\(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [263]

Optimal result

Integrand size = 35, antiderivative size = 283 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=-\frac {a^3 \left (2 A (2 c-3 d) d-B \left (6 c^2-12 c d+7 d^2\right )\right ) x}{2 d^4}+\frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^4 (c+d) \sqrt {c^2-d^2} f}-\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{2 d^3 (c+d) f}+\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{2 d^2 (c+d) f}+\frac {a (B c-A d) \cos (e+f x) (a+a \sin (e+f x))^2}{d (c+d) f (c+d \sin (e+f x))} \] Output:

-1/2*a^3*(2*A*(2*c-3*d)*d-B*(6*c^2-12*c*d+7*d^2))*x/d^4+2*a^3*(c-d)^2*(A*d 
*(2*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))/d^4/(c+d)/(c^2-d^2)^(1/2)/f-1/2*a^3*(4*A*c*d-B*(6*c^2-3*c*d-5*d^2)) 
*cos(f*x+e)/d^3/(c+d)/f+1/2*(2*A*d-B*(3*c+d))*cos(f*x+e)*(a^3+a^3*sin(f*x+ 
e))/d^2/(c+d)/f+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))
 

Mathematica [A] (verified)

Time = 6.76 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {a^3 (1+\sin (e+f x))^3 \left (2 \left (2 A d (-2 c+3 d)+B \left (6 c^2-12 c d+7 d^2\right )\right ) (e+f x)-\frac {8 (c-d)^2 \left (-A d (2 c+3 d)+B \left (3 c^2+3 c d-d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}-4 d (-2 B c+A d+3 B d) \cos (e+f x)+\frac {4 (c-d)^2 d (B c-A d) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}-B d^2 \sin (2 (e+f x))\right )}{4 d^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.80 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3454, 25, 3042, 3455, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\int \frac {d \left (2 A (c-3 d) d-B \left (3 c^2-3 d c+2 d^2\right )\right ) a^3+(c+d) \left (2 A (2 c-3 d) d-B \left (6 c^2-12 d c+7 d^2\right )\right ) \sin (e+f x) a^3}{c+d \sin (e+f x)}dx}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\int \frac {d \left (2 A (c-3 d) d-B \left (3 c^2-3 d c+2 d^2\right )\right ) a^3+(c+d) \left (2 A (2 c-3 d) d-B \left (6 c^2-12 d c+7 d^2\right )\right ) \sin (e+f x) a^3}{c+d \sin (e+f x)}dx}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\frac {a^3 x (c+d) \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{d}-\frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\frac {a^3 x (c+d) \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{d}-\frac {2 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\frac {a^3 x (c+d) \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{d}-\frac {4 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\frac {8 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}+\frac {a^3 x (c+d) \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{d}}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a (B c-A d) \cos (e+f x) (a \sin (e+f x)+a)^2}{d f (c+d) (c+d \sin (e+f x))}-\frac {\frac {\frac {\frac {a^3 x (c+d) \left (2 A d (2 c-3 d)-B \left (6 c^2-12 c d+7 d^2\right )\right )}{d}-\frac {4 a^3 (c-d)^2 \left (A d (2 c+3 d)-B \left (3 c^2+3 c d-d^2\right )\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}}{d}+\frac {a^3 \left (4 A c d-B \left (6 c^2-3 c d-5 d^2\right )\right ) \cos (e+f x)}{d f}}{2 d}-\frac {(2 A d-B (3 c+d)) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{2 d f}}{d (c+d)}\)

Input:

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

Output:

(a*(B*c - A*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^2)/(d*(c + d)*f*(c + d*Si 
n[e + f*x])) - ((((a^3*(c + d)*(2*A*(2*c - 3*d)*d - B*(6*c^2 - 12*c*d + 7* 
d^2))*x)/d - (4*a^3*(c - d)^2*(A*d*(2*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])])/(d*Sqrt[c^2 - d^ 
2]*f))/d + (a^3*(4*A*c*d - B*(6*c^2 - 3*c*d - 5*d^2))*Cos[e + f*x])/(d*f)) 
/(2*d) - ((2*A*d - B*(3*c + d))*Cos[e + f*x]*(a^3 + a^3*Sin[e + f*x]))/(2* 
d*f))/(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 3455

