\(\int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx\) [16]

Optimal result

Integrand size = 31, antiderivative size = 153 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx=\frac {(A b-2 a B) x}{b^3}-\frac {2 a \left (a^2 A b-2 A b^3-2 a^3 B+3 a b^2 B\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2} f}-\frac {B \cos (e+f x)}{b^2 f}+\frac {a^2 (A b-a B) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 3.82 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx=\frac {(A b-2 a B) (e+f x)+\frac {2 a \left (-a^2 A b+2 A b^3+2 a^3 B-3 a b^2 B\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-b B \cos (e+f x)+\frac {a^2 b (A b-a B) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}}{b^3 f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.24, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3467, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3214

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

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 3467

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ 
(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 
*(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2))   Int[(c + d* 
Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c 
- 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* 
a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 
1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} 
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ 
n, -1]
 

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 = 212, normalized size of antiderivative = 1.39

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 B b}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+2 \left (A b -2 B a \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b^{3}}-\frac {2 a \left (\frac {-\frac {b^{2} \left (A b -B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a^{2}-b^{2}}-\frac {b a \left (A b -B a \right )}{a^{2}-b^{2}}}{a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (A \,a^{2} b -2 A \,b^{3}-2 B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{3}}}{f}\)	\(212\)
default	\(\frac {\frac {-\frac {2 B b}{1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}+2 \left (A b -2 B a \right ) \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{b^{3}}-\frac {2 a \left (\frac {-\frac {b^{2} \left (A b -B a \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a^{2}-b^{2}}-\frac {b a \left (A b -B a \right )}{a^{2}-b^{2}}}{a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (A \,a^{2} b -2 A \,b^{3}-2 B \,a^{3}+3 B a \,b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{b^{3}}}{f}\)	\(212\)
risch	\(\frac {x A}{b^{2}}-\frac {2 x B a}{b^{3}}-\frac {B \,{\mathrm e}^{i \left (f x +e \right )}}{2 b^{2} f}-\frac {B \,{\mathrm e}^{-i \left (f x +e \right )}}{2 b^{2} f}+\frac {2 i a^{2} \left (-A b +B a \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{b^{3} \left (a^{2}-b^{2}\right ) f \left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{2}}-\frac {2 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}-\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{3}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f b}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{2}}+\frac {2 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) A}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {2 a^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f \,b^{3}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) B}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f b}\)	\(800\)

Input:

int(sin(f*x+e)^2*(A+B*sin(f*x+e))/(a+b*sin(f*x+e))^2,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 359 vs. \(2 (152) = 304\).

Time = 0.19 (sec) , antiderivative size = 806, normalized size of antiderivative = 5.27 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

integrate(sin(f*x+e)^2*(A+B*sin(f*x+e))/(a+b*sin(f*x+e))^2,x, algorithm="f 
ricas")
 

Output:

[-1/2*(2*(2*B*a^6 - A*a^5*b - 4*B*a^4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A* 
a*b^5)*f*x - (2*B*a^5 - A*a^4*b - 3*B*a^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - 
 A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(- 
((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2 - 2*(a*cos( 
f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x + e 
)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*(2*B*a^5*b - A*a^4*b^2 - 3*B*a^ 
3*b^3 + A*a^2*b^4 + B*a*b^5)*cos(f*x + e) + 2*((2*B*a^5*b - A*a^4*b^2 - 4* 
B*a^3*b^3 + 2*A*a^2*b^4 + 2*B*a*b^5 - A*b^6)*f*x + (B*a^4*b^2 - 2*B*a^2*b^ 
4 + B*b^6)*cos(f*x + e))*sin(f*x + e))/((a^4*b^4 - 2*a^2*b^6 + b^8)*f*sin( 
f*x + e) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*f), -((2*B*a^6 - A*a^5*b - 4*B*a^ 
4*b^2 + 2*A*a^3*b^3 + 2*B*a^2*b^4 - A*a*b^5)*f*x + (2*B*a^5 - A*a^4*b - 3* 
B*a^3*b^2 + 2*A*a^2*b^3 + (2*B*a^4*b - A*a^3*b^2 - 3*B*a^2*b^3 + 2*A*a*b^4 
)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b 
^2)*cos(f*x + e))) + (2*B*a^5*b - A*a^4*b^2 - 3*B*a^3*b^3 + A*a^2*b^4 + B* 
a*b^5)*cos(f*x + e) + ((2*B*a^5*b - A*a^4*b^2 - 4*B*a^3*b^3 + 2*A*a^2*b^4 
+ 2*B*a*b^5 - A*b^6)*f*x + (B*a^4*b^2 - 2*B*a^2*b^4 + B*b^6)*cos(f*x + e)) 
*sin(f*x + e))/((a^4*b^4 - 2*a^2*b^6 + b^8)*f*sin(f*x + e) + (a^5*b^3 - 2* 
a^3*b^5 + a*b^7)*f)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(sin(f*x+e)^2*(A+B*sin(f*x+e))/(a+b*sin(f*x+e))^2,x, algorithm="m 
axima")
 

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*b^2-4*a^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. 356 vs. \(2 (152) = 304\).

