\(\int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx\) [328]

Optimal result

Integrand size = 20, antiderivative size = 197 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\frac {\left (2 a^2+b^2+c^2\right ) \arctan \left (\frac {c+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2-c^2}}\right )}{\left (a^2-b^2-c^2\right )^{5/2} e}+\frac {c \cos (d+e x)-b \sin (d+e x)}{2 \left (a^2-b^2-c^2\right ) e (a+b \cos (d+e x)+c \sin (d+e x))^2}+\frac {3 (a c \cos (d+e x)-a b \sin (d+e x))}{2 \left (a^2-b^2-c^2\right )^2 e (a+b \cos (d+e x)+c \sin (d+e x))} \] Output:

(2*a^2+b^2+c^2)*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)^(5/2)/e+1/2*(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))^2+3/2*(a*c*cos(e*x+d)-a*b*sin(e*x+d))/(a^2-b^2-c^ 
2)^2/e/(a+b*cos(e*x+d)+c*sin(e*x+d))
 

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.02 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\frac {-\frac {2 \left (2 a^2+b^2+c^2\right ) \text {arctanh}\left (\frac {c+(a-b) \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2+c^2}}\right )}{\left (-a^2+b^2+c^2\right )^{5/2}}+\frac {a c+\left (b^2+c^2\right ) \sin (d+e x)}{b \left (-a^2+b^2+c^2\right ) (a+b \cos (d+e x)+c \sin (d+e x))^2}-\frac {c \left (2 a^2+b^2+c^2\right )+3 a \left (b^2+c^2\right ) \sin (d+e x)}{b \left (-a^2+b^2+c^2\right )^2 (a+b \cos (d+e x)+c \sin (d+e x))}}{2 e} \] Input:

Integrate[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-3),x]
 

Output:

((-2*(2*a^2 + b^2 + c^2)*ArcTanh[(c + (a - b)*Tan[(d + e*x)/2])/Sqrt[-a^2 
+ b^2 + c^2]])/(-a^2 + b^2 + c^2)^(5/2) + (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])^2) - (c*(2*a^ 
2 + b^2 + c^2) + 3*a*(b^2 + c^2)*Sin[d + e*x])/(b*(-a^2 + b^2 + c^2)^2*(a 
+ b*Cos[d + e*x] + c*Sin[d + e*x])))/(2*e)
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {3042, 3608, 25, 3042, 3632, 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.

\(\displaystyle \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3608

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3632

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3603

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(-3),x]
 

Output:

(c*Cos[d + e*x] - b*Sin[d + e*x])/(2*(a^2 - b^2 - c^2)*e*(a + b*Cos[d + e* 
x] + c*Sin[d + e*x])^2) + ((2*(2*a^2 + b^2 + c^2)*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) 
+ (3*(a*c*Cos[d + e*x] - a*b*Sin[d + e*x]))/((a^2 - b^2 - c^2)*e*(a + b*Co 
s[d + e*x] + c*Sin[d + e*x])))/(2*(a^2 - b^2 - c^2))
 

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 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]
 

rule 3632

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(610\) vs. \(2(188)=376\).

Time = 1.51 (sec) , antiderivative size = 611, normalized size of antiderivative = 3.10

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (4 a^{3} b -7 a^{2} b^{2}-5 a^{2} c^{2}+2 a \,b^{3}+2 a b \,c^{2}+b^{4}+3 b^{2} c^{2}+2 c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{\left (a -b \right ) \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right )}+\frac {c \left (4 a^{4}-12 a^{3} b +13 a^{2} b^{2}+7 a^{2} c^{2}-6 a \,b^{3}-6 a b \,c^{2}+b^{4}-b^{2} c^{2}-2 c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}-\frac {\left (4 a^{4} b -5 b^{2} a^{3}-11 a^{3} c^{2}-3 a^{2} b^{3}+3 a^{2} b \,c^{2}+5 b^{4} a +7 a \,b^{2} c^{2}+2 a \,c^{4}-b^{5}+b^{3} c^{2}+2 b \,c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}+\frac {c \left (4 a^{4}-3 a^{2} b^{2}-a^{2} c^{2}-b^{4}-b^{2} c^{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} b +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}+c^{2}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}}{e}\)	\(611\)
default	\(\frac {\frac {-\frac {\left (4 a^{3} b -7 a^{2} b^{2}-5 a^{2} c^{2}+2 a \,b^{3}+2 a b \,c^{2}+b^{4}+3 b^{2} c^{2}+2 c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{\left (a -b \right ) \left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right )}+\frac {c \left (4 a^{4}-12 a^{3} b +13 a^{2} b^{2}+7 a^{2} c^{2}-6 a \,b^{3}-6 a b \,c^{2}+b^{4}-b^{2} c^{2}-2 c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}-\frac {\left (4 a^{4} b -5 b^{2} a^{3}-11 a^{3} c^{2}-3 a^{2} b^{3}+3 a^{2} b \,c^{2}+5 b^{4} a +7 a \,b^{2} c^{2}+2 a \,c^{4}-b^{5}+b^{3} c^{2}+2 b \,c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}+\frac {c \left (4 a^{4}-3 a^{2} b^{2}-a^{2} c^{2}-b^{4}-b^{2} c^{2}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \left (a^{2}-2 b a +b^{2}\right )}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a -\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} b +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a +b \right )^{2}}+\frac {\left (2 a^{2}+b^{2}+c^{2}\right ) \arctan \left (\frac {2 \left (a -b \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-b^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}-2 a^{2} c^{2}+b^{4}+2 b^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-b^{2}-c^{2}}}}{e}\)	\(611\)
risch	\(\text {Expression too large to display}\)	\(1181\)

