\(\int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\) [39]

Optimal result

Integrand size = 33, antiderivative size = 181 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=-\frac {2 a \left (a^2 c-2 b^2 c+a b d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2 f}+\frac {2 c^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d)^2 \sqrt {c^2-d^2} f}+\frac {a^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.23 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {-\frac {2 a \left (a^2 c-2 b^2 c+a b d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2}+\frac {2 c^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d)^2 \sqrt {c^2-d^2}}-\frac {a^2 \cos (e+f x)}{(a-b) (a+b) (-b c+a d) (a+b \sin (e+f x))}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 3535, 25, 3042, 3480, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3535

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

((-2*a*(a^2*c - 2*b^2*c + a*b*d)*ArcTan[(2*b + 2*a*Tan[(e + f*x)/2])/(2*Sq 
rt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*(b*c - a*d)*f) + (2*(a^2 - b^2)*c^2*ArcT 
an[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((b*c - a*d)*Sqrt[c^ 
2 - d^2]*f))/((a^2 - b^2)*(b*c - a*d)) + (a^2*Cos[e + f*x])/((a^2 - b^2)*( 
b*c - a*d)*f*(a + b*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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ 
(b*c - a*d)   Int[1/(c + d*Sin[e + f*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]
 

rule 3535

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[(-(A*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S 
in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m 
+ 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin 
[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n 
+ 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d 
*(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 
- d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) || 
 !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 
 0])))
 

Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-\frac {2 a \left (\frac {\frac {b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a^{2}-b^{2}}+\frac {a \left (a d -b c \right )}{a^{2}-b^{2}}}{a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (a^{2} c +a b d -2 b^{2} c \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}+\frac {8 c^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (4 a^{2} d^{2}-8 a b c d +4 b^{2} c^{2}\right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(244\)
default	\(\frac {-\frac {2 a \left (\frac {\frac {b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a^{2}-b^{2}}+\frac {a \left (a d -b c \right )}{a^{2}-b^{2}}}{a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (a^{2} c +a b d -2 b^{2} c \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{a^{2} d^{2}-2 a b c d +b^{2} c^{2}}+\frac {8 c^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (4 a^{2} d^{2}-8 a b c d +4 b^{2} c^{2}\right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(244\)
risch	\(\frac {2 i a^{2} \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{b \left (a^{2}-b^{2}\right ) \left (a d -b c \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 {c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right )^{2} f}-\frac {c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right )^{2} f}+\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 ) c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}+\frac {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 d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}-\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 ) b^{2} c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}-\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 ) c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}-\frac {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 d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}+\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 ) b^{2} c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}\)	\(794\)

Input:

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

Output:

1/f*(-2*a/(a^2*d^2-2*a*b*c*d+b^2*c^2)*((b*(a*d-b*c)/(a^2-b^2)*tan(1/2*f*x+ 
1/2*e)+a*(a*d-b*c)/(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^2*c+a*b*d-2*b^2*c)/(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)))+8*c^2/(4*a^2*d^2-8*a*b*c*d+4*b^2*c^2)/(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 625 vs. \(2 (171) = 342\).

Time = 59.94 (sec) , antiderivative size = 2837, normalized size of antiderivative = 15.67 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(sin(f*x+e)^2/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),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*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.61 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} c^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} - \frac {{\left (a^{3} c - 2 \, a b^{2} c + a^{2} b d\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} c^{2} - b^{4} c^{2} - 2 \, a^{3} b c d + 2 \, a b^{3} c d + a^{4} d^{2} - a^{2} b^{2} d^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{2}}{{\left (a^{2} b c - b^{3} c - a^{3} d + a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f} \] Input:

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

Output:

2*((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^2/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(c^2 - 
d^2)) - (a^3*c - 2*a*b^2*c + a^2*b*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sg 
n(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/((a^2*b^2*c^2 
 - b^4*c^2 - 2*a^3*b*c*d + 2*a*b^3*c*d + a^4*d^2 - a^2*b^2*d^2)*sqrt(a^2 - 
 b^2)) + (a*b*tan(1/2*f*x + 1/2*e) + a^2)/((a^2*b*c - b^3*c - a^3*d + a*b^ 
2*d)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)))/f
 

Mupad [B] (verification not implemented)