Time = 0.21 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.33 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx=\frac {\frac {2 \, {\left (2 \, B a^{4} - A a^{3} b - 3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{3} - b^{5}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (B a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - A a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - A a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, B a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - A a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, B b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, B a^{3} - A a^{2} b - B a b^{2}\right )}}{{\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )} {\left (a^{2} b^{2} - b^{4}\right )}} - \frac {{\left (2 \, B a - A b\right )} {\left (f x + e\right )}}{b^{3}}}{f} \] Input:

integrate(sin(f*x+e)^2*(A+B*sin(f*x+e))/(a+b*sin(f*x+e))^2,x, algorithm="g 
iac")
 

Output:

(2*(2*B*a^4 - A*a^3*b - 3*B*a^2*b^2 + 2*A*a*b^3)*(pi*floor(1/2*(f*x + e)/p 
i + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/(( 
a^2*b^3 - b^5)*sqrt(a^2 - b^2)) - 2*(B*a^2*b*tan(1/2*f*x + 1/2*e)^3 - A*a* 
b^2*tan(1/2*f*x + 1/2*e)^3 + 2*B*a^3*tan(1/2*f*x + 1/2*e)^2 - A*a^2*b*tan( 
1/2*f*x + 1/2*e)^2 - B*a*b^2*tan(1/2*f*x + 1/2*e)^2 + 3*B*a^2*b*tan(1/2*f* 
x + 1/2*e) - A*a*b^2*tan(1/2*f*x + 1/2*e) - 2*B*b^3*tan(1/2*f*x + 1/2*e) + 
 2*B*a^3 - A*a^2*b - B*a*b^2)/((a*tan(1/2*f*x + 1/2*e)^4 + 2*b*tan(1/2*f*x 
 + 1/2*e)^3 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)*( 
a^2*b^2 - b^4)) - (2*B*a - A*b)*(f*x + e)/b^3)/f
 

Mupad [B] (verification not implemented)

Time = 39.92 (sec) , antiderivative size = 3718, normalized size of antiderivative = 24.30 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(A*a^2*b - 2*B*a^3 + B*a*b^2))/(b^2*(a^2 - b^2)) - (2*tan(e/2 + (f*x)/ 
2)^3*(B*a^2 - A*a*b))/(b*(a^2 - b^2)) + (2*tan(e/2 + (f*x)/2)*(2*B*b^2 - 3 
*B*a^2 + A*a*b))/(b*(a^2 - b^2)) + (2*tan(e/2 + (f*x)/2)^2*(A*a^2*b - 2*B* 
a^3 + B*a*b^2))/(b^2*(a^2 - b^2)))/(f*(a + 2*b*tan(e/2 + (f*x)/2) + 2*a*ta 
n(e/2 + (f*x)/2)^2 + a*tan(e/2 + (f*x)/2)^4 + 2*b*tan(e/2 + (f*x)/2)^3)) + 
 (log(tan(e/2 + (f*x)/2) + 1i)*(A*b - 2*B*a)*1i)/(b^3*f) - (log(tan(e/2 + 
(f*x)/2) - 1i)*(A*b*1i - B*a*2i))/(b^3*f) - (a*atan(((a*(-(a + b)^3*(a - b 
)^3)^(1/2)*((32*(A^2*a^2*b^8 - 2*A^2*a^4*b^6 + A^2*a^6*b^4 + 4*B^2*a^4*b^6 
 - 8*B^2*a^6*b^4 + 4*B^2*a^8*b^2 - 4*A*B*a^3*b^7 + 8*A*B*a^5*b^5 - 4*A*B*a 
^7*b^3))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(e/2 + (f*x)/2)*(2*A^2*a*b^1 
0 - 9*A^2*a^3*b^8 + 8*A^2*a^5*b^6 - 2*A^2*a^7*b^4 + 8*B^2*a^3*b^8 - 29*B^2 
*a^5*b^6 + 28*B^2*a^7*b^4 - 8*B^2*a^9*b^2 - 8*A*B*a^2*b^9 + 32*A*B*a^4*b^7 
 - 30*A*B*a^6*b^5 + 8*A*B*a^8*b^3))/(b^10 - 2*a^2*b^8 + a^4*b^6) + (a*(-(a 
 + b)^3*(a - b)^3)^(1/2)*((32*tan(e/2 + (f*x)/2)*(4*A*a^2*b^11 - 6*A*a^4*b 
^9 + 2*A*a^6*b^7 - 6*B*a^3*b^10 + 10*B*a^5*b^8 - 4*B*a^7*b^6))/(b^10 - 2*a 
^2*b^8 + a^4*b^6) - (32*(A*a^3*b^9 + 2*B*a^2*b^10 - 3*B*a^4*b^8 + B*a^6*b^ 
6 - A*a*b^11))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (a*((32*(a^2*b^12 - 2*a^4*b^1 
0 + a^6*b^8))/(b^9 - 2*a^2*b^7 + a^4*b^5) + (32*tan(e/2 + (f*x)/2)*(3*a*b^ 
14 - 8*a^3*b^12 + 7*a^5*b^10 - 2*a^7*b^8))/(b^10 - 2*a^2*b^8 + a^4*b^6))*( 
-(a + b)^3*(a - b)^3)^(1/2)*(2*A*b^3 + 2*B*a^3 - A*a^2*b - 3*B*a*b^2))/...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.63 \[ \int \frac {\sin ^2(e+f x) (A+B \sin (e+f x))}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}-\cos \left (f x +e \right ) a^{2} b +\cos \left (f x +e \right ) b^{3}-a^{3} f x +a \,b^{2} f x}{b^{2} f \left (a^{2}-b^{2}\right )} \] Input:

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

Output:

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