\(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx\) [284]

Optimal result

Integrand size = 35, antiderivative size = 381 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=-\frac {2 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^3 (c-d)^4 (c+d) \sqrt {c^2-d^2} f}-\frac {d \left (B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )+A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )\right ) \cos (e+f x)}{15 a^3 (c-d)^4 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 (c+d \sin (e+f x))}-\frac {(2 A c+3 B c-9 A d+4 B d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 (c+d \sin (e+f x))}-\frac {\left (B \left (3 c^2-23 c d-15 d^2\right )+A \left (2 c^2-12 c d+45 d^2\right )\right ) \cos (e+f x)}{15 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) (c+d \sin (e+f x))} \] Output:

-2*d^2*(A*d*(4*c+3*d)-B*(3*c^2+3*c*d+d^2))*arctan((d+c*tan(1/2*f*x+1/2*e)) 
/(c^2-d^2)^(1/2))/a^3/(c-d)^4/(c+d)/(c^2-d^2)^(1/2)/f-1/15*d*(B*(3*c^3-23* 
c^2*d-63*c*d^2-22*d^3)+A*(2*c^3-12*c^2*d+43*c*d^2+72*d^3))*cos(f*x+e)/a^3/ 
(c-d)^4/(c+d)/f/(c+d*sin(f*x+e))-1/5*(A-B)*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x 
+e))^3/(c+d*sin(f*x+e))-1/15*(2*A*c-9*A*d+3*B*c+4*B*d)*cos(f*x+e)/a/(c-d)^ 
2/f/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))-1/15*(B*(3*c^2-23*c*d-15*d^2)+A*(2 
*c^2-12*c*d+45*d^2))*cos(f*x+e)/(c-d)^3/f/(a^3+a^3*sin(f*x+e))/(c+d*sin(f* 
x+e))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(1253\) vs. \(2(381)=762\).

Time = 12.52 (sec) , antiderivative size = 1253, normalized size of antiderivative = 3.29 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

Integrate[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x] 
)^2),x]
 

Output:

(2*d^2*(3*B*c^2 - 4*A*c*d + 3*B*c*d - 3*A*d^2 + B*d^2)*ArcTan[(Sec[(e + f* 
x)/2]*(d*Cos[(e + f*x)/2] + c*Sin[(e + f*x)/2]))/Sqrt[c^2 - d^2]]*(Cos[(e 
+ f*x)/2] + Sin[(e + f*x)/2])^6)/((c - d)^4*(c + d)*Sqrt[c^2 - d^2]*f*(a + 
 a*Sin[e + f*x])^3) + ((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(60*B*c^4*Cos 
[(e + f*x)/2] - 80*A*c^3*d*Cos[(e + f*x)/2] - 390*B*c^3*d*Cos[(e + f*x)/2] 
 + 540*A*c^2*d^2*Cos[(e + f*x)/2] - 1090*B*c^2*d^2*Cos[(e + f*x)/2] + 1430 
*A*c*d^3*Cos[(e + f*x)/2] - 885*B*c*d^3*Cos[(e + f*x)/2] + 735*A*d^4*Cos[( 
e + f*x)/2] - 320*B*d^4*Cos[(e + f*x)/2] - 40*A*c^4*Cos[(3*(e + f*x))/2] - 
 60*B*c^4*Cos[(3*(e + f*x))/2] + 196*A*c^3*d*Cos[(3*(e + f*x))/2] + 304*B* 
c^3*d*Cos[(3*(e + f*x))/2] - 476*A*c^2*d^2*Cos[(3*(e + f*x))/2] + 1076*B*c 
^2*d^2*Cos[(3*(e + f*x))/2] - 1546*A*c*d^3*Cos[(3*(e + f*x))/2] + 1181*B*c 
*d^3*Cos[(3*(e + f*x))/2] - 969*A*d^4*Cos[(3*(e + f*x))/2] + 334*B*d^4*Cos 
[(3*(e + f*x))/2] + 60*B*c^2*d^2*Cos[(5*(e + f*x))/2] - 90*A*c*d^3*Cos[(5* 
(e + f*x))/2] + 15*B*c*d^3*Cos[(5*(e + f*x))/2] - 15*A*d^4*Cos[(5*(e + f*x 
))/2] + 30*B*d^4*Cos[(5*(e + f*x))/2] + 4*A*c^3*d*Cos[(7*(e + f*x))/2] + 6 
*B*c^3*d*Cos[(7*(e + f*x))/2] - 24*A*c^2*d^2*Cos[(7*(e + f*x))/2] - 46*B*c 
^2*d^2*Cos[(7*(e + f*x))/2] + 86*A*c*d^3*Cos[(7*(e + f*x))/2] - 111*B*c*d^ 
3*Cos[(7*(e + f*x))/2] + 129*A*d^4*Cos[(7*(e + f*x))/2] - 44*B*d^4*Cos[(7* 
(e + f*x))/2] + 80*A*c^4*Sin[(e + f*x)/2] + 60*B*c^4*Sin[(e + f*x)/2] - 34 
0*A*c^3*d*Sin[(e + f*x)/2] - 440*B*c^3*d*Sin[(e + f*x)/2] + 820*A*c^2*d...
 