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n 
 + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 
) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*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] && 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 = 2.52 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.43

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\frac {-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A \,d^{2}-2 B c d +3 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-2 B c d +3 B \,d^{2}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (4 A c d -6 A \,d^{2}-6 B \,c^{2}+12 B c d -7 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}+\frac {\frac {-\frac {d^{2} \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A \,c^{3} d -A \,c^{2} d^{2}-4 A c \,d^{3}+3 A \,d^{4}-3 B \,c^{4}+3 B \,c^{3} d +4 B \,c^{2} d^{2}-5 B c \,d^{3}+B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}\right )}{f}\)	\(406\)
default	\(\frac {2 a^{3} \left (-\frac {\frac {-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A \,d^{2}-2 B c d +3 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-2 B c d +3 B \,d^{2}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (4 A c d -6 A \,d^{2}-6 B \,c^{2}+12 B c d -7 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{4}}+\frac {\frac {-\frac {d^{2} \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A \,c^{3} d -A \,c^{2} d^{2}-4 A c \,d^{3}+3 A \,d^{4}-3 B \,c^{4}+3 B \,c^{3} d +4 B \,c^{2} d^{2}-5 B c \,d^{3}+B \,d^{4}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{4}}\right )}{f}\)	\(406\)
risch	\(\text {Expression too large to display}\)	\(1089\)

Input:

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

Output:

2/f*a^3*(-1/d^4*((-1/2*B*d^2*tan(1/2*f*x+1/2*e)^3+(A*d^2-2*B*c*d+3*B*d^2)* 
tan(1/2*f*x+1/2*e)^2+1/2*B*d^2*tan(1/2*f*x+1/2*e)+A*d^2-2*B*c*d+3*B*d^2)/( 
1+tan(1/2*f*x+1/2*e)^2)^2+1/2*(4*A*c*d-6*A*d^2-6*B*c^2+12*B*c*d-7*B*d^2)*a 
rctan(tan(1/2*f*x+1/2*e)))+1/d^4*((-d^2*(A*c^2*d-2*A*c*d^2+A*d^3-B*c^3+2*B 
*c^2*d-B*c*d^2)/(c+d)/c*tan(1/2*f*x+1/2*e)-d*(A*c^2*d-2*A*c*d^2+A*d^3-B*c^ 
3+2*B*c^2*d-B*c*d^2)/(c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e) 
+c)+(2*A*c^3*d-A*c^2*d^2-4*A*c*d^3+3*A*d^4-3*B*c^4+3*B*c^3*d+4*B*c^2*d^2-5 
*B*c*d^3+B*d^4)/(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))))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 1027, normalized size of antiderivative = 3.63 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^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. 571 vs. \(2 (271) = 542\).

Time = 0.25 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.02 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{3} c^{4} - 2 \, A a^{3} c^{3} d - 3 \, B a^{3} c^{3} d + A a^{3} c^{2} d^{2} - 4 \, B a^{3} c^{2} d^{2} + 4 \, A a^{3} c d^{3} + 5 \, B a^{3} c d^{3} - 3 \, A a^{3} d^{4} - B a^{3} d^{4}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{4} + d^{5}\right )} \sqrt {c^{2} - d^{2}}} - \frac {4 \, {\left (B a^{3} c^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, B a^{3} c^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, A a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B a^{3} c d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a^{3} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B a^{3} c^{4} - A a^{3} c^{3} d - 2 \, B a^{3} c^{3} d + 2 \, A a^{3} c^{2} d^{2} + B a^{3} c^{2} d^{2} - A a^{3} c d^{3}\right )}}{{\left (c^{2} d^{3} + c d^{4}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}} - \frac {{\left (6 \, B a^{3} c^{2} - 4 \, A a^{3} c d - 12 \, B a^{3} c d + 6 \, A a^{3} d^{2} + 7 \, B a^{3} d^{2}\right )} {\left (f x + e\right )}}{d^{4}} - \frac {2 \, {\left (B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, B a^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, A a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, B a^{3} c - 2 \, A a^{3} d - 6 \, B a^{3} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{3}}}{2 \, f} \] Input:

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

Output:

-1/2*(4*(3*B*a^3*c^4 - 2*A*a^3*c^3*d - 3*B*a^3*c^3*d + A*a^3*c^2*d^2 - 4*B 
*a^3*c^2*d^2 + 4*A*a^3*c*d^3 + 5*B*a^3*c*d^3 - 3*A*a^3*d^4 - B*a^3*d^4)*(p 
i*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)))/((c*d^4 + d^5)*sqrt(c^2 - d^2)) - 4*(B*a^3*c^3*d*tan( 
1/2*f*x + 1/2*e) - A*a^3*c^2*d^2*tan(1/2*f*x + 1/2*e) - 2*B*a^3*c^2*d^2*ta 
n(1/2*f*x + 1/2*e) + 2*A*a^3*c*d^3*tan(1/2*f*x + 1/2*e) + B*a^3*c*d^3*tan( 
1/2*f*x + 1/2*e) - A*a^3*d^4*tan(1/2*f*x + 1/2*e) + B*a^3*c^4 - A*a^3*c^3* 
d - 2*B*a^3*c^3*d + 2*A*a^3*c^2*d^2 + B*a^3*c^2*d^2 - A*a^3*c*d^3)/((c^2*d 
^3 + c*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)) - ( 
6*B*a^3*c^2 - 4*A*a^3*c*d - 12*B*a^3*c*d + 6*A*a^3*d^2 + 7*B*a^3*d^2)*(f*x 
 + e)/d^4 - 2*(B*a^3*d*tan(1/2*f*x + 1/2*e)^3 + 4*B*a^3*c*tan(1/2*f*x + 1/ 
2*e)^2 - 2*A*a^3*d*tan(1/2*f*x + 1/2*e)^2 - 6*B*a^3*d*tan(1/2*f*x + 1/2*e) 
^2 - B*a^3*d*tan(1/2*f*x + 1/2*e) + 4*B*a^3*c - 2*A*a^3*d - 6*B*a^3*d)/((t 
an(1/2*f*x + 1/2*e)^2 + 1)^2*d^3))/f
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 47.63 (sec) , antiderivative size = 11993, normalized size of antiderivative = 42.38 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] 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*(A*a^3*d^3 - 3*B*a^3*c^3 - A*a^3*c*d^2 + 2*A*a^3*c^2*d + 2*B*a^3*c*d 
^2 + 3*B*a^3*c^2*d))/(d^3*(c + d)) + (2*tan(e/2 + (f*x)/2)^4*(A*a^3*d^3 - 
3*B*a^3*c^3 - B*a^3*d^3 - A*a^3*c*d^2 + 2*A*a^3*c^2*d + B*a^3*c*d^2 + 3*B* 
a^3*c^2*d))/(d^3*(c + d)) + (2*tan(e/2 + (f*x)/2)^2*(2*A*a^3*d^3 - 6*B*a^3 
*c^3 + B*a^3*d^3 - 2*A*a^3*c*d^2 + 4*A*a^3*c^2*d + 5*B*a^3*c*d^2 + 6*B*a^3 
*c^2*d))/(d^3*(c + d)) + (4*tan(e/2 + (f*x)/2)^3*(A*a^3*d^3 - 3*B*a^3*c^3 
- A*a^3*c*d^2 + 2*A*a^3*c^2*d + 2*B*a^3*c*d^2 + 3*B*a^3*c^2*d))/(c*d^2*(c 
+ d)) + (tan(e/2 + (f*x)/2)^5*(2*A*a^3*d^3 - 3*B*a^3*c^3 - 4*A*a^3*c*d^2 + 
 2*A*a^3*c^2*d - 2*B*a^3*c*d^2 + 3*B*a^3*c^2*d))/(c*d^2*(c + d)) + (tan(e/ 
2 + (f*x)/2)*(2*A*a^3*d^3 - 9*B*a^3*c^3 + 6*A*a^3*c^2*d + 10*B*a^3*c*d^2 + 
 9*B*a^3*c^2*d))/(c*d^2*(c + d)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + 3*c*tan 
(e/2 + (f*x)/2)^2 + 3*c*tan(e/2 + (f*x)/2)^4 + c*tan(e/2 + (f*x)/2)^6 + 4* 
