\(\int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx\) [327]

Optimal result

Integrand size = 20, antiderivative size = 121 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx=\frac {2 a \arctan \left (\frac {c+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{3/2} e}+\frac {c \cos (d+e x)-b \sin (d+e x)}{\left (a^2-b^2-c^2\right ) e (a+b \cos (d+e x)+c \sin (d+e x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx=\frac {\frac {2 a \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{3/2}}+\frac {a c+\left (b^2+c^2\right ) \sin (d+e x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (d+e x)+c \sin (d+e x))}}{e} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3608, 25, 27, 3042, 3603, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3608

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 3603

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(-1), x_Symbol] :> Module[{f = FreeFactors[Tan[(d + e*x)/2], x]}, Simp[2*(f 
/e)   Subst[Int[1/(a + b + 2*c*f*x + (a - b)*f^2*x^2), x], x, Tan[(d + e*x) 
/2]/f], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0]
 

rule 3608

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_), x_Symbol] :> Simp[((-c)*Cos[d + e*x] + b*Sin[d + e*x])*((a + b*Cos[d 
+ e*x] + c*Sin[d + e*x])^(n + 1)/(e*(n + 1)*(a^2 - b^2 - c^2))), x] + Simp[ 
1/((n + 1)*(a^2 - b^2 - c^2))   Int[(a*(n + 1) - b*(n + 2)*Cos[d + e*x] - c 
*(n + 2)*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n + 1), x], x 
] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && LtQ[n, -1] && 
NeQ[n, -3/2]
 

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.68

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 \left (b a -b^{2}-c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}+\frac {2 a c}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} b +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b}+\frac {2 a \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}}{e}\)	\(203\)
default	\(\frac {\frac {-\frac {2 \left (b a -b^{2}-c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}+\frac {2 a c}{a^{3}-a^{2} b -a \,b^{2}-a \,c^{2}+b^{3}+b \,c^{2}}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} b +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b}+\frac {2 a \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{2}-b^{2}-c^{2}\right )^{\frac {3}{2}}}}{e}\)	\(203\)
risch	\(-\frac {2 i \left (-i a \,{\mathrm e}^{i \left (e x +d \right )}-i b +c \right )}{\left (-a^{2}+b^{2}+c^{2}\right ) e \left (c \,{\mathrm e}^{2 i \left (e x +d \right )}+i b \,{\mathrm e}^{2 i \left (e x +d \right )}-c +2 i a \,{\mathrm e}^{i \left (e x +d \right )}+i b \right )}-\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}+i a^{2} b -i b^{3}-i b \,c^{2}+b a \sqrt {-a^{2}+b^{2}+c^{2}}-a^{2} c +b^{2} c +c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right ) e}+\frac {a \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i a c \sqrt {-a^{2}+b^{2}+c^{2}}-i a^{2} b +i b^{3}+i b \,c^{2}+b a \sqrt {-a^{2}+b^{2}+c^{2}}+a^{2} c -b^{2} c -c^{3}}{\left (b^{2}+c^{2}\right ) \sqrt {-a^{2}+b^{2}+c^{2}}}\right )}{\sqrt {-a^{2}+b^{2}+c^{2}}\, \left (a^{2}-b^{2}-c^{2}\right ) e}\)	\(373\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (116) = 232\).

Time = 0.13 (sec) , antiderivative size = 819, normalized size of antiderivative = 6.77 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/2*((a*b*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*sqrt(-a^2 + b^2 + c^2)*l 
og((a^2*b^2 - 2*b^4 - c^4 - (a^2 + 3*b^2)*c^2 - (2*a^2*b^2 - b^4 - 2*a^2*c 
^2 + c^4)*cos(e*x + d)^2 - 2*(a*b^3 + a*b*c^2)*cos(e*x + d) - 2*(a*b^2*c + 
 a*c^3 - (b*c^3 - (2*a^2*b - b^3)*c)*cos(e*x + d))*sin(e*x + d) - 2*(2*a*b 
*c*cos(e*x + d)^2 - a*b*c + (b^2*c + c^3)*cos(e*x + d) - (b^3 + b*c^2 + (a 
*b^2 - a*c^2)*cos(e*x + d))*sin(e*x + d))*sqrt(-a^2 + b^2 + c^2))/(2*a*b*c 
os(e*x + d) + (b^2 - c^2)*cos(e*x + d)^2 + a^2 + c^2 + 2*(b*c*cos(e*x + d) 
 + a*c)*sin(e*x + d))) - 2*(c^3 - (a^2 - b^2)*c)*cos(e*x + d) - 2*(a^2*b - 
 b^3 - b*c^2)*sin(e*x + d))/((a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - 2*(a^2*b - 
 b^3)*c^2)*e*cos(e*x + d) + (c^5 - 2*(a^2 - b^2)*c^3 + (a^4 - 2*a^2*b^2 + 
b^4)*c)*e*sin(e*x + d) + (a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 - 2*(a^3 - a*b^2 
)*c^2)*e), ((a*b*cos(e*x + d) + a*c*sin(e*x + d) + a^2)*sqrt(a^2 - b^2 - c 
^2)*arctan(-(a*b*cos(e*x + d) + a*c*sin(e*x + d) + b^2 + c^2)*sqrt(a^2 - b 
^2 - c^2)/((c^3 - (a^2 - b^2)*c)*cos(e*x + d) + (a^2*b - b^3 - b*c^2)*sin( 
e*x + d))) - (c^3 - (a^2 - b^2)*c)*cos(e*x + d) - (a^2*b - b^3 - b*c^2)*si 
n(e*x + d))/((a^4*b - 2*a^2*b^3 + b^5 + b*c^4 - 2*(a^2*b - b^3)*c^2)*e*cos 
(e*x + d) + (c^5 - 2*(a^2 - b^2)*c^3 + (a^4 - 2*a^2*b^2 + b^4)*c)*e*sin(e* 
x + d) + (a^5 - 2*a^3*b^2 + a*b^4 + a*c^4 - 2*(a^3 - a*b^2)*c^2)*e)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^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(c^2+b^2-a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.77 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - b^{2} - c^{2}}}\right )\right )} a}{{\left (a^{2} - b^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {a b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - a c}{{\left (a^{3} - a^{2} b - a b^{2} + b^{3} - a c^{2} + b c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a + b\right )}}\right )}}{e} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 15.63 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.61 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx=\frac {2\,a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )+\frac {2\,\left (-a^2\,c+b^2\,c+c^3\right )}{-a^2+b^2+c^2}}{2\,\sqrt {-a^2+b^2+c^2}}\right )}{e\,{\left (-a^2+b^2+c^2\right )}^{3/2}}-\frac {\frac {2\,a\,c}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}+\frac {2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (b^2-a\,b+c^2\right )}{\left (a-b\right )\,\left (-a^2+b^2+c^2\right )}}{e\,\left (\left (a-b\right )\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+a+b\right )} \] Input:

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

Output:

(2*a*atanh((tan(d/2 + (e*x)/2)*(2*a - 2*b) + (2*(b^2*c - a^2*c + c^3))/(b^ 
2 - a^2 + c^2))/(2*(b^2 - a^2 + c^2)^(1/2))))/(e*(b^2 - a^2 + c^2)^(3/2)) 
- ((2*a*c)/((a - b)*(b^2 - a^2 + c^2)) + (2*tan(d/2 + (e*x)/2)*(b^2 - a*b 
+ c^2))/((a - b)*(b^2 - a^2 + c^2)))/(e*(a + b + tan(d/2 + (e*x)/2)^2*(a - 
 b) + 2*c*tan(d/2 + (e*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 485, normalized size of antiderivative = 4.01 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \cos \left (e x +d \right ) a b c +2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) \sin \left (e x +d \right ) a \,c^{2}+2 \sqrt {a^{2}-b^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +c}{\sqrt {a^{2}-b^{2}-c^{2}}}\right ) a^{2} c +\cos \left (e x +d \right ) a^{2} b^{2}+\cos \left (e x +d \right ) a^{2} c^{2}-\cos \left (e x +d \right ) b^{4}-2 \cos \left (e x +d \right ) b^{2} c^{2}-\cos \left (e x +d \right ) c^{4}+a^{3} b -a \,b^{3}-a b \,c^{2}}{c e \left (\cos \left (e x +d \right ) a^{4} b -2 \cos \left (e x +d \right ) a^{2} b^{3}-2 \cos \left (e x +d \right ) a^{2} b \,c^{2}+\cos \left (e x +d \right ) b^{5}+2 \cos \left (e x +d \right ) b^{3} c^{2}+\cos \left (e x +d \right ) b \,c^{4}+\sin \left (e x +d \right ) a^{4} c -2 \sin \left (e x +d \right ) a^{2} b^{2} c -2 \sin \left (e x +d \right ) a^{2} c^{3}+\sin \left (e x +d \right ) b^{4} c +2 \sin \left (e x +d \right ) b^{2} c^{3}+\sin \left (e x +d \right ) c^{5}+a^{5}-2 a^{3} b^{2}-2 a^{3} c^{2}+a \,b^{4}+2 a \,b^{2} c^{2}+a \,c^{4}\right )} \] Input:

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

Output:

(2*sqrt(a**2 - b**2 - c**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b 
+ c)/sqrt(a**2 - b**2 - c**2))*cos(d + e*x)*a*b*c + 2*sqrt(a**2 - b**2 - c 
**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b + c)/sqrt(a**2 - b**2 - 
 c**2))*sin(d + e*x)*a*c**2 + 2*sqrt(a**2 - b**2 - c**2)*atan((tan((d + e* 
x)/2)*a - tan((d + e*x)/2)*b + c)/sqrt(a**2 - b**2 - c**2))*a**2*c + cos(d 
 + e*x)*a**2*b**2 + cos(d + e*x)*a**2*c**2 - cos(d + e*x)*b**4 - 2*cos(d + 
 e*x)*b**2*c**2 - cos(d + e*x)*c**4 + a**3*b - a*b**3 - a*b*c**2)/(c*e*(co 
s(d + e*x)*a**4*b - 2*cos(d + e*x)*a**2*b**3 - 2*cos(d + e*x)*a**2*b*c**2 
+ cos(d + e*x)*b**5 + 2*cos(d + e*x)*b**3*c**2 + cos(d + e*x)*b*c**4 + sin 
(d + e*x)*a**4*c - 2*sin(d + e*x)*a**2*b**2*c - 2*sin(d + e*x)*a**2*c**3 + 
 sin(d + e*x)*b**4*c + 2*sin(d + e*x)*b**2*c**3 + sin(d + e*x)*c**5 + a**5 
 - 2*a**3*b**2 - 2*a**3*c**2 + a*b**4 + 2*a*b**2*c**2 + a*c**4))