\(\int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx\) [719]

Optimal result

Integrand size = 25, antiderivative size = 290 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {2 b^2 \left (a b c-3 a^2 d+2 b^2 d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^3 f}+\frac {2 d^2 \left (3 b c^2-a c d-2 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d)^3 \left (c^2-d^2\right )^{3/2} f}+\frac {d \left (a^2 d^2+b^2 \left (c^2-2 d^2\right )\right ) \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d)^2 \left (c^2-d^2\right ) f (c+d \sin (e+f x))}+\frac {b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x)) (c+d \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 2.47 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\frac {\frac {2 b^2 \left (a b c-3 a^2 d+2 b^2 d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {2 d^2 \left (-3 b c^2+a c d+2 b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac {b^3 (b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}+\frac {d^3 (b c-a d) \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{(b c-a d)^3 f} \] Input:

Integrate[1/((a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 1.62 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.22, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3281, 25, 3042, 3534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[1/((a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])^2),x]
 

Output:

(b^2*Cos[e + f*x])/((a^2 - b^2)*(b*c - a*d)*f*(a + b*Sin[e + f*x])*(c + d* 
Sin[e + f*x])) + (((2*b^2*(a*b*c - 3*a^2*d + 2*b^2*d)*(c^2 - d^2)*ArcTan[( 
2*b + 2*a*Tan[(e + f*x)/2])/(2*Sqrt[a^2 - b^2])])/(Sqrt[a^2 - b^2]*(b*c - 
a*d)*f) + (2*(a^2 - b^2)*d^2*(3*b*c^2 - a*c*d - 2*b*d^2)*ArcTan[(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))/ 
((b*c - a*d)*(c^2 - d^2)) + (d*(a^2*d^2 + b^2*(c^2 - 2*d^2))*Cos[e + f*x]) 
/((b*c - a*d)*(c^2 - d^2)*f*(c + d*Sin[e + f*x])))/((a^2 - b^2)*(b*c - a*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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[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*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 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 3534

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x 
]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*Sin[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[(m + 1)*(b*c - a* 
d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - 
a*b*B + a^2*C) + (m + 1)*(b*c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A 
*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, A, B, 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 = 7.65 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{2}-d^{2}\right )}+\frac {d \left (a d -b c \right )}{c^{2}-d^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (a c d -3 b \,c^{2}+2 b \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{3}}+\frac {2 b^{2} \left (\frac {\frac {b^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {b \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (3 a^{2} d -a b c -2 b^{2} d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{3}}}{f}\)	\(331\)
default	\(\frac {\frac {2 d^{2} \left (\frac {\frac {d^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c \left (c^{2}-d^{2}\right )}+\frac {d \left (a d -b c \right )}{c^{2}-d^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (a c d -3 b \,c^{2}+2 b \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{3}}+\frac {2 b^{2} \left (\frac {\frac {b^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {b \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (3 a^{2} d -a b c -2 b^{2} d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{3}}}{f}\)	\(331\)
risch	\(\text {Expression too large to display}\)	\(1547\)

Input:

int(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(2*d^2/(a*d-b*c)^3*((d^2*(a*d-b*c)/c/(c^2-d^2)*tan(1/2*f*x+1/2*e)+d*(a 
*d-b*c)/(c^2-d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(a*c* 
d-3*b*c^2+2*b*d^2)/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d) 
/(c^2-d^2)^(1/2)))+2*b^2/(a*d-b*c)^3*((b^2*(a*d-b*c)/a/(a^2-b^2)*tan(1/2*f 
*x+1/2*e)+b*(a*d-b*c)/(a^2-b^2))/(tan(1/2*f*x+1/2*e)^2*a+2*b*tan(1/2*f*x+1 
/2*e)+a)+(3*a^2*d-a*b*c-2*b^2*d)/(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 [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*sin(f*x+e))**2/(c+d*sin(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="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. 975 vs. \(2 (280) = 560\).