Time = 51.16 (sec) , antiderivative size = 23933, normalized size of antiderivative = 132.23 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((2*a^2)/((a^2 - b^2)*(a*d - b*c)) + (2*a*b*tan(e/2 + (f*x)/2))/((a^2 - 
b^2)*(a*d - b*c)))/(f*(a + 2*b*tan(e/2 + (f*x)/2) + a*tan(e/2 + (f*x)/2)^2 
)) - (c^2*atan(((c^2*(d^2 - c^2)^(1/2)*((32*(2*a^4*b^4*c^6 - a^2*b^6*c^6 - 
 a^6*b^2*c^6 + a^8*c^4*d^2 - 3*a^3*b^5*c^5*d + 2*a^5*b^3*c^5*d + 2*a^7*b*c 
^3*d^3 + 8*a^4*b^4*c^4*d^2 - 5*a^5*b^3*c^3*d^3 + a^6*b^2*c^2*d^4 - 6*a^6*b 
^2*c^4*d^2))/(a^7*d^3 - b^7*c^3 + 2*a^2*b^5*c^3 - a^4*b^3*c^3 + a^3*b^4*d^ 
3 - 2*a^5*b^2*d^3 - 3*a^2*b^5*c*d^2 - 6*a^3*b^4*c^2*d + 6*a^4*b^3*c*d^2 + 
3*a^5*b^2*c^2*d + 3*a*b^6*c^2*d - 3*a^6*b*c*d^2) + (32*tan(e/2 + (f*x)/2)* 
(2*a^7*b*c^6 - 2*a*b^7*c^6 - 2*a^8*c^5*d + 9*a^3*b^5*c^6 - 8*a^5*b^3*c^6 + 
 2*a^8*c^3*d^3 + 2*a^2*b^6*c^5*d - 13*a^4*b^4*c^5*d + 2*a^6*b^2*c*d^5 + 10 
*a^6*b^2*c^5*d + 4*a^7*b*c^2*d^4 - 4*a^7*b*c^4*d^2 - 8*a^3*b^5*c^4*d^2 + 1 
6*a^4*b^4*c^3*d^3 - 10*a^5*b^3*c^2*d^4 + 13*a^5*b^3*c^4*d^2 - 13*a^6*b^2*c 
^3*d^3))/(a^7*d^3 - b^7*c^3 + 2*a^2*b^5*c^3 - a^4*b^3*c^3 + a^3*b^4*d^3 - 
2*a^5*b^2*d^3 - 3*a^2*b^5*c*d^2 - 6*a^3*b^4*c^2*d + 6*a^4*b^3*c*d^2 + 3*a^ 
5*b^2*c^2*d + 3*a*b^6*c^2*d - 3*a^6*b*c*d^2) + (c^2*(d^2 - c^2)^(1/2)*((32 
*(a*b^9*c^7 - a^3*b^7*c^7 + a^10*c^2*d^5 - 3*a^2*b^8*c^6*d + 2*a^4*b^6*c^6 
*d + a^6*b^4*c^6*d - a^7*b^3*c*d^6 - 4*a^9*b*c^3*d^4 + a^3*b^7*c^5*d^2 + 6 
*a^4*b^6*c^4*d^3 - 9*a^5*b^5*c^3*d^4 + 3*a^5*b^5*c^5*d^2 + 5*a^6*b^4*c^2*d 
^5 - 12*a^6*b^4*c^4*d^3 + 13*a^7*b^3*c^3*d^4 - 4*a^7*b^3*c^5*d^2 - 6*a^8*b 
^2*c^2*d^5 + 6*a^8*b^2*c^4*d^3 + a^9*b*c*d^6))/(a^7*d^3 - b^7*c^3 + 2*a...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1480, normalized size of antiderivative = 8.18 \[ \int \frac {\sin ^2(e+f x)}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*s 
in(e + f*x)*a**3*b*c**3 + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b 
)/sqrt(a**2 - b**2))*sin(e + f*x)*a**3*b*c*d**2 - 2*sqrt(a**2 - b**2)*atan 
((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**2*b**2*c**2*d 
 + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*si 
n(e + f*x)*a**2*b**2*d**3 + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(e + f*x)*a*b**3*c**3 - 4*sqrt(a**2 - b**2)*atan 
((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a*b**3*c*d**2 - 
2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4* 
c**3 + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2) 
)*a**4*c*d**2 - 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a** 
2 - b**2))*a**3*b*c**2*d + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + 
b)/sqrt(a**2 - b**2))*a**3*b*d**3 + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x 
)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**2*c**3 - 4*sqrt(a**2 - b**2)*atan(( 
tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**2*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)*a**4 
*b*c**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**2*b**3*c**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**5*c**2 + 2*sqrt(c**2 - d**2) 
*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a**5*c**2 - 4*sqrt(c*...