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 409, normalized size of antiderivative = 1.07, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3457, 25, 3042, 3457, 25, 3042, 3457, 25, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {-\frac {\int -\frac {\left (B \left (3 c^2-17 d c-7 d^2\right )+A \left (2 c^2-8 d c+27 d^2\right )\right ) a^2+2 d (2 A c+3 B c-9 A d+4 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\left (B \left (3 c^2-17 d c-7 d^2\right )+A \left (2 c^2-8 d c+27 d^2\right )\right ) a^2+2 d (2 A c+3 B c-9 A d+4 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (B \left (3 c^2-17 d c-7 d^2\right )+A \left (2 c^2-8 d c+27 d^2\right )\right ) a^2+2 d (2 A c+3 B c-9 A d+4 B d) \sin (e+f x) a^2}{(\sin (e+f x) a+a) (c+d \sin (e+f x))^2}dx}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {-\frac {\int -\frac {2 d^2 (A c+24 B c-36 A d+11 B d) a^3+d \left (B \left (3 c^2-23 d c-15 d^2\right )+A \left (2 c^2-12 d c+45 d^2\right )\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {2 d^2 (A c+24 B c-36 A d+11 B d) a^3+d \left (B \left (3 c^2-23 d c-15 d^2\right )+A \left (2 c^2-12 d c+45 d^2\right )\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {2 d^2 (A c+24 B c-36 A d+11 B d) a^3+d \left (B \left (3 c^2-23 d c-15 d^2\right )+A \left (2 c^2-12 d c+45 d^2\right )\right ) \sin (e+f x) a^3}{(c+d \sin (e+f x))^2}dx}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {\frac {\frac {-\frac {\int \frac {15 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 d c+d^2\right )\right )}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {-\frac {15 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {-\frac {15 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {\frac {-\frac {30 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {\frac {\frac {60 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {\frac {-\frac {30 a^3 d^2 \left (A d (4 c+3 d)-B \left (3 c^2+3 c d+d^2\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a^3 d \left (A \left (2 c^3-12 c^2 d+43 c d^2+72 d^3\right )+B \left (3 c^3-23 c^2 d-63 c d^2-22 d^3\right )\right ) \cos (e+f x)}{f \left (c^2-d^2\right ) (c+d \sin (e+f x))}}{a^2 (c-d)}-\frac {a^2 \left (A \left (2 c^2-12 c d+45 d^2\right )+B \left (3 c^2-23 c d-15 d^2\right )\right ) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))}}{3 a^2 (c-d)}-\frac {a (2 A c-9 A d+3 B c+4 B d) \cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2 (c+d \sin (e+f x))}}{5 a^2 (c-d)}-\frac {(A-B) \cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 (c+d \sin (e+f x))}\)

Input:

Int[(A + B*Sin[e + f*x])/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^2),x 
]
 

Output:

-1/5*((A - B)*Cos[e + f*x])/((c - d)*f*(a + a*Sin[e + f*x])^3*(c + d*Sin[e 
 + f*x])) + (-1/3*(a*(2*A*c + 3*B*c - 9*A*d + 4*B*d)*Cos[e + f*x])/((c - d 
)*f*(a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])) + (-((a^2*(B*(3*c^2 - 23* 
c*d - 15*d^2) + A*(2*c^2 - 12*c*d + 45*d^2))*Cos[e + f*x])/((c - d)*f*(a + 
 a*Sin[e + f*x])*(c + d*Sin[e + f*x]))) + ((-30*a^3*d^2*(A*d*(4*c + 3*d) - 
 B*(3*c^2 + 3*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^3*d*(B*(3*c^3 - 23*c^2*d - 63*c*d^2 - 
 22*d^3) + A*(2*c^3 - 12*c^2*d + 43*c*d^2 + 72*d^3))*Cos[e + f*x])/((c^2 - 
 d^2)*f*(c + d*Sin[e + f*x])))/(a^2*(c - d)))/(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 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 = 2.51 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.93

