\(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx\) [507]

Optimal result

Integrand size = 31, antiderivative size = 185 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx=\frac {\left (2 a^2 A+A c^2-3 a c C\right ) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{\left (a^2-c^2\right )^{5/2} e}-\frac {B}{2 c e (a+c \sin (d+e x))^2}+\frac {(A c-a C) \cos (d+e x)}{2 \left (a^2-c^2\right ) e (a+c \sin (d+e x))^2}+\frac {\left (3 a A c-a^2 C-2 c^2 C\right ) \cos (d+e x)}{2 \left (a^2-c^2\right )^2 e (a+c \sin (d+e x))} \] Output:

(2*A*a^2+A*c^2-3*C*a*c)*arctan((c+a*tan(1/2*e*x+1/2*d))/(a^2-c^2)^(1/2))/( 
a^2-c^2)^(5/2)/e-1/2*B/c/e/(a+c*sin(e*x+d))^2+1/2*(A*c-C*a)*cos(e*x+d)/(a^ 
2-c^2)/e/(a+c*sin(e*x+d))^2+1/2*(3*A*a*c-C*a^2-2*C*c^2)*cos(e*x+d)/(a^2-c^ 
2)^2/e/(a+c*sin(e*x+d))
 

Mathematica [A] (verified)

Time = 1.14 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.94 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx=\frac {\frac {2 \left (2 a^2 A+A c^2-3 a c C\right ) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{\left (a^2-c^2\right )^{5/2}}+\frac {B \left (-a^2+c^2\right )+c (A c-a C) \cos (d+e x)}{(a-c) c (a+c) (a+c \sin (d+e x))^2}-\frac {\left (-3 a A c+a^2 C+2 c^2 C\right ) \cos (d+e x)}{(a-c)^2 (a+c)^2 (a+c \sin (d+e x))}}{2 e} \] Input:

Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^3,x]
 

Output:

((2*(2*a^2*A + A*c^2 - 3*a*c*C)*ArcTan[(c + a*Tan[(d + e*x)/2])/Sqrt[a^2 - 
 c^2]])/(a^2 - c^2)^(5/2) + (B*(-a^2 + c^2) + c*(A*c - a*C)*Cos[d + e*x])/ 
((a - c)*c*(a + c)*(a + c*Sin[d + e*x])^2) - ((-3*a*A*c + a^2*C + 2*c^2*C) 
*Cos[d + e*x])/((a - c)^2*(a + c)^2*(a + c*Sin[d + e*x])))/(2*e)
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.11, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4876, 3042, 3147, 17, 3233, 25, 3042, 3233, 25, 27, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3}dx\)

\(\Big \downarrow \) 4876

\(\displaystyle \int \frac {A+C \sin (d+e x)}{(a+c \sin (d+e x))^3}dx+B \int \frac {\cos (d+e x)}{(a+c \sin (d+e x))^3}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin (d+e x)}{(a+c \sin (d+e x))^3}dx+B \int \frac {\cos (d+e x)}{(a+c \sin (d+e x))^3}dx\)

\(\Big \downarrow \) 3147

\(\displaystyle \int \frac {A+C \sin (d+e x)}{(a+c \sin (d+e x))^3}dx+\frac {B \int \frac {1}{(a+c \sin (d+e x))^3}d(c \sin (d+e x))}{c e}\)

\(\Big \downarrow \) 17

\(\displaystyle \int \frac {A+C \sin (d+e x)}{(a+c \sin (d+e x))^3}dx-\frac {B}{2 c e (a+c \sin (d+e x))^2}\)

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^3,x]
 

Output:

-1/2*B/(c*e*(a + c*Sin[d + e*x])^2) + ((A*c - a*C)*Cos[d + e*x])/(2*(a^2 - 
 c^2)*e*(a + c*Sin[d + e*x])^2) + ((2*(2*a^2*A + A*c^2 - 3*a*c*C)*ArcTan[( 
2*c + 2*a*Tan[(d + e*x)/2])/(2*Sqrt[a^2 - c^2])])/((a^2 - c^2)^(3/2)*e) + 
((3*a*A*c - a^2*C - 2*c^2*C)*Cos[d + e*x])/((a^2 - c^2)*e*(a + c*Sin[d + e 
*x])))/(2*(a^2 - c^2))
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