Input:

int(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^3,x,method=_RETURNVERBOSE)
 

Output:

1/e*(2*(-1/2*(4*a^3*b-7*a^2*b^2-5*a^2*c^2+2*a*b^3+2*a*b*c^2+b^4+3*b^2*c^2+ 
2*c^4)/(a-b)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)*tan(1/2*e*x+1/2*d 
)^3+1/2*c*(4*a^4-12*a^3*b+13*a^2*b^2+7*a^2*c^2-6*a*b^3-6*a*b*c^2+b^4-b^2*c 
^2-2*c^4)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-2*a*b+b^2)*tan( 
1/2*e*x+1/2*d)^2-1/2*(4*a^4*b-5*a^3*b^2-11*a^3*c^2-3*a^2*b^3+3*a^2*b*c^2+5 
*a*b^4+7*a*b^2*c^2+2*a*c^4-b^5+b^3*c^2+2*b*c^4)/(a^4-2*a^2*b^2-2*a^2*c^2+b 
^4+2*b^2*c^2+c^4)/(a^2-2*a*b+b^2)*tan(1/2*e*x+1/2*d)+1/2*c*(4*a^4-3*a^2*b^ 
2-a^2*c^2-b^4-b^2*c^2)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/(a^2-2* 
a*b+b^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+(2*a^2+b^2+c^2)/(a^4-2*a^2*b^2-2*a^2*c^2+b^4+2*b^2*c^2+c^4)/( 
a^2-b^2-c^2)^(1/2)*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. 878 vs. \(2 (188) = 376\).

Time = 0.17 (sec) , antiderivative size = 1947, normalized size of antiderivative = 9.88 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^3,x, algorithm="fricas")
 

Output:

[1/4*(6*a*b*c^3 - 12*(a*b*c^3 - (a^3*b - a*b^3)*c)*cos(e*x + d)^2 - (2*a^4 
 + a^2*b^2 + c^4 + (3*a^2 + b^2)*c^2 + (2*a^2*b^2 + b^4 - 2*a^2*c^2 - c^4) 
*cos(e*x + d)^2 + 2*(2*a^3*b + a*b^3 + a*b*c^2)*cos(e*x + d) + 2*(a*c^3 + 
(2*a^3 + a*b^2)*c + (b*c^3 + (2*a^2*b + b^3)*c)*cos(e*x + d))*sin(e*x + d) 
)*sqrt(-a^2 + b^2 + c^2)*log(-(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*cos(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))) - 6*(a^3*b - a*b^3)*c + 
2*(c^5 - (5*a^2 - 2*b^2)*c^3 + (4*a^4 - 5*a^2*b^2 + b^4)*c)*cos(e*x + d) - 
 2*(4*a^4*b - 5*a^2*b^3 + b^5 + b*c^4 - (5*a^2*b - 2*b^3)*c^2 + 3*(a^3*b^2 
 - a*b^4 - a^3*c^2 + a*c^4)*cos(e*x + d))*sin(e*x + d))/((a^6*b^2 - 3*a^4* 
b^4 + 3*a^2*b^6 - b^8 + c^8 - (3*a^2 - 2*b^2)*c^6 + 3*(a^4 - a^2*b^2)*c^4 
- (a^6 - 3*a^2*b^4 + 2*b^6)*c^2)*e*cos(e*x + d)^2 + 2*(a^7*b - 3*a^5*b^3 + 
 3*a^3*b^5 - a*b^7 - a*b*c^6 + 3*(a^3*b - a*b^3)*c^4 - 3*(a^5*b - 2*a^3*b^ 
3 + a*b^5)*c^2)*e*cos(e*x + d) + (a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6 - 
c^8 + (2*a^2 - 3*b^2)*c^6 + 3*(a^2*b^2 - b^4)*c^4 - (2*a^6 - 3*a^4*b^2 + b 
^6)*c^2)*e - 2*((b*c^7 - 3*(a^2*b - b^3)*c^5 + 3*(a^4*b - 2*a^2*b^3 + b...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^3,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 856 vs. \(2 (188) = 376\).

