\(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx\) [256]

Optimal result

Integrand size = 35, antiderivative size = 198 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=-\frac {a^2 (2 B c-A d-2 B d) x}{d^3}-\frac {2 a^2 (c-d) \left (A d (c+2 d)-B \left (2 c^2+2 c d-d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 (c+d) \sqrt {c^2-d^2} f}+\frac {a^2 (A d-B (2 c+d)) \cos (e+f x)}{d^2 (c+d) f}+\frac {(B c-A d) \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{d (c+d) f (c+d \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 6.16 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.97 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {a^2 (1+\sin (e+f x))^2 \left ((-2 B c+A d+2 B d) (e+f x)+\frac {2 (c-d) \left (-A d (c+2 d)+B \left (2 c^2+2 c d-d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d) \sqrt {c^2-d^2}}-B d \cos (e+f x)-\frac {d (-c+d) (-B c+A d) \cos (e+f x)}{(c+d) (c+d \sin (e+f x))}\right )}{d^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.371, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3454

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

-((((a^2*(c + d)*(2*B*c - A*d - 2*B*d)*x)/d + (2*a^2*(c - d)*(A*d*(c + 2*d 
) - B*(2*c^2 + 2*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^2*(A*d - B*(2*c + d))*Cos[e + f 
*x])/(d*f))/(d*(c + d))) + ((B*c - A*d)*Cos[e + f*x]*(a^2 + a^2*Sin[e + f* 
x]))/(d*(c + d)*f*(c + d*Sin[e + f*x]))
 

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 = 1.02 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.27

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d -2 B c +2 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{3}}-\frac {\frac {-\frac {d^{2} \left (A c d -A \,d^{2}-B \,c^{2}+B c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A c d -A \,d^{2}-B \,c^{2}+B c d \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (A \,c^{2} d +A c \,d^{2}-2 A \,d^{3}-2 B \,c^{3}+3 B c \,d^{2}-B \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{3}}\right )}{f}\)	\(251\)
default	\(\frac {2 a^{2} \left (\frac {-\frac {B d}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+\left (A d -2 B c +2 B d \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{3}}-\frac {\frac {-\frac {d^{2} \left (A c d -A \,d^{2}-B \,c^{2}+B c d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}-\frac {d \left (A c d -A \,d^{2}-B \,c^{2}+B c d \right )}{c +d}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (A \,c^{2} d +A c \,d^{2}-2 A \,d^{3}-2 B \,c^{3}+3 B c \,d^{2}-B \,d^{3}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}}{d^{3}}\right )}{f}\)	\(251\)
risch	\(\frac {a^{2} x A}{d^{2}}-\frac {2 a^{2} x B c}{d^{3}}+\frac {2 a^{2} x B}{d^{2}}-\frac {B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}}{2 d^{2} f}-\frac {B \,a^{2} {\mathrm e}^{-i \left (f x +e \right )}}{2 d^{2} f}+\frac {2 i a^{2} \left (-A c d +A \,d^{2}+B \,c^{2}-B c d \right ) \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d^{3} \left (c +d \right ) f \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) A c}{\left (c +d \right )^{2} f \,d^{2}}+\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) A}{\left (c +d \right )^{2} f d}-\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B \,c^{2}}{\left (c +d \right )^{2} f \,d^{3}}-\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B c}{\left (c +d \right )^{2} f \,d^{2}}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c -\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B}{\left (c +d \right )^{2} f d}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) A c}{\left (c +d \right )^{2} f \,d^{2}}-\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) A}{\left (c +d \right )^{2} f d}+\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B \,c^{2}}{\left (c +d \right )^{2} f \,d^{3}}+\frac {2 \sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B c}{\left (c +d \right )^{2} f \,d^{2}}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right ) B}{\left (c +d \right )^{2} f d}\)	\(788\)

Input:

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

Output:

2/f*a^2*(1/d^3*(-B*d/(1+tan(1/2*f*x+1/2*e)^2)+(A*d-2*B*c+2*B*d)*arctan(tan 
(1/2*f*x+1/2*e)))-1/d^3*((-d^2*(A*c*d-A*d^2-B*c^2+B*c*d)/(c+d)/c*tan(1/2*f 
*x+1/2*e)-d*(A*c*d-A*d^2-B*c^2+B*c*d)/(c+d))/(tan(1/2*f*x+1/2*e)^2*c+2*d*t 
an(1/2*f*x+1/2*e)+c)+(A*c^2*d+A*c*d^2-2*A*d^3-2*B*c^3+3*B*c*d^2-B*d^3)/(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 = 0.12 (sec) , antiderivative size = 731, normalized size of antiderivative = 3.69 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+a*sin(f*x+e))^2*(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. 480 vs. \(2 (194) = 388\).

