\(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx\) [255]

Optimal result

Integrand size = 35, antiderivative size = 171 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=-\frac {a^2 \left (2 A (c-2 d) d-B \left (2 c^2-4 c d+3 d^2\right )\right ) x}{2 d^3}-\frac {2 a^2 (c-d)^2 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^3 \sqrt {c^2-d^2} f}+\frac {a^2 (2 B c-2 A d-3 B d) \cos (e+f x)}{2 d^2 f}-\frac {B \cos (e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{2 d f} \] Output:

-1/2*a^2*(2*A*(c-2*d)*d-B*(2*c^2-4*c*d+3*d^2))*x/d^3-2*a^2*(c-d)^2*(-A*d+B 
*c)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/d^3/(c^2-d^2)^(1/2)/f 
+1/2*a^2*(-2*A*d+2*B*c-3*B*d)*cos(f*x+e)/d^2/f-1/2*B*cos(f*x+e)*(a^2+a^2*s 
in(f*x+e))/d/f
 

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.04 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {a^2 (1+\sin (e+f x))^2 \left (2 \left (2 A d (-c+2 d)+B \left (2 c^2-4 c d+3 d^2\right )\right ) (e+f x)-\frac {8 (c-d)^2 (B c-A d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}-4 d (-B c+A d+2 B d) \cos (e+f x)-B d^2 \sin (2 (e+f x))\right )}{4 d^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3455

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

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 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 = 0.75 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.37