Time = 0.16 (sec) , antiderivative size = 856, normalized size of antiderivative = 4.35 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^3,x, algorithm="giac")
 

Output:

-((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)))*(2*a^2 + 
 b^2 + c^2)/((a^4 - 2*a^2*b^2 + b^4 - 2*a^2*c^2 + 2*b^2*c^2 + c^4)*sqrt(a^ 
2 - b^2 - c^2)) + (4*a^4*b*tan(1/2*e*x + 1/2*d)^3 - 11*a^3*b^2*tan(1/2*e*x 
 + 1/2*d)^3 + 9*a^2*b^3*tan(1/2*e*x + 1/2*d)^3 - a*b^4*tan(1/2*e*x + 1/2*d 
)^3 - b^5*tan(1/2*e*x + 1/2*d)^3 - 5*a^3*c^2*tan(1/2*e*x + 1/2*d)^3 + 7*a^ 
2*b*c^2*tan(1/2*e*x + 1/2*d)^3 + a*b^2*c^2*tan(1/2*e*x + 1/2*d)^3 - 3*b^3* 
c^2*tan(1/2*e*x + 1/2*d)^3 + 2*a*c^4*tan(1/2*e*x + 1/2*d)^3 - 2*b*c^4*tan( 
1/2*e*x + 1/2*d)^3 - 4*a^4*c*tan(1/2*e*x + 1/2*d)^2 + 12*a^3*b*c*tan(1/2*e 
*x + 1/2*d)^2 - 13*a^2*b^2*c*tan(1/2*e*x + 1/2*d)^2 + 6*a*b^3*c*tan(1/2*e* 
x + 1/2*d)^2 - b^4*c*tan(1/2*e*x + 1/2*d)^2 - 7*a^2*c^3*tan(1/2*e*x + 1/2* 
d)^2 + 6*a*b*c^3*tan(1/2*e*x + 1/2*d)^2 + b^2*c^3*tan(1/2*e*x + 1/2*d)^2 + 
 2*c^5*tan(1/2*e*x + 1/2*d)^2 + 4*a^4*b*tan(1/2*e*x + 1/2*d) - 5*a^3*b^2*t 
an(1/2*e*x + 1/2*d) - 3*a^2*b^3*tan(1/2*e*x + 1/2*d) + 5*a*b^4*tan(1/2*e*x 
 + 1/2*d) - b^5*tan(1/2*e*x + 1/2*d) - 11*a^3*c^2*tan(1/2*e*x + 1/2*d) + 3 
*a^2*b*c^2*tan(1/2*e*x + 1/2*d) + 7*a*b^2*c^2*tan(1/2*e*x + 1/2*d) + b^3*c 
^2*tan(1/2*e*x + 1/2*d) + 2*a*c^4*tan(1/2*e*x + 1/2*d) + 2*b*c^4*tan(1/2*e 
*x + 1/2*d) - 4*a^4*c + 3*a^2*b^2*c + b^4*c + a^2*c^3 + b^2*c^3)/((a^6 - 2 
*a^5*b - a^4*b^2 + 4*a^3*b^3 - a^2*b^4 - 2*a*b^5 + b^6 - 2*a^4*c^2 + 4*a^3 
*b*c^2 - 4*a*b^3*c^2 + 2*b^4*c^2 + a^2*c^4 - 2*a*b*c^4 + b^2*c^4)*(a*ta...
 

Mupad [B] (verification not implemented)