method	result	size
derivativedivides	\(\frac {-\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (4 A c d +3 A \,d^{2}-3 B \,c^{2}-3 B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{4}}-\frac {-8 A +8 B}{2 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +8 A d +2 B c -6 B d}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -12 A d -6 B c +10 B d \right )}{3 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-4 A c d +6 A \,d^{2}-3 B \,d^{2}\right )}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} f}\)	\(355\)
default	\(\frac {-\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (4 A c d +3 A \,d^{2}-3 B \,c^{2}-3 B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{4}}-\frac {-8 A +8 B}{2 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {2 \left (4 A -4 B \right )}{5 \left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}-\frac {-4 A c +8 A d +2 B c -6 B d}{\left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (8 A c -12 A d -6 B c +10 B d \right )}{3 \left (c -d \right )^{3} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 \left (A \,c^{2}-4 A c d +6 A \,d^{2}-3 B \,d^{2}\right )}{\left (c -d \right )^{4} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}}{a^{3} f}\)	\(355\)
risch	\(\text {Expression too large to display}\)	\(1830\)

Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x,method=_RETUR 
NVERBOSE)
 

Output:

2/f/a^3*(-d^2/(c-d)^4*((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)+(4*A*c*d+3*A*d 
^2-3*B*c^2-3*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)))-1/4*(-8*A+8*B)/(c-d)^2/(tan(1/2*f*x+1/2*e) 
+1)^4-1/5*(4*A-4*B)/(c-d)^2/(tan(1/2*f*x+1/2*e)+1)^5-1/2*(-4*A*c+8*A*d+2*B 
*c-6*B*d)/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^2-1/3*(8*A*c-12*A*d-6*B*c+10*B*d) 
/(c-d)^3/(tan(1/2*f*x+1/2*e)+1)^3-(A*c^2-4*A*c*d+6*A*d^2-3*B*d^2)/(c-d)^4/ 
(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. 2201 vs. \(2 (368) = 736\).

Time = 0.23 (sec) , antiderivative size = 4486, normalized size of antiderivative = 11.77 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algori 
thm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

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

Leaf count of result is larger than twice the leaf count of optimal. 743 vs. \(2 (368) = 736\).

Time = 0.32 (sec) , antiderivative size = 743, normalized size of antiderivative = 1.95 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x, algori 
thm="giac")
 

Output:

2/15*(15*(3*B*c^2*d^2 - 4*A*c*d^3 + 3*B*c*d^3 - 3*A*d^4 + B*d^4)*(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^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)*sqrt(c^2 - d^2)) + 15*(B*c*d^4*tan(1/2*f*x + 1/2*e) - 
 A*d^5*tan(1/2*f*x + 1/2*e) + B*c^2*d^3 - A*c*d^4)/((a^3*c^6 - 3*a^3*c^5*d 
 + 2*a^3*c^4*d^2 + 2*a^3*c^3*d^3 - 3*a^3*c^2*d^4 + a^3*c*d^5)*(c*tan(1/2*f 
*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) - (15*A*c^2*tan(1/2*f*x + 1 
/2*e)^4 - 60*A*c*d*tan(1/2*f*x + 1/2*e)^4 + 90*A*d^2*tan(1/2*f*x + 1/2*e)^ 
4 - 45*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 - 150*A*c*d*tan(1/2*f*x + 1/2*e)^3 - 60*B*c* 
d*tan(1/2*f*x + 1/2*e)^3 + 300*A*d^2*tan(1/2*f*x + 1/2*e)^3 - 135*B*d^2*ta 
n(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 - 190*A*c*d*tan(1/2*f*x + 1/2*e)^2 - 100*B*c*d*tan(1/2*f*x 
+ 1/2*e)^2 + 420*A*d^2*tan(1/2*f*x + 1/2*e)^2 - 185*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) - 1 
10*A*c*d*tan(1/2*f*x + 1/2*e) - 80*B*c*d*tan(1/2*f*x + 1/2*e) + 270*A*d^2* 
tan(1/2*f*x + 1/2*e) - 115*B*d^2*tan(1/2*f*x + 1/2*e) + 7*A*c^2 + 3*B*c^2 
- 34*A*c*d - 16*B*c*d + 72*A*d^2 - 32*B*d^2)/((a^3*c^4 - 4*a^3*c^3*d + 6*a 
^3*c^2*d^2 - 4*a^3*c*d^3 + a^3*d^4)*(tan(1/2*f*x + 1/2*e) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 41.92 (sec) , antiderivative size = 1349, normalized size of antiderivative = 3.54 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

int((A + B*sin(e + f*x))/((a + a*sin(e + f*x))^3*(c + d*sin(e + f*x))^2),x 
)
 

Output:

