \left (c^{2}+2 c d +d^{2}\right ) c^{2}}-\frac {d^{2} \left (7 A \,c^{3} d +11 A \,c^{2} d^{2}-16 A c \,d^{3}-2 A \,d^{4}-13 B \,c^{4}-5 B \,c^{3} d +22 B \,c^{2} d^{2}-4 B c \,d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{2}+2 c d +d^{2}\right )}-\frac {d \left (2 A \,c^{3} d +4 A \,c^{2} d^{2}-5 A c \,d^{3}-A \,d^{4}-4 B \,c^{4}-2 B \,c^{3} d +7 B \,c^{2} d^{2}-B c \,d^{3}\right )}{2 \left (c^{2}+2 c d +d^{2}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{2}}+\frac {\left (2 A \,c^{3} d +4 A \,c^{2} d^{2}+A c \,d^{3}-7 A \,d^{4}-6 B \,c^{4}-6 B \,c^{3} d +9 B \,c^{2} d^{2}+9 B c \,d^{3}-6 B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}\right )}{f}\)	\(586\)
risch	\(\text {Expression too large to display}\)	\(1507\)

Input:

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

Output:

2/f*a^3*(1/d^4*(-B*d/(1+tan(1/2*f*x+1/2*e)^2)+(A*d-3*B*c+3*B*d)*arctan(tan 
(1/2*f*x+1/2*e)))-1/d^4*((-1/2*d^2*(A*c^3*d+5*A*c^2*d^2-4*A*c*d^3-2*A*d^4- 
3*B*c^4-3*B*c^3*d+6*B*c^2*d^2)/(c^2+2*c*d+d^2)/c*tan(1/2*f*x+1/2*e)^3-1/2* 
d*(2*A*c^5*d+4*A*c^4*d^2-A*c^3*d^3+7*A*c^2*d^4-10*A*c*d^5-2*A*d^6-4*B*c^6- 
2*B*c^5*d-B*c^4*d^2-5*B*c^3*d^3+14*B*c^2*d^4-2*B*c*d^5)/(c^2+2*c*d+d^2)/c^ 
2*tan(1/2*f*x+1/2*e)^2-1/2*d^2*(7*A*c^3*d+11*A*c^2*d^2-16*A*c*d^3-2*A*d^4- 
13*B*c^4-5*B*c^3*d+22*B*c^2*d^2-4*B*c*d^3)/c/(c^2+2*c*d+d^2)*tan(1/2*f*x+1 
/2*e)-1/2*d*(2*A*c^3*d+4*A*c^2*d^2-5*A*c*d^3-A*d^4-4*B*c^4-2*B*c^3*d+7*B*c 
^2*d^2-B*c*d^3)/(c^2+2*c*d+d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1 
/2*e)+c)^2+1/2*(2*A*c^3*d+4*A*c^2*d^2+A*c*d^3-7*A*d^4-6*B*c^4-6*B*c^3*d+9* 
B*c^2*d^2+9*B*c*d^3-6*B*d^4)/(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. 790 vs. \(2 (294) = 588\).

Time = 0.17 (sec) , antiderivative size = 1670, normalized size of antiderivative = 5.48 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*(4*(3*B*a^3*c^3*d^2 - (A - 3*B)*a^3*c^2*d^3 - (2*A + 3*B)*a^3*c*d^4 
- (A + 3*B)*a^3*d^5)*f*x*cos(f*x + e)^2 + 4*(B*a^3*c^2*d^3 + 2*B*a^3*c*d^4 
 + B*a^3*d^5)*cos(f*x + e)^3 - 4*(3*B*a^3*c^5 - (A - 3*B)*a^3*c^4*d - 2*A* 
a^3*c^3*d^2 - 2*A*a^3*c^2*d^3 - (2*A + 3*B)*a^3*c*d^4 - (A + 3*B)*a^3*d^5) 
*f*x - (6*B*a^3*c^5 - 2*(A - 6*B)*a^3*c^4*d - 3*(2*A - 3*B)*a^3*c^3*d^2 - 
3*(3*A - 2*B)*a^3*c^2*d^3 - 3*(2*A - B)*a^3*c*d^4 - (7*A + 6*B)*a^3*d^5 - 
(6*B*a^3*c^3*d^2 - 2*(A - 6*B)*a^3*c^2*d^3 - 3*(2*A - B)*a^3*c*d^4 - (7*A 
+ 6*B)*a^3*d^5)*cos(f*x + e)^2 + 2*(6*B*a^3*c^4*d - 2*(A - 6*B)*a^3*c^3*d^ 
2 - 3*(2*A - B)*a^3*c^2*d^3 - (7*A + 6*B)*a^3*c*d^4)*sin(f*x + e))*sqrt(-( 
c - d)/(c + d))*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c 
^2 - d^2 + 2*((c^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x 
+ e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c 
^2 - d^2)) - 2*(6*B*a^3*c^4*d - 2*(A - 3*B)*a^3*c^3*d^2 - (4*A + 3*B)*a^3* 
c^2*d^3 + 5*(A + B)*a^3*c*d^4 + (A + 2*B)*a^3*d^5)*cos(f*x + e) - 2*(4*(3* 
B*a^3*c^4*d - (A - 3*B)*a^3*c^3*d^2 - (2*A + 3*B)*a^3*c^2*d^3 - (A + 3*B)* 
a^3*c*d^4)*f*x + (9*B*a^3*c^3*d^2 - 3*(A - 3*B)*a^3*c^2*d^3 - (3*A + 4*B)* 
a^3*c*d^4 + 2*(3*A + B)*a^3*d^5)*cos(f*x + e))*sin(f*x + e))/((c^2*d^6 + 2 
*c*d^7 + d^8)*f*cos(f*x + e)^2 - 2*(c^3*d^5 + 2*c^2*d^6 + c*d^7)*f*sin(f*x 
 + e) - (c^4*d^4 + 2*c^3*d^5 + 2*c^2*d^6 + 2*c*d^7 + d^8)*f), -1/2*(2*(3*B 
*a^3*c^3*d^2 - (A - 3*B)*a^3*c^2*d^3 - (2*A + 3*B)*a^3*c*d^4 - (A + 3*B...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Leaf count of result is larger than twice the leaf count of optimal. 953 vs. \(2 (294) = 588\).