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 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 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

rule 3233

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

rule 4876

Int[(u_)*((v_) + (d_.)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_.)), x_Symbol] : 
> With[{e = FreeFactors[Sin[c*(a + b*x)], x]}, Int[ActivateTrig[u*v], x] + 
Simp[d   Int[ActivateTrig[u]*Cos[c*(a + b*x)]^n, x], x] /; FunctionOfQ[Sin[ 
c*(a + b*x)]/e, u, x]] /; FreeQ[{a, b, c, d}, x] &&  !FreeQ[v, x] && Intege 
rQ[(n - 1)/2] && NonsumQ[u] && (EqQ[F, Cos] || EqQ[F, cos])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(372\) vs. \(2(174)=348\).

Time = 0.62 (sec) , antiderivative size = 373, normalized size of antiderivative = 2.02

method	result	size
parts	\(\frac {\frac {\frac {c \left (5 A \,a^{2} c -2 c^{3} A -3 C \,a^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{a \left (a^{4}-2 a^{2} c^{2}+c^{4}\right )}+\frac {\left (4 A \,a^{4} c +7 A \,a^{2} c^{3}-2 A \,c^{5}-2 C \,a^{5}-5 C \,a^{3} c^{2}-2 C a \,c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) a^{2}}+\frac {c \left (11 A \,a^{2} c -2 c^{3} A -5 C \,a^{3}-4 C a \,c^{2}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a \left (a^{4}-2 a^{2} c^{2}+c^{4}\right )}+\frac {2 \left (4 A \,a^{2} c -c^{3} A -2 C \,a^{3}-C a \,c^{2}\right )}{2 a^{4}-4 a^{2} c^{2}+2 c^{4}}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2} A +A \,c^{2}-3 a c C \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-c^{2}}}}{e}-\frac {B}{2 c e \left (a +c \sin \left (e x +d \right )\right )^{2}}\)	\(373\)
derivativedivides	\(\frac {\frac {\frac {\left (5 A \,a^{2} c^{2}-2 A \,c^{4}+2 B \,a^{4}-4 B \,a^{2} c^{2}+2 B \,c^{4}-3 C \,a^{3} c \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) a}+\frac {\left (4 A \,a^{4} c +7 A \,a^{2} c^{3}-2 A \,c^{5}+2 B \,a^{4} c -4 B \,a^{2} c^{3}+2 B \,c^{5}-2 C \,a^{5}-5 C \,a^{3} c^{2}-2 C a \,c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) a^{2}}+\frac {\left (11 A \,a^{2} c^{2}-2 A \,c^{4}+2 B \,a^{4}-4 B \,a^{2} c^{2}+2 B \,c^{4}-5 C \,a^{3} c -4 C a \,c^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a \left (a^{4}-2 a^{2} c^{2}+c^{4}\right )}+\frac {2 \left (4 A \,a^{2} c -c^{3} A -2 C \,a^{3}-C a \,c^{2}\right )}{2 a^{4}-4 a^{2} c^{2}+2 c^{4}}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2} A +A \,c^{2}-3 a c C \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-c^{2}}}}{e}\)	\(419\)
default	\(\frac {\frac {\frac {\left (5 A \,a^{2} c^{2}-2 A \,c^{4}+2 B \,a^{4}-4 B \,a^{2} c^{2}+2 B \,c^{4}-3 C \,a^{3} c \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) a}+\frac {\left (4 A \,a^{4} c +7 A \,a^{2} c^{3}-2 A \,c^{5}+2 B \,a^{4} c -4 B \,a^{2} c^{3}+2 B \,c^{5}-2 C \,a^{5}-5 C \,a^{3} c^{2}-2 C a \,c^{4}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) a^{2}}+\frac {\left (11 A \,a^{2} c^{2}-2 A \,c^{4}+2 B \,a^{4}-4 B \,a^{2} c^{2}+2 B \,c^{4}-5 C \,a^{3} c -4 C a \,c^{3}\right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a \left (a^{4}-2 a^{2} c^{2}+c^{4}\right )}+\frac {2 \left (4 A \,a^{2} c -c^{3} A -2 C \,a^{3}-C a \,c^{2}\right )}{2 a^{4}-4 a^{2} c^{2}+2 c^{4}}}{\left (\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2} A +A \,c^{2}-3 a c C \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{4}-2 a^{2} c^{2}+c^{4}\right ) \sqrt {a^{2}-c^{2}}}}{e}\)	\(419\)
risch	\(-\frac {-3 i A a \,c^{3}+3 i c^{3} A a \,{\mathrm e}^{2 i \left (e x +d \right )}+2 c^{2} A \,a^{2} {\mathrm e}^{3 i \left (e x +d \right )}+c^{4} A \,{\mathrm e}^{3 i \left (e x +d \right )}-2 i C \,a^{4} {\mathrm e}^{2 i \left (e x +d \right )}-2 i c^{4} C \,{\mathrm e}^{2 i \left (e x +d \right )}+6 i c A \,a^{3} {\mathrm e}^{2 i \left (e x +d \right )}-3 c^{3} C a \,{\mathrm e}^{3 i \left (e x +d \right )}+2 B \,a^{4} {\mathrm e}^{2 i \left (e x +d \right )}-4 c^{2} B \,a^{2} {\mathrm e}^{2 i \left (e x +d \right )}+2 c^{4} B \,{\mathrm e}^{2 i \left (e x +d \right )}+2 i C \,c^{4}-10 A \,a^{2} c^{2} {\mathrm e}^{i \left (e x +d \right )}+A \,c^{4} {\mathrm e}^{i \left (e x +d \right )}-5 i c^{2} C \,a^{2} {\mathrm e}^{2 i \left (e x +d \right )}+i C \,a^{2} c^{2}+4 C \,a^{3} c \,{\mathrm e}^{i \left (e x +d \right )}+5 C a \,c^{3} {\mathrm e}^{i \left (e x +d \right )}}{\left (-i c \,{\mathrm e}^{2 i \left (e x +d \right )}+i c +2 \,{\mathrm e}^{i \left (e x +d \right )} a \right )^{2} \left (a^{2}-c^{2}\right )^{2} e c}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a -a^{2}+c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) a^{2} A}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a -a^{2}+c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) A \,c^{2}}{2 \sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}+\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a -a^{2}+c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) a c C}{2 \sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a +a^{2}-c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) a^{2} A}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a +a^{2}-c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) A \,c^{2}}{2 \sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}-\frac {3 \ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a +a^{2}-c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) a c C}{2 \sqrt {-a^{2}+c^{2}}\, \left (a +c \right )^{2} \left (a -c \right )^{2} e}\)	\(813\)

Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x,method=_RETURNVERBO 
SE)
 

Output:

1/e*(2*(1/2*c*(5*A*a^2*c-2*A*c^3-3*C*a^3)/a/(a^4-2*a^2*c^2+c^4)*tan(1/2*e* 
x+1/2*d)^3+1/2*(4*A*a^4*c+7*A*a^2*c^3-2*A*c^5-2*C*a^5-5*C*a^3*c^2-2*C*a*c^ 
4)/(a^4-2*a^2*c^2+c^4)/a^2*tan(1/2*e*x+1/2*d)^2+1/2*c*(11*A*a^2*c-2*A*c^3- 
5*C*a^3-4*C*a*c^2)/a/(a^4-2*a^2*c^2+c^4)*tan(1/2*e*x+1/2*d)+1/2*(4*A*a^2*c 
-A*c^3-2*C*a^3-C*a*c^2)/(a^4-2*a^2*c^2+c^4))/(tan(1/2*e*x+1/2*d)^2*a+2*c*t 
an(1/2*e*x+1/2*d)+a)^2+(2*A*a^2+A*c^2-3*C*a*c)/(a^4-2*a^2*c^2+c^4)/(a^2-c^ 
2)^(1/2)*arctan(1/2*(2*a*tan(1/2*e*x+1/2*d)+2*c)/(a^2-c^2)^(1/2)))-1/2*B/c 
/e/(a+c*sin(e*x+d))^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 397 vs. \(2 (173) = 346\).