(2*d^2*atan(((d^2*(3*B*c^2 - 3*A*d^2 + B*d^2 - 4*A*c*d + 3*B*c*d)*(2*a^3*d 
^6 - 6*a^3*c*d^5 + 2*a^3*c^5*d + 4*a^3*c^2*d^4 + 4*a^3*c^3*d^3 - 6*a^3*c^4 
*d^2))/(a^3*(c + d)^(3/2)*(c - d)^(9/2)) + (2*c*d^2*tan(e/2 + (f*x)/2)*(3* 
B*c^2 - 3*A*d^2 + B*d^2 - 4*A*c*d + 3*B*c*d)*(a^3*c^5 + a^3*d^5 - 3*a^3*c* 
d^4 - 3*a^3*c^4*d + 2*a^3*c^2*d^3 + 2*a^3*c^3*d^2))/(a^3*(c + d)^(3/2)*(c 
- d)^(9/2)))/(2*B*d^4 - 6*A*d^4 + 6*B*c^2*d^2 - 8*A*c*d^3 + 6*B*c*d^3))*(3 
*B*c^2 - 3*A*d^2 + B*d^2 - 4*A*c*d + 3*B*c*d))/(a^3*f*(c + d)^(3/2)*(c - d 
)^(9/2)) - ((2*(7*A*c^4 + 15*A*d^4 + 3*B*c^4 + 38*A*c^2*d^2 - 48*B*c^2*d^2 
 + 72*A*c*d^3 - 27*A*c^3*d - 47*B*c*d^3 - 13*B*c^3*d))/(15*(c + d)*(c - d) 
*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + (4*tan(e/2 + (f*x)/2)^3*(5*A*c^4 + 15* 
A*d^4 + 3*B*c^4 + 19*A*c^2*d^2 - 45*B*c^2*d^2 + 84*A*c*d^3 - 18*A*c^3*d - 
52*B*c*d^3 - 11*B*c^3*d))/(3*c*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) + 
(2*tan(e/2 + (f*x)/2)*(20*A*c^5 + 15*A*d^5 + 15*B*c^5 + 346*A*c^2*d^3 + 10 
6*A*c^3*d^2 - 286*B*c^2*d^3 - 221*B*c^3*d^2 + 219*A*c*d^4 - 76*A*c^4*d - 7 
9*B*c*d^4 - 59*B*c^4*d))/(15*c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - 
d^3)) + (2*tan(e/2 + (f*x)/2)^5*(2*A*c^5 + 5*A*d^5 + B*c^5 + 24*A*c^2*d^3 
+ 4*A*c^3*d^2 - 16*B*c^2*d^3 - 13*B*c^3*d^2 + 13*A*c*d^4 - 6*A*c^4*d - 11* 
B*c*d^4 - 3*B*c^4*d))/(c*(c + d)*(c - d)*(3*c*d^2 - 3*c^2*d + c^3 - d^3)) 
+ (2*tan(e/2 + (f*x)/2)^4*(11*A*c^5 + 30*A*d^5 + 3*B*c^5 + 162*A*c^2*d^3 + 
 4*A*c^3*d^2 - 139*B*c^2*d^3 - 84*B*c^3*d^2 + 135*A*c*d^4 - 27*A*c^4*d ...
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 6828, normalized size of antiderivative = 17.92 \[ \int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^2,x)
 

Output:

(2*( - 300*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d** 
2))*tan((e + f*x)/2)**7*a*c**3*d**3 - 345*sqrt(c**2 - d**2)*atan((tan((e + 
 f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**7*a*c**2*d**4 - 90*sq 
rt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + 
f*x)/2)**7*a*c*d**5 + 225*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/ 
sqrt(c**2 - d**2))*tan((e + f*x)/2)**7*b*c**4*d**2 + 315*sqrt(c**2 - d**2) 
*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**7*b*c* 
*3*d**3 + 165*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - 
d**2))*tan((e + f*x)/2)**7*b*c**2*d**4 + 30*sqrt(c**2 - d**2)*atan((tan((e 
 + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**7*b*c*d**5 - 1500*s 
qrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + 
 f*x)/2)**6*a*c**3*d**3 - 2325*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c 
+ d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**6*a*c**2*d**4 - 1140*sqrt(c**2 - 
 d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)** 
6*a*c*d**5 - 180*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 
 - d**2))*tan((e + f*x)/2)**6*a*d**6 + 1125*sqrt(c**2 - d**2)*atan((tan((e 
 + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**6*b*c**4*d**2 + 202 
5*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*tan(( 
e + f*x)/2)**6*b*c**3*d**3 + 1455*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2) 
*c + d)/sqrt(c**2 - d**2))*tan((e + f*x)/2)**6*b*c**2*d**4 + 480*sqrt(c...