Time = 0.28 (sec) , antiderivative size = 953, normalized size of antiderivative = 3.12 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

((6*B*a^3*c^4 - 2*A*a^3*c^3*d + 6*B*a^3*c^3*d - 4*A*a^3*c^2*d^2 - 9*B*a^3* 
c^2*d^2 - A*a^3*c*d^3 - 9*B*a^3*c*d^3 + 7*A*a^3*d^4 + 6*B*a^3*d^4)*(pi*flo 
or(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d)/sq 
rt(c^2 - d^2)))/((c^2*d^4 + 2*c*d^5 + d^6)*sqrt(c^2 - d^2)) - 2*B*a^3/((ta 
n(1/2*f*x + 1/2*e)^2 + 1)*d^3) - (3*B*a^3*c^5*d*tan(1/2*f*x + 1/2*e)^3 - A 
*a^3*c^4*d^2*tan(1/2*f*x + 1/2*e)^3 + 3*B*a^3*c^4*d^2*tan(1/2*f*x + 1/2*e) 
^3 - 5*A*a^3*c^3*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*B*a^3*c^3*d^3*tan(1/2*f*x 
+ 1/2*e)^3 + 4*A*a^3*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*A*a^3*c*d^5*tan(1/ 
2*f*x + 1/2*e)^3 + 4*B*a^3*c^6*tan(1/2*f*x + 1/2*e)^2 - 2*A*a^3*c^5*d*tan( 
1/2*f*x + 1/2*e)^2 + 2*B*a^3*c^5*d*tan(1/2*f*x + 1/2*e)^2 - 4*A*a^3*c^4*d^ 
2*tan(1/2*f*x + 1/2*e)^2 + B*a^3*c^4*d^2*tan(1/2*f*x + 1/2*e)^2 + A*a^3*c^ 
3*d^3*tan(1/2*f*x + 1/2*e)^2 + 5*B*a^3*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 - 7* 
A*a^3*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 - 14*B*a^3*c^2*d^4*tan(1/2*f*x + 1/2* 
e)^2 + 10*A*a^3*c*d^5*tan(1/2*f*x + 1/2*e)^2 + 2*B*a^3*c*d^5*tan(1/2*f*x + 
 1/2*e)^2 + 2*A*a^3*d^6*tan(1/2*f*x + 1/2*e)^2 + 13*B*a^3*c^5*d*tan(1/2*f* 
x + 1/2*e) - 7*A*a^3*c^4*d^2*tan(1/2*f*x + 1/2*e) + 5*B*a^3*c^4*d^2*tan(1/ 
2*f*x + 1/2*e) - 11*A*a^3*c^3*d^3*tan(1/2*f*x + 1/2*e) - 22*B*a^3*c^3*d^3* 
tan(1/2*f*x + 1/2*e) + 16*A*a^3*c^2*d^4*tan(1/2*f*x + 1/2*e) + 4*B*a^3*c^2 
*d^4*tan(1/2*f*x + 1/2*e) + 2*A*a^3*c*d^5*tan(1/2*f*x + 1/2*e) + 4*B*a^3*c 
^6 - 2*A*a^3*c^5*d + 2*B*a^3*c^5*d - 4*A*a^3*c^4*d^2 - 7*B*a^3*c^4*d^2 ...
 