Time = 0.23 (sec) , antiderivative size = 480, normalized size of antiderivative = 2.42 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, B a^{2} c^{3} - A a^{2} c^{2} d - A a^{2} c d^{2} - 3 \, B a^{2} c d^{2} + 2 \, A a^{2} d^{3} + B a^{2} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c d^{3} + d^{4}\right )} \sqrt {c^{2} - d^{2}}} - \frac {2 \, {\left (B a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - A a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - B a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + A a^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, B a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + B a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A a^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, B a^{2} c^{3} - A a^{2} c^{2} d + A a^{2} c d^{2}\right )}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )} {\left (c^{2} d^{2} + c d^{3}\right )}} - \frac {{\left (2 \, B a^{2} c - A a^{2} d - 2 \, B a^{2} d\right )} {\left (f x + e\right )}}{d^{3}}}{f} \] Input:

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

Output:

(2*(2*B*a^2*c^3 - A*a^2*c^2*d - A*a^2*c*d^2 - 3*B*a^2*c*d^2 + 2*A*a^2*d^3 
+ B*a^2*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2* 
f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((c*d^3 + d^4)*sqrt(c^2 - d^2)) - 2*(B 
*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 - A*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^3 - B 
*a^2*c*d^2*tan(1/2*f*x + 1/2*e)^3 + A*a^2*d^3*tan(1/2*f*x + 1/2*e)^3 + 2*B 
*a^2*c^3*tan(1/2*f*x + 1/2*e)^2 - A*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^2 + A*a 
^2*c*d^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*a^2*c^2*d*tan(1/2*f*x + 1/2*e) - A*a 
^2*c*d^2*tan(1/2*f*x + 1/2*e) + B*a^2*c*d^2*tan(1/2*f*x + 1/2*e) + A*a^2*d 
^3*tan(1/2*f*x + 1/2*e) + 2*B*a^2*c^3 - A*a^2*c^2*d + A*a^2*c*d^2)/((c*tan 
(1/2*f*x + 1/2*e)^4 + 2*d*tan(1/2*f*x + 1/2*e)^3 + 2*c*tan(1/2*f*x + 1/2*e 
)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)*(c^2*d^2 + c*d^3)) - (2*B*a^2*c - A*a^ 
2*d - 2*B*a^2*d)*(f*x + e)/d^3)/f
 

Mupad [B] (verification not implemented)