Time = 0.53 (sec) , antiderivative size = 975, normalized size of antiderivative = 3.36 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x, algorithm="giac")
 

Output:

2*((a*b^3*c - 3*a^2*b^2*d + 2*b^4*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn 
(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/((a^2*b^3*c^3 
- b^5*c^3 - 3*a^3*b^2*c^2*d + 3*a*b^4*c^2*d + 3*a^4*b*c*d^2 - 3*a^2*b^3*c* 
d^2 - a^5*d^3 + a^3*b^2*d^3)*sqrt(a^2 - b^2)) + (3*b*c^2*d^2 - a*c*d^3 - 2 
*b*d^4)*(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)))/((b^3*c^5 - 3*a*b^2*c^4*d + 3*a^2*b*c^3*d^2 
 - b^3*c^3*d^2 - a^3*c^2*d^3 + 3*a*b^2*c^2*d^3 - 3*a^2*b*c*d^4 + a^3*d^5)* 
sqrt(c^2 - d^2)) + (b^4*c^4*tan(1/2*f*x + 1/2*e)^3 - b^4*c^2*d^2*tan(1/2*f 
*x + 1/2*e)^3 + a^4*d^4*tan(1/2*f*x + 1/2*e)^3 - a^2*b^2*d^4*tan(1/2*f*x + 
 1/2*e)^3 + a*b^3*c^4*tan(1/2*f*x + 1/2*e)^2 + 2*b^4*c^3*d*tan(1/2*f*x + 1 
/2*e)^2 - a*b^3*c^2*d^2*tan(1/2*f*x + 1/2*e)^2 + a^4*c*d^3*tan(1/2*f*x + 1 
/2*e)^2 - a^2*b^2*c*d^3*tan(1/2*f*x + 1/2*e)^2 - 2*b^4*c*d^3*tan(1/2*f*x + 
 1/2*e)^2 + 2*a^3*b*d^4*tan(1/2*f*x + 1/2*e)^2 - 2*a*b^3*d^4*tan(1/2*f*x + 
 1/2*e)^2 + b^4*c^4*tan(1/2*f*x + 1/2*e) + 2*a*b^3*c^3*d*tan(1/2*f*x + 1/2 
*e) - b^4*c^2*d^2*tan(1/2*f*x + 1/2*e) + 2*a^3*b*c*d^3*tan(1/2*f*x + 1/2*e 
) - 4*a*b^3*c*d^3*tan(1/2*f*x + 1/2*e) + a^4*d^4*tan(1/2*f*x + 1/2*e) - a^ 
2*b^2*d^4*tan(1/2*f*x + 1/2*e) + a*b^3*c^4 - a*b^3*c^2*d^2 + a^4*c*d^3 - a 
^2*b^2*c*d^3)/((a^3*b^2*c^5 - a*b^4*c^5 - 2*a^4*b*c^4*d + 2*a^2*b^3*c^4*d 
+ a^5*c^3*d^2 - 2*a^3*b^2*c^3*d^2 + a*b^4*c^3*d^2 + 2*a^4*b*c^2*d^3 - 2*a^ 
2*b^3*c^2*d^3 - a^5*c*d^4 + a^3*b^2*c*d^4)*(a*c*tan(1/2*f*x + 1/2*e)^4 ...
 

Mupad [B] (verification not implemented)

Time = 32.74 (sec) , antiderivative size = 71320, normalized size of antiderivative = 245.93 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*sin(e + f*x))^2*(c + d*sin(e + f*x))^2),x)
 

Output:

((2*(a^3*d^3 + b^3*c^3 - a*b^2*d^3 - b^3*c*d^2))/((a^2*d^2 + b^2*c^2 - 2*a 
*b*c*d)*(a^2*c^2 - a^2*d^2 - b^2*c^2 + b^2*d^2)) + (2*tan(e/2 + (f*x)/2)^3 
*(a^4*d^4 + b^4*c^4 - a^2*b^2*d^4 - b^4*c^2*d^2))/(a*c*(a^2*d^2 + b^2*c^2 
- 2*a*b*c*d)*(a^2*c^2 - a^2*d^2 - b^2*c^2 + b^2*d^2)) + (2*tan(e/2 + (f*x) 
/2)*(a^4*d^4 + b^4*c^4 - a^2*b^2*d^4 - b^4*c^2*d^2 - 4*a*b^3*c*d^3 + 2*a*b 
^3*c^3*d + 2*a^3*b*c*d^3))/(a*c*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d)*(a^2*c^2 - 
 a^2*d^2 - b^2*c^2 + b^2*d^2)) + (2*tan(e/2 + (f*x)/2)^2*(a*c + 2*b*d)*(a^ 
3*d^3 + b^3*c^3 - a*b^2*d^3 - b^3*c*d^2))/(a*c*(a^2*d^2 + b^2*c^2 - 2*a*b* 
c*d)*(a^2*c^2 - a^2*d^2 - b^2*c^2 + b^2*d^2)))/(f*(a*c + tan(e/2 + (f*x)/2 
)^3*(2*a*d + 2*b*c) + tan(e/2 + (f*x)/2)^2*(2*a*c + 4*b*d) + tan(e/2 + (f* 
x)/2)*(2*a*d + 2*b*c) + a*c*tan(e/2 + (f*x)/2)^4)) - (b^2*atan(((b^2*(-(a 
+ b)^3*(a - b)^3)^(1/2)*((32*(4*a*b^10*c^4*d^7 - 8*a*b^10*c^6*d^5 + 4*a*b^ 
10*c^8*d^3 + a^3*b^8*c^10*d + 4*a^4*b^7*c*d^10 - 8*a^6*b^5*c*d^10 + 4*a^8* 
b^3*c*d^10 + a^10*b*c^3*d^8 - 4*a^2*b^9*c^3*d^8 + 8*a^2*b^9*c^5*d^6 - 7*a^ 
2*b^9*c^7*d^4 + 4*a^2*b^9*c^9*d^2 - 4*a^3*b^8*c^2*d^9 + 21*a^3*b^8*c^6*d^5 
 - 22*a^3*b^8*c^8*d^3 - 18*a^4*b^7*c^5*d^6 + 26*a^4*b^7*c^7*d^4 - 8*a^4*b^ 
7*c^9*d^2 + 8*a^5*b^6*c^2*d^9 - 18*a^5*b^6*c^4*d^7 - 8*a^5*b^6*c^6*d^5 + 2 
2*a^5*b^6*c^8*d^3 + 21*a^6*b^5*c^3*d^8 - 8*a^6*b^5*c^5*d^6 - 15*a^6*b^5*c^ 
7*d^4 - 7*a^7*b^4*c^2*d^9 + 26*a^7*b^4*c^4*d^7 - 15*a^7*b^4*c^6*d^5 - 22*a 
^8*b^3*c^3*d^8 + 22*a^8*b^3*c^5*d^6 + 4*a^9*b^2*c^2*d^9 - 8*a^9*b^2*c^4...
 

Reduce [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 7318, normalized size of antiderivative = 25.23 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x)
 

Output:

(6*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
e + f*x)**2*a**3*b**3*c**4*d**3 - 12*sqrt(a**2 - b**2)*atan((tan((e + f*x) 
/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a**3*b**3*c**2*d**5 + 6*sqrt 
(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x 
)**2*a**3*b**3*d**7 + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sq 
rt(a**2 - b**2))*sin(e + f*x)**2*a**2*b**4*c**5*d**2 - 8*sqrt(a**2 - b**2) 
*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a**2*b** 
4*c**3*d**4 + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 
- b**2))*sin(e + f*x)**2*a**2*b**4*c*d**6 - 2*sqrt(a**2 - b**2)*atan((tan( 
(e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a*b**5*c**6*d + 6*s 
qrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + 
f*x)**2*a*b**5*c**2*d**5 - 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + 
b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a*b**5*d**7 - 4*sqrt(a**2 - b**2)*at 
an((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*b**6*c**5*d 
**2 + 8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2)) 
*sin(e + f*x)**2*b**6*c**3*d**4 - 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/ 
2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*b**6*c*d**6 + 6*sqrt(a**2 - b 
**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**4*b* 
*2*c**4*d**3 - 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a** 
2 - b**2))*sin(e + f*x)*a**4*b**2*c**2*d**5 + 6*sqrt(a**2 - b**2)*atan(...