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {\left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}-\frac {\frac {-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A \,d^{2}-B c d +2 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-B c d +2 B \,d^{2}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (2 A c d -4 A \,d^{2}-2 B \,c^{2}+4 B c d -3 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{3}}\right )}{f}\)	\(235\)
default	\(\frac {2 a^{2} \left (\frac {\left (A \,c^{2} d -2 A c \,d^{2}+A \,d^{3}-B \,c^{3}+2 B \,c^{2} d -B c \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d^{3} \sqrt {c^{2}-d^{2}}}-\frac {\frac {-\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2}+\left (A \,d^{2}-B c d +2 B \,d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\frac {B \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2}+A \,d^{2}-B c d +2 B \,d^{2}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{2}}+\frac {\left (2 A c d -4 A \,d^{2}-2 B \,c^{2}+4 B c d -3 B \,d^{2}\right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2}}{d^{3}}\right )}{f}\)	\(235\)
risch	\(-\frac {a^{2} x A c}{d^{2}}+\frac {2 a^{2} x A}{d}+\frac {a^{2} x B \,c^{2}}{d^{3}}-\frac {2 a^{2} x B c}{d^{2}}+\frac {3 a^{2} x B}{2 d}-\frac {a^{2} {\mathrm e}^{i \left (f x +e \right )} A}{2 d f}+\frac {a^{2} {\mathrm e}^{i \left (f x +e \right )} B c}{2 d^{2} f}-\frac {a^{2} {\mathrm e}^{i \left (f x +e \right )} B}{d f}-\frac {a^{2} {\mathrm e}^{-i \left (f x +e \right )} A}{2 d f}+\frac {a^{2} {\mathrm e}^{-i \left (f x +e \right )} B c}{2 d^{2} f}-\frac {a^{2} {\mathrm e}^{-i \left (f x +e \right )} B}{d f}+\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 ) 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 ) A}{\left (c +d \right ) 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 ) B \,c^{2}}{\left (c +d \right ) f \,d^{3}}+\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 c}{\left (c +d \right ) 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 ) A c}{\left (c +d \right ) 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 ) A}{\left (c +d \right ) 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 ) B \,c^{2}}{\left (c +d \right ) f \,d^{3}}-\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 c}{\left (c +d \right ) f \,d^{2}}-\frac {a^{2} B \sin \left (2 f x +2 e \right )}{4 f d}\)	\(703\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 452, normalized size of antiderivative = 2.64 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\left [-\frac {B a^{2} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, B a^{2} c^{2} - 2 \, {\left (A + 2 \, B\right )} a^{2} c d + {\left (4 \, A + 3 \, B\right )} a^{2} d^{2}\right )} f x - {\left (B a^{2} c^{2} - {\left (A + B\right )} a^{2} c d + A a^{2} d^{2}\right )} \sqrt {-\frac {c - d}{c + d}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) - 2 \, {\left (B a^{2} c d - {\left (A + 2 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}, -\frac {B a^{2} d^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, B a^{2} c^{2} - 2 \, {\left (A + 2 \, B\right )} a^{2} c d + {\left (4 \, A + 3 \, B\right )} a^{2} d^{2}\right )} f x - 2 \, {\left (B a^{2} c^{2} - {\left (A + B\right )} a^{2} c d + A a^{2} d^{2}\right )} \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right ) - 2 \, {\left (B a^{2} c d - {\left (A + 2 \, B\right )} a^{2} d^{2}\right )} \cos \left (f x + e\right )}{2 \, d^{3} f}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorith 