Time = 44.03 (sec) , antiderivative size = 8706, normalized size of antiderivative = 43.97 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*(A*a^2*d^2 + 2*B*a^2*c^2 - A*a^2*c*d))/(d^2*(c + d)) + (2*tan(e/2 + 
(f*x)/2)^2*(A*a^2*d^2 + 2*B*a^2*c^2 - A*a^2*c*d))/(d^2*(c + d)) + (2*tan(e 
/2 + (f*x)/2)*(A*a^2*d^2 + 3*B*a^2*c^2 - A*a^2*c*d + B*a^2*c*d))/(c*d*(c + 
 d)) + (2*tan(e/2 + (f*x)/2)^3*(A*a^2*d^2 + B*a^2*c^2 - A*a^2*c*d - B*a^2* 
c*d))/(c*d*(c + d)))/(f*(c + 2*d*tan(e/2 + (f*x)/2) + 2*c*tan(e/2 + (f*x)/ 
2)^2 + c*tan(e/2 + (f*x)/2)^4 + 2*d*tan(e/2 + (f*x)/2)^3)) - (atan((((B*a^ 
2*c*2i - a^2*d*(A + 2*B)*1i)*((32*(A^2*a^4*c^2*d^6 + 2*A^2*a^4*c^3*d^5 + A 
^2*a^4*c^4*d^4 + 4*B^2*a^4*c^2*d^6 - 8*B^2*a^4*c^4*d^4 + 4*B^2*a^4*c^6*d^2 
 + 4*A*B*a^4*c^2*d^6 + 4*A*B*a^4*c^3*d^5 - 4*A*B*a^4*c^4*d^4 - 4*A*B*a^4*c 
^5*d^3))/(2*c*d^6 + d^7 + c^2*d^5) + ((B*a^2*c*2i - a^2*d*(A + 2*B)*1i)*(( 
((32*(c^2*d^10 + 2*c^3*d^9 + c^4*d^8))/(2*c*d^6 + d^7 + c^2*d^5) + (32*tan 
(e/2 + (f*x)/2)*(3*c*d^12 + 6*c^2*d^11 + c^3*d^10 - 4*c^4*d^9 - 2*c^5*d^8) 
)/(2*c*d^7 + d^8 + c^2*d^6))*(B*a^2*c*2i - a^2*d*(A + 2*B)*1i))/d^3 - (32* 
(A*a^2*c*d^9 + 2*B*a^2*c*d^9 - A*a^2*c^3*d^7 + B*a^2*c^2*d^8 - 2*B*a^2*c^3 
*d^7 - B*a^2*c^4*d^6))/(2*c*d^6 + d^7 + c^2*d^5) + (32*tan(e/2 + (f*x)/2)* 
(4*A*a^2*c*d^10 + 2*B*a^2*c*d^10 + 2*A*a^2*c^2*d^9 - 4*A*a^2*c^3*d^8 - 2*A 
*a^2*c^4*d^7 - 4*B*a^2*c^2*d^9 - 6*B*a^2*c^3*d^8 + 4*B*a^2*c^4*d^7 + 4*B*a 
^2*c^5*d^6))/(2*c*d^7 + d^8 + c^2*d^6)))/d^3 + (32*tan(e/2 + (f*x)/2)*(8*A 
^2*a^4*c^2*d^7 + 4*A^2*a^4*c^3*d^6 - 4*A^2*a^4*c^4*d^5 - 2*A^2*a^4*c^5*d^4 
 + 6*B^2*a^4*c^2*d^7 - 29*B^2*a^4*c^3*d^6 - 4*B^2*a^4*c^4*d^5 + 28*B^2*...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 872, normalized size of antiderivative = 4.40 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(a**2*( - 2*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**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*d**3 + 4*sqrt(c**2 - d**2)*atan((ta 
n((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b*c**2*d + 4*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**2 - 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d* 
*2))*sin(e + f*x)*b*d**3 - 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + 
d)/sqrt(c**2 - d**2))*a*c**2*d - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2 
)*c + d)/sqrt(c**2 - d**2))*a*c*d**2 + 4*sqrt(c**2 - d**2)*atan((tan((e + 
f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c**3 + 4*sqrt(c**2 - d**2)*atan((tan(( 
e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c**2*d - 2*sqrt(c**2 - d**2)*atan( 
(tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*b*c*d**2 - cos(e + f*x)*sin(e 
+ f*x)*b*c**2*d**2 - 2*cos(e + f*x)*sin(e + f*x)*b*c*d**3 - cos(e + f*x)*s 
in(e + f*x)*b*d**4 + cos(e + f*x)*a*c**2*d**2 - cos(e + f*x)*a*d**4 - 2*co 
s(e + f*x)*b*c**3*d - 2*cos(e + f*x)*b*c**2*d**2 + sin(e + f*x)*a*c**2*d** 
2*f*x + 2*sin(e + f*x)*a*c*d**3*f*x + sin(e + f*x)*a*d**4*f*x - 2*sin(e + 
f*x)*b*c**3*d*f*x - 2*sin(e + f*x)*b*c**2*d**2*f*x - sin(e + f*x)*b*c**2*d 
**2 + 2*sin(e + f*x)*b*c*d**3*f*x - 2*sin(e + f*x)*b*c*d**3 + 2*sin(e + f* 
x)*b*d**4*f*x - sin(e + f*x)*b*d**4 + a*c**3*d*f*x + 2*a*c**2*d**2*f*x + a 
*c*d**3*f*x - 2*b*c**4*f*x - 2*b*c**3*d*f*x - b*c**3*d + 2*b*c**2*d**2*...