Mupad [B] (verification not implemented)

Time = 45.75 (sec) , antiderivative size = 13891, normalized size of antiderivative = 45.54 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((A*a^3*d^4 + 6*B*a^3*c^4 + 5*A*a^3*c*d^3 - 2*A*a^3*c^3*d + B*a^3*c*d^3 
+ 6*B*a^3*c^3*d - 4*A*a^3*c^2*d^2 - 5*B*a^3*c^2*d^2)/(d^3*(c + d)^2) + (4* 
tan(e/2 + (f*x)/2)^3*(A*a^3*d^4 + 6*B*a^3*c^4 + 5*A*a^3*c*d^3 - 2*A*a^3*c^ 
3*d + B*a^3*c*d^3 + 6*B*a^3*c^3*d - 4*A*a^3*c^2*d^2 - 5*B*a^3*c^2*d^2))/(c 
*d^2*(c + d)^2) + (tan(e/2 + (f*x)/2)^5*(2*A*a^3*d^4 + 3*B*a^3*c^4 + 4*A*a 
^3*c*d^3 - A*a^3*c^3*d + 3*B*a^3*c^3*d - 5*A*a^3*c^2*d^2 - 6*B*a^3*c^2*d^2 
))/(c*d^2*(c + d)^2) + (2*tan(e/2 + (f*x)/2)^2*(A*a^3*d^6 + 6*B*a^3*c^6 + 
5*A*a^3*c*d^5 - 2*A*a^3*c^5*d + B*a^3*c*d^5 + 6*B*a^3*c^5*d - 3*A*a^3*c^2* 
d^4 + 3*A*a^3*c^3*d^3 - 4*A*a^3*c^4*d^2 - 3*B*a^3*c^2*d^4 + 11*B*a^3*c^3*d 
^3 + 3*B*a^3*c^4*d^2))/(c^2*d^3*(c + d)^2) + (tan(e/2 + (f*x)/2)^4*(2*A*a^ 
3*d^6 + 6*B*a^3*c^6 + 10*A*a^3*c*d^5 - 2*A*a^3*c^5*d + 2*B*a^3*c*d^5 + 6*B 
*a^3*c^5*d - 7*A*a^3*c^2*d^4 + A*a^3*c^3*d^3 - 4*A*a^3*c^4*d^2 - 14*B*a^3* 
c^2*d^4 + 5*B*a^3*c^3*d^3 + 3*B*a^3*c^4*d^2))/(c^2*d^3*(c + d)^2) + (tan(e 
/2 + (f*x)/2)*(2*A*a^3*d^4 + 21*B*a^3*c^4 + 16*A*a^3*c*d^3 - 7*A*a^3*c^3*d 
 + 4*B*a^3*c*d^3 + 21*B*a^3*c^3*d - 11*A*a^3*c^2*d^2 - 14*B*a^3*c^2*d^2))/ 
(c*d^2*(c + d)^2))/(f*(tan(e/2 + (f*x)/2)^2*(3*c^2 + 4*d^2) + tan(e/2 + (f 
*x)/2)^4*(3*c^2 + 4*d^2) + c^2*tan(e/2 + (f*x)/2)^6 + c^2 + 8*c*d*tan(e/2 
+ (f*x)/2)^3 + 4*c*d*tan(e/2 + (f*x)/2)^5 + 4*c*d*tan(e/2 + (f*x)/2))) - ( 
atan((((B*a^3*c*3i - a^3*d*(A + 3*B)*1i)*((8*(4*A^2*a^6*c^2*d^9 + 16*A^2*a 
^6*c^3*d^8 + 24*A^2*a^6*c^4*d^7 + 16*A^2*a^6*c^5*d^6 + 4*A^2*a^6*c^6*d^...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 2339, normalized size of antiderivative = 7.67 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(a**3*( - 8*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d* 
*2))*sin(e + f*x)**2*a*c**3*d**3 - 24*sqrt(c**2 - d**2)*atan((tan((e + f*x 
)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**2*a*c**2*d**4 - 28*sqrt(c**2 
- d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**2*a 
*c*d**5 + 24*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d 
**2))*sin(e + f*x)**2*b*c**4*d**2 + 48*sqrt(c**2 - d**2)*atan((tan((e + f* 
x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**2*b*c**3*d**3 + 12*sqrt(c**2 
 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**2* 
b*c**2*d**4 - 24*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 
 - d**2))*sin(e + f*x)**2*b*c*d**5 - 16*sqrt(c**2 - d**2)*atan((tan((e + f 
*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*a*c**4*d**2 - 48*sqrt(c**2 - 
 d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*a*c** 
3*d**3 - 56*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d* 
*2))*sin(e + f*x)*a*c**2*d**4 + 48*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2 
)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b*c**5*d + 96*sqrt(c**2 - d**2)*a 
tan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b*c**4*d**2 + 
 24*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin 
(e + f*x)*b*c**3*d**3 - 48*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d) 
/sqrt(c**2 - d**2))*sin(e + f*x)*b*c**2*d**4 - 8*sqrt(c**2 - d**2)*atan((t 
an((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*c**5*d - 24*sqrt(c**2 - d**...