m="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.78 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {\frac {{\left (2 \, B a^{2} c^{2} - 2 \, A a^{2} c d - 4 \, B a^{2} c d + 4 \, A a^{2} d^{2} + 3 \, B a^{2} d^{2}\right )} {\left (f x + e\right )}}{d^{3}} - \frac {4 \, {\left (B a^{2} c^{3} - A a^{2} c^{2} d - 2 \, B a^{2} c^{2} d + 2 \, A a^{2} c d^{2} + B a^{2} c d^{2} - A a^{2} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} d^{3}} + \frac {2 \, {\left (B a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, B a^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, A a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, B a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, B a^{2} c - 2 \, A a^{2} d - 4 \, B a^{2} d\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 1\right )}^{2} d^{2}}}{2 \, f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 42.97 (sec) , antiderivative size = 7371, normalized size of antiderivative = 43.11 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan(((((8*(16*A^2*a^4*c^2*d^6 - 16*A^2*a^4*c^3*d^5 + 4*A^2*a^4*c^4*d^4 + 
 9*B^2*a^4*c^2*d^6 - 24*B^2*a^4*c^3*d^5 + 28*B^2*a^4*c^4*d^4 - 16*B^2*a^4* 
c^5*d^3 + 4*B^2*a^4*c^6*d^2 + 24*A*B*a^4*c^2*d^6 - 44*A*B*a^4*c^3*d^5 + 32 
*A*B*a^4*c^4*d^4 - 8*A*B*a^4*c^5*d^3))/d^5 + ((((32*c^2*d^3 + (8*tan(e/2 + 
 (f*x)/2)*(12*c*d^10 - 8*c^3*d^8))/d^6)*(B*a^2*c^2*1i + (a^2*d^2*(4*A + 3* 
B)*1i)/2 - (a^2*d*(2*A*c + 4*B*c)*1i)/2))/d^3 - (8*(8*A*a^2*c*d^8 + 6*B*a^ 
2*c*d^8 - 8*A*a^2*c^2*d^7 - 8*B*a^2*c^2*d^7 + 2*B*a^2*c^3*d^6))/d^5 + (8*t 
an(e/2 + (f*x)/2)*(8*A*a^2*c*d^9 - 16*A*a^2*c^2*d^8 + 8*A*a^2*c^3*d^7 - 8* 
B*a^2*c^2*d^8 + 16*B*a^2*c^3*d^7 - 8*B*a^2*c^4*d^6))/d^6)*(B*a^2*c^2*1i + 
(a^2*d^2*(4*A + 3*B)*1i)/2 - (a^2*d*(2*A*c + 4*B*c)*1i)/2))/d^3 + (8*tan(e 
/2 + (f*x)/2)*(32*A^2*a^4*c^4*d^5 - 32*A^2*a^4*c^3*d^6 - 16*A^2*a^4*c^2*d^ 
7 - 8*A^2*a^4*c^5*d^4 - 48*B^2*a^4*c^2*d^7 + 43*B^2*a^4*c^3*d^6 + 8*B^2*a^ 
4*c^4*d^5 - 44*B^2*a^4*c^5*d^4 + 32*B^2*a^4*c^6*d^3 - 8*B^2*a^4*c^7*d^2 + 
28*A^2*a^4*c*d^8 + 18*B^2*a^4*c*d^8 - 80*A*B*a^4*c^2*d^7 + 8*A*B*a^4*c^3*d 
^6 + 76*A*B*a^4*c^4*d^5 - 64*A*B*a^4*c^5*d^4 + 16*A*B*a^4*c^6*d^3 + 48*A*B 
*a^4*c*d^8))/d^6)*(B*a^2*c^2*1i + (a^2*d^2*(4*A + 3*B)*1i)/2 - (a^2*d*(2*A 
*c + 4*B*c)*1i)/2)*1i)/d^3 + (((8*(16*A^2*a^4*c^2*d^6 - 16*A^2*a^4*c^3*d^5 
 + 4*A^2*a^4*c^4*d^4 + 9*B^2*a^4*c^2*d^6 - 24*B^2*a^4*c^3*d^5 + 28*B^2*a^4 
*c^4*d^4 - 16*B^2*a^4*c^5*d^3 + 4*B^2*a^4*c^6*d^2 + 24*A*B*a^4*c^2*d^6 - 4 
4*A*B*a^4*c^3*d^5 + 32*A*B*a^4*c^4*d^4 - 8*A*B*a^4*c^5*d^3))/d^5 + (((8...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.34 \[ \int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{c+d \sin (e+f x)} \, dx=\frac {a^{2} \left (4 a \,d^{3} e +2 b \,c^{3} e +3 b \,d^{3} e -\cos \left (f x +e \right ) \sin \left (f x +e \right ) b c \,d^{2}-2 a \,c^{2} d f x +2 a c \,d^{2} f x -2 b \,c^{2} d f x -b c \,d^{2} f x -2 \cos \left (f x +e \right ) b c \,d^{2}+4 a \,d^{3} f x +2 b \,c^{3} f x +3 b \,d^{3} f x -\cos \left (f x +e \right ) \sin \left (f x +e \right ) b \,d^{3}-2 \cos \left (f x +e \right ) a c \,d^{2}+2 \cos \left (f x +e \right ) b \,c^{2} d -b c \,d^{2} e -4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) a \,d^{2}-4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) b \,c^{2}-2 a \,c^{2} d e +2 a c \,d^{2} e -2 b \,c^{2} d e +4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) a c d +4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) b c d -2 \cos \left (f x +e \right ) a \,d^{3}-4 \cos \left (f x +e \right ) b \,d^{3}\right )}{2 d^{3} f \left (c +d \right )} \] Input:

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

Output:

(a**2*(4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2) 
)*a*c*d - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d* 
*2))*a*d**2 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 
- d**2))*b*c**2 + 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c 
**2 - d**2))*b*c*d - cos(e + f*x)*sin(e + f*x)*b*c*d**2 - cos(e + f*x)*sin 
(e + f*x)*b*d**3 - 2*cos(e + f*x)*a*c*d**2 - 2*cos(e + f*x)*a*d**3 + 2*cos 
(e + f*x)*b*c**2*d - 2*cos(e + f*x)*b*c*d**2 - 4*cos(e + f*x)*b*d**3 - 2*a 
*c**2*d*e - 2*a*c**2*d*f*x + 2*a*c*d**2*e + 2*a*c*d**2*f*x + 4*a*d**3*e + 
4*a*d**3*f*x + 2*b*c**3*e + 2*b*c**3*f*x - 2*b*c**2*d*e - 2*b*c**2*d*f*x - 
 b*c*d**2*e - b*c*d**2*f*x + 3*b*d**3*e + 3*b*d**3*f*x))/(2*d**3*f*(c + d) 
)