d*tan(e/2 + (f*x)/2)^3 + 2*d*tan(e/2 + (f*x)/2)^5)) - (atan(((((8*(36*A^2* 
a^6*c^2*d^9 + 24*A^2*a^6*c^3*d^8 - 44*A^2*a^6*c^4*d^7 - 16*A^2*a^6*c^5*d^6 
 + 16*A^2*a^6*c^6*d^5 + 49*B^2*a^6*c^2*d^9 - 70*B^2*a^6*c^3*d^8 - 59*B^2*a 
^6*c^4*d^7 + 144*B^2*a^6*c^5*d^6 - 24*B^2*a^6*c^6*d^5 - 72*B^2*a^6*c^7*d^4 
 + 36*B^2*a^6*c^8*d^3 + 84*A*B*a^6*c^2*d^9 - 32*A*B*a^6*c^3*d^8 - 148*A*B* 
a^6*c^4*d^7 + 88*A*B*a^6*c^5*d^6 + 72*A*B*a^6*c^6*d^5 - 48*A*B*a^6*c^7*d^4 
))/(2*c*d^9 + d^10 + c^2*d^8) + (8*tan(e/2 + (f*x)/2)*(144*A^2*a^6*c^2*d^1 
0 - 164*A^2*a^6*c^3*d^9 - 136*A^2*a^6*c^4*d^8 + 136*A^2*a^6*c^5*d^7 + 3...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 1223, normalized size of antiderivative = 4.32 \[ \int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^2,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)*a*c**2*d**2 + 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c 
+ d)/sqrt(c**2 - d**2))*sin(e + f*x)*a*c*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)*a*d**4 - 12*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 + 16*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - 
 d**2))*sin(e + f*x)*b*c*d**3 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2) 
*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b*d**4 + 8*sqrt(c**2 - d**2)*atan( 
(tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*c**3*d + 4*sqrt(c**2 - d**2) 
*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*c**2*d**2 - 12*sqrt(c* 
*2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a*c*d**3 - 12* 
sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c**4 
+ 16*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b* 
c**2*d**2 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - 
d**2))*b*c*d**3 - cos(e + f*x)*sin(e + f*x)**2*b*c**2*d**3 - 2*cos(e + f*x 
)*sin(e + f*x)**2*b*c*d**4 - cos(e + f*x)*sin(e + f*x)**2*b*d**5 - 2*cos(e 
 + f*x)*sin(e + f*x)*a*c**2*d**3 - 4*cos(e + f*x)*sin(e + f*x)*a*c*d**4 - 
2*cos(e + f*x)*sin(e + f*x)*a*d**5 + 3*cos(e + f*x)*sin(e + f*x)*b*c**3*d* 
*2 - 9*cos(e + f*x)*sin(e + f*x)*b*c*d**4 - 6*cos(e + f*x)*sin(e + f*x)*b* 
d**5 - 4*cos(e + f*x)*a*c**3*d**2 - 2*cos(e + f*x)*a*c**2*d**3 - 2*cos(...