Time = 0.11 (sec) , antiderivative size = 880, normalized size of antiderivative = 4.76 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x, algorithm="f 
ricas")
 

Output:

[1/4*(2*B*a^6 - 6*B*a^4*c^2 + 6*B*a^2*c^4 - 2*B*c^6 + 2*(C*a^4*c^2 - 3*A*a 
^3*c^3 + C*a^2*c^4 + 3*A*a*c^5 - 2*C*c^6)*cos(e*x + d)*sin(e*x + d) + (2*A 
*a^4*c - 3*C*a^3*c^2 + 3*A*a^2*c^3 - 3*C*a*c^4 + A*c^5 - (2*A*a^2*c^3 - 3* 
C*a*c^4 + A*c^5)*cos(e*x + d)^2 + 2*(2*A*a^3*c^2 - 3*C*a^2*c^3 + A*a*c^4)* 
sin(e*x + d))*sqrt(-a^2 + c^2)*log(((2*a^2 - c^2)*cos(e*x + d)^2 - 2*a*c*s 
in(e*x + d) - a^2 - c^2 + 2*(a*cos(e*x + d)*sin(e*x + d) + c*cos(e*x + d)) 
*sqrt(-a^2 + c^2))/(c^2*cos(e*x + d)^2 - 2*a*c*sin(e*x + d) - a^2 - c^2)) 
+ 2*(2*C*a^5*c - 4*A*a^4*c^2 - C*a^3*c^3 + 5*A*a^2*c^4 - C*a*c^5 - A*c^6)* 
cos(e*x + d))/((a^6*c^3 - 3*a^4*c^5 + 3*a^2*c^7 - c^9)*e*cos(e*x + d)^2 - 
2*(a^7*c^2 - 3*a^5*c^4 + 3*a^3*c^6 - a*c^8)*e*sin(e*x + d) - (a^8*c - 2*a^ 
6*c^3 + 2*a^2*c^7 - c^9)*e), 1/2*(B*a^6 - 3*B*a^4*c^2 + 3*B*a^2*c^4 - B*c^ 
6 + (C*a^4*c^2 - 3*A*a^3*c^3 + C*a^2*c^4 + 3*A*a*c^5 - 2*C*c^6)*cos(e*x + 
d)*sin(e*x + d) + (2*A*a^4*c - 3*C*a^3*c^2 + 3*A*a^2*c^3 - 3*C*a*c^4 + A*c 
^5 - (2*A*a^2*c^3 - 3*C*a*c^4 + A*c^5)*cos(e*x + d)^2 + 2*(2*A*a^3*c^2 - 3 
*C*a^2*c^3 + A*a*c^4)*sin(e*x + d))*sqrt(a^2 - c^2)*arctan(-(a*sin(e*x + d 
) + c)/(sqrt(a^2 - c^2)*cos(e*x + d))) + (2*C*a^5*c - 4*A*a^4*c^2 - C*a^3* 
c^3 + 5*A*a^2*c^4 - C*a*c^5 - A*c^6)*cos(e*x + d))/((a^6*c^3 - 3*a^4*c^5 + 
 3*a^2*c^7 - c^9)*e*cos(e*x + d)^2 - 2*(a^7*c^2 - 3*a^5*c^4 + 3*a^3*c^6 - 
a*c^8)*e*sin(e*x + d) - (a^8*c - 2*a^6*c^3 + 2*a^2*c^7 - c^9)*e)]
 

Sympy [F(-1)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))**3,x)
                                                                                    
                                                                                    
 

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,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*c^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. 571 vs. \(2 (173) = 346\).

Time = 0.31 (sec) , antiderivative size = 571, normalized size of antiderivative = 3.09 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x, algorithm="g 
iac")
 

Output:

((2*A*a^2 - 3*C*a*c + A*c^2)*(pi*floor(1/2*(e*x + d)/pi + 1/2)*sgn(a) + ar 
ctan((a*tan(1/2*e*x + 1/2*d) + c)/sqrt(a^2 - c^2)))/((a^4 - 2*a^2*c^2 + c^ 
4)*sqrt(a^2 - c^2)) + (2*B*a^5*tan(1/2*e*x + 1/2*d)^3 - 3*C*a^4*c*tan(1/2* 
e*x + 1/2*d)^3 + 5*A*a^3*c^2*tan(1/2*e*x + 1/2*d)^3 - 4*B*a^3*c^2*tan(1/2* 
e*x + 1/2*d)^3 - 2*A*a*c^4*tan(1/2*e*x + 1/2*d)^3 + 2*B*a*c^4*tan(1/2*e*x 
+ 1/2*d)^3 - 2*C*a^5*tan(1/2*e*x + 1/2*d)^2 + 4*A*a^4*c*tan(1/2*e*x + 1/2* 
d)^2 + 2*B*a^4*c*tan(1/2*e*x + 1/2*d)^2 - 5*C*a^3*c^2*tan(1/2*e*x + 1/2*d) 
^2 + 7*A*a^2*c^3*tan(1/2*e*x + 1/2*d)^2 - 4*B*a^2*c^3*tan(1/2*e*x + 1/2*d) 
^2 - 2*C*a*c^4*tan(1/2*e*x + 1/2*d)^2 - 2*A*c^5*tan(1/2*e*x + 1/2*d)^2 + 2 
*B*c^5*tan(1/2*e*x + 1/2*d)^2 + 2*B*a^5*tan(1/2*e*x + 1/2*d) - 5*C*a^4*c*t 
an(1/2*e*x + 1/2*d) + 11*A*a^3*c^2*tan(1/2*e*x + 1/2*d) - 4*B*a^3*c^2*tan( 
1/2*e*x + 1/2*d) - 4*C*a^2*c^3*tan(1/2*e*x + 1/2*d) - 2*A*a*c^4*tan(1/2*e* 
x + 1/2*d) + 2*B*a*c^4*tan(1/2*e*x + 1/2*d) - 2*C*a^5 + 4*A*a^4*c - C*a^3* 
c^2 - A*a^2*c^3)/((a^6 - 2*a^4*c^2 + a^2*c^4)*(a*tan(1/2*e*x + 1/2*d)^2 + 
2*c*tan(1/2*e*x + 1/2*d) + a)^2))/e
 

Mupad [B] (verification not implemented)

Time = 17.91 (sec) , antiderivative size = 557, normalized size of antiderivative = 3.01 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )\,\left (2\,a^4\,c-4\,a^2\,c^3+2\,c^5\right )}{2\,{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )}{{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,c^2+c^4\right )}{2\,A\,a^2-3\,C\,a\,c+A\,c^2}\right )\,\left (2\,A\,a^2-3\,C\,a\,c+A\,c^2\right )}{e\,{\left (a+c\right )}^{5/2}\,{\left (a-c\right )}^{5/2}}-\frac {\frac {2\,C\,a^3-4\,A\,a^2\,c+C\,a\,c^2+A\,c^3}{a^4-2\,a^2\,c^2+c^4}-\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (2\,B\,a^4-2\,A\,c^4+2\,B\,c^4+5\,A\,a^2\,c^2-4\,B\,a^2\,c^2-3\,C\,a^3\,c\right )}{a\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (2\,A\,c^4-2\,B\,a^4-2\,B\,c^4-11\,A\,a^2\,c^2+4\,B\,a^2\,c^2+4\,C\,a\,c^3+5\,C\,a^3\,c\right )}{a\,\left (a^4-2\,a^2\,c^2+c^4\right )}+\frac {{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,A\,c^5+2\,C\,a^5-2\,B\,c^5-7\,A\,a^2\,c^3+4\,B\,a^2\,c^3+5\,C\,a^3\,c^2-4\,A\,a^4\,c-2\,B\,a^4\,c+2\,C\,a\,c^4\right )}{a^2\,\left (a^4-2\,a^2\,c^2+c^4\right )}}{e\,\left ({\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (2\,a^2+4\,c^2\right )+a^2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4+a^2+4\,a\,c\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3+4\,a\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )} \] Input:

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + c*sin(d + e*x))^3,x)
 

Output:

(atan(((((2*A*a^2 + A*c^2 - 3*C*a*c)*(2*a^4*c + 2*c^5 - 4*a^2*c^3))/(2*(a 
+ c)^(5/2)*(a - c)^(5/2)*(a^4 + c^4 - 2*a^2*c^2)) + (a*tan(d/2 + (e*x)/2)* 
(2*A*a^2 + A*c^2 - 3*C*a*c))/((a + c)^(5/2)*(a - c)^(5/2)))*(a^4 + c^4 - 2 
*a^2*c^2))/(2*A*a^2 + A*c^2 - 3*C*a*c))*(2*A*a^2 + A*c^2 - 3*C*a*c))/(e*(a 
 + c)^(5/2)*(a - c)^(5/2)) - ((A*c^3 + 2*C*a^3 - 4*A*a^2*c + C*a*c^2)/(a^4 
 + c^4 - 2*a^2*c^2) - (tan(d/2 + (e*x)/2)^3*(2*B*a^4 - 2*A*c^4 + 2*B*c^4 + 
 5*A*a^2*c^2 - 4*B*a^2*c^2 - 3*C*a^3*c))/(a*(a^4 + c^4 - 2*a^2*c^2)) + (ta 
n(d/2 + (e*x)/2)*(2*A*c^4 - 2*B*a^4 - 2*B*c^4 - 11*A*a^2*c^2 + 4*B*a^2*c^2 
 + 4*C*a*c^3 + 5*C*a^3*c))/(a*(a^4 + c^4 - 2*a^2*c^2)) + (tan(d/2 + (e*x)/ 
2)^2*(2*A*c^5 + 2*C*a^5 - 2*B*c^5 - 7*A*a^2*c^3 + 4*B*a^2*c^3 + 5*C*a^3*c^ 
2 - 4*A*a^4*c - 2*B*a^4*c + 2*C*a*c^4))/(a^2*(a^4 + c^4 - 2*a^2*c^2)))/(e* 
(tan(d/2 + (e*x)/2)^2*(2*a^2 + 4*c^2) + a^2*tan(d/2 + (e*x)/2)^4 + a^2 + 4 
*a*c*tan(d/2 + (e*x)/2)^3 + 4*a*c*tan(d/2 + (e*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 418, normalized size of antiderivative = 2.26 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^3} \, dx=\frac {4 \sqrt {a^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a +c}{\sqrt {a^{2}-c^{2}}}\right ) \sin \left (e x +d \right )^{2} a^{2} c^{3}+8 \sqrt {a^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a +c}{\sqrt {a^{2}-c^{2}}}\right ) \sin \left (e x +d \right ) a^{3} c^{2}+4 \sqrt {a^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a +c}{\sqrt {a^{2}-c^{2}}}\right ) a^{4} c +2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3} c^{3}-2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,c^{5}+2 \cos \left (e x +d \right ) a^{4} c^{2}-2 \cos \left (e x +d \right ) a^{2} c^{4}+\sin \left (e x +d \right )^{2} a^{2} c^{4}-\sin \left (e x +d \right )^{2} c^{6}+2 \sin \left (e x +d \right ) a^{3} c^{3}-2 \sin \left (e x +d \right ) a \,c^{5}-a^{5} b +a^{4} c^{2}+2 a^{3} b \,c^{2}-a^{2} c^{4}-a b \,c^{4}}{2 a c e \left (\sin \left (e x +d \right )^{2} a^{4} c^{2}-2 \sin \left (e x +d \right )^{2} a^{2} c^{4}+\sin \left (e x +d \right )^{2} c^{6}+2 \sin \left (e x +d \right ) a^{5} c -4 \sin \left (e x +d \right ) a^{3} c^{3}+2 \sin \left (e x +d \right ) a \,c^{5}+a^{6}-2 a^{4} c^{2}+a^{2} c^{4}\right )} \] Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^3,x)
 

Output:

(4*sqrt(a**2 - c**2)*atan((tan((d + e*x)/2)*a + c)/sqrt(a**2 - c**2))*sin( 
d + e*x)**2*a**2*c**3 + 8*sqrt(a**2 - c**2)*atan((tan((d + e*x)/2)*a + c)/ 
sqrt(a**2 - c**2))*sin(d + e*x)*a**3*c**2 + 4*sqrt(a**2 - c**2)*atan((tan( 
(d + e*x)/2)*a + c)/sqrt(a**2 - c**2))*a**4*c + 2*cos(d + e*x)*sin(d + e*x 
)*a**3*c**3 - 2*cos(d + e*x)*sin(d + e*x)*a*c**5 + 2*cos(d + e*x)*a**4*c** 
2 - 2*cos(d + e*x)*a**2*c**4 + sin(d + e*x)**2*a**2*c**4 - sin(d + e*x)**2 
*c**6 + 2*sin(d + e*x)*a**3*c**3 - 2*sin(d + e*x)*a*c**5 - a**5*b + a**4*c 
**2 + 2*a**3*b*c**2 - a**2*c**4 - a*b*c**4)/(2*a*c*e*(sin(d + e*x)**2*a**4 
*c**2 - 2*sin(d + e*x)**2*a**2*c**4 + sin(d + e*x)**2*c**6 + 2*sin(d + e*x 
)*a**5*c - 4*sin(d + e*x)*a**3*c**3 + 2*sin(d + e*x)*a*c**5 + a**6 - 2*a** 
4*c**2 + a**2*c**4))