Time = 18.91 (sec) , antiderivative size = 700, normalized size of antiderivative = 3.55 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx=-\frac {\frac {-4\,a^4\,c+3\,a^2\,b^2\,c+a^2\,c^3+b^4\,c+b^2\,c^3}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (4\,a^4\,b-5\,a^3\,b^2-11\,a^3\,c^2-3\,a^2\,b^3+3\,a^2\,b\,c^2+5\,a\,b^4+7\,a\,b^2\,c^2+2\,a\,c^4-b^5+b^3\,c^2+2\,b\,c^4\right )}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (-4\,a^4\,c+12\,a^3\,b\,c-13\,a^2\,b^2\,c-7\,a^2\,c^3+6\,a\,b^3\,c+6\,a\,b\,c^3-b^4\,c+b^2\,c^3+2\,c^5\right )}{{\left (a-b\right )}^2\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (4\,a^3\,b-7\,a^2\,b^2-5\,a^2\,c^2+2\,a\,b^3+2\,a\,b\,c^2+b^4+3\,b^2\,c^2+2\,c^4\right )}{\left (a-b\right )\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}}{e\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (4\,a\,c-4\,b\,c\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2+4\,c^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2+\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (4\,a\,c+4\,b\,c\right )\right )}-\frac {\mathrm {atanh}\left (\frac {a^4\,c-2\,a^2\,b^2\,c-2\,a^2\,c^3+b^4\,c+2\,b^2\,c^3+c^5}{{\left (-a^2+b^2+c^2\right )}^{5/2}}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^4-2\,a^2\,b^2-2\,a^2\,c^2+b^4+2\,b^2\,c^2+c^4\right )}{2\,{\left (-a^2+b^2+c^2\right )}^{5/2}}\right )\,\left (2\,a^2+b^2+c^2\right )}{e\,{\left (-a^2+b^2+c^2\right )}^{5/2}} \] Input:

int(1/(a + b*cos(d + e*x) + c*sin(d + e*x))^3,x)
 

Output:

- ((b^4*c - 4*a^4*c + a^2*c^3 + b^2*c^3 + 3*a^2*b^2*c)/((a - b)^2*(a^4 + b 
^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)) + (tan(d/2 + (e*x)/2)*(5*a* 
b^4 + 4*a^4*b + 2*a*c^4 + 2*b*c^4 - b^5 - 3*a^2*b^3 - 5*a^3*b^2 - 11*a^3*c 
^2 + b^3*c^2 + 7*a*b^2*c^2 + 3*a^2*b*c^2))/((a - b)^2*(a^4 + b^4 + c^4 - 2 
*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^2)) + (tan(d/2 + (e*x)/2)^2*(2*c^5 - b^4*c 
- 4*a^4*c - 7*a^2*c^3 + b^2*c^3 - 13*a^2*b^2*c + 6*a*b*c^3 + 6*a*b^3*c + 1 
2*a^3*b*c))/((a - b)^2*(a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*a^2*c^2 + 2*b^2*c^ 
2)) + (tan(d/2 + (e*x)/2)^3*(2*a*b^3 + 4*a^3*b + b^4 + 2*c^4 - 7*a^2*b^2 - 
 5*a^2*c^2 + 3*b^2*c^2 + 2*a*b*c^2))/((a - b)*(a^4 + b^4 + c^4 - 2*a^2*b^2 
 - 2*a^2*c^2 + 2*b^2*c^2)))/(e*(2*a*b + tan(d/2 + (e*x)/2)^3*(4*a*c - 4*b* 
c) + tan(d/2 + (e*x)/2)^2*(2*a^2 - 2*b^2 + 4*c^2) + tan(d/2 + (e*x)/2)^4*( 
a^2 - 2*a*b + b^2) + a^2 + b^2 + tan(d/2 + (e*x)/2)*(4*a*c + 4*b*c))) - (a 
tanh((a^4*c + b^4*c + c^5 - 2*a^2*c^3 + 2*b^2*c^3 - 2*a^2*b^2*c)/(b^2 - a^ 
2 + c^2)^(5/2) + (tan(d/2 + (e*x)/2)*(2*a - 2*b)*(a^4 + b^4 + c^4 - 2*a^2* 
b^2 - 2*a^2*c^2 + 2*b^2*c^2))/(2*(b^2 - a^2 + c^2)^(5/2)))*(2*a^2 + b^2 + 
c^2))/(e*(b^2 - a^2 + c^2)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 5132, normalized size of antiderivative = 26.05 \[ \int \frac {1}{(a+b \cos (d+e x)+c \sin (d+e x))^3} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*cos(e*x+d)+c*sin(e*x+d))^3,x)
 

Output:

(16*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)*sin(d + e*x)*a**3*b*c**2 - 16 
*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)*sin(d + e*x)*a**2*b**2*c**2 + 8* 
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)*sin(d + e*x)*a*b**3*c**2 + 8*sqrt 
(a**2 - b**2 - c**2)*atan((tan((d + e*x)/2)*a - tan((d + e*x)/2)*b + c)/sq 
rt(a**2 - b**2 - c**2))*cos(d + e*x)*sin(d + e*x)*a*b*c**4 - 8*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)*sin(d + e*x)*b**4*c**2 - 8*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)*sin(d + e*x)*b**2*c**4 + 16*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**4*b*c - 16*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**3*b**2*c + 8*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**2*b**3*c + 8*s 
qrt(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**2*b*c**3 - 8*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...