\(\int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx\) [506]

Optimal result

Integrand size = 31, antiderivative size = 118 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\frac {2 (a A-c C) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{\left (a^2-c^2\right )^{3/2} e}-\frac {B}{c e (a+c \sin (d+e x))}+\frac {(A c-a C) \cos (d+e x)}{\left (a^2-c^2\right ) e (a+c \sin (d+e x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\frac {\frac {2 (a A-c C) \arctan \left (\frac {c+a \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-c^2}}\right )}{\left (a^2-c^2\right )^{3/2}}+\frac {B \left (a^2-c^2\right )-c (A c-a C) \cos (d+e x)}{c (-a+c) (a+c) (a+c \sin (d+e x))}}{e} \] Input:

Integrate[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4876, 3042, 3147, 17, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4876

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[(A + B*Cos[d + e*x] + C*Sin[d + e*x])/(a + c*Sin[d + e*x])^2,x]
 

Output:

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

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 [A] (verified)

Time = 0.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (A \,c^{2}+B \,a^{2}-B \,c^{2}-a c C \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a \left (a^{2}-c^{2}\right )}+\frac {2 \left (A c -a C \right )}{a^{2}-c^{2}}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a}+\frac {2 \left (A a -C c \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{2}-c^{2}\right )^{\frac {3}{2}}}}{e}\)	\(155\)
default	\(\frac {\frac {\frac {2 \left (A \,c^{2}+B \,a^{2}-B \,c^{2}-a c C \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{a \left (a^{2}-c^{2}\right )}+\frac {2 \left (A c -a C \right )}{a^{2}-c^{2}}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a}+\frac {2 \left (A a -C c \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{2}-c^{2}\right )^{\frac {3}{2}}}}{e}\)	\(155\)
parts	\(\frac {\frac {\frac {2 c \left (A c -a C \right ) \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{\left (a^{2}-c^{2}\right ) a}+\frac {2 \left (A c -a C \right )}{a^{2}-c^{2}}}{\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2} a +2 c \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+a}+\frac {2 \left (A a -C c \right ) \arctan \left (\frac {2 a \tan \left (\frac {e x}{2}+\frac {d}{2}\right )+2 c}{2 \sqrt {a^{2}-c^{2}}}\right )}{\left (a^{2}-c^{2}\right )^{\frac {3}{2}}}}{e}-\frac {B}{c e \left (a +c \sin \left (e x +d \right )\right )}\)	\(164\)
risch	\(-\frac {2 i \left (-i B \,a^{2} {\mathrm e}^{i \left (e x +d \right )}+i B \,c^{2} {\mathrm e}^{i \left (e x +d \right )}+i A \,c^{2}+A a c \,{\mathrm e}^{i \left (e x +d \right )}-i a c C -C \,a^{2} {\mathrm e}^{i \left (e x +d \right )}\right )}{c e \left (a^{2}-c^{2}\right ) \left (-i c \,{\mathrm e}^{2 i \left (e x +d \right )}+i c +2 \,{\mathrm e}^{i \left (e x +d \right )} a \right )}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a -a^{2}+c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) A a}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right ) \left (a -c \right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a -a^{2}+c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) C c}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right ) \left (a -c \right ) e}+\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a +a^{2}-c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) A a}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right ) \left (a -c \right ) e}-\frac {\ln \left ({\mathrm e}^{i \left (e x +d \right )}+\frac {i \sqrt {-a^{2}+c^{2}}\, a +a^{2}-c^{2}}{\sqrt {-a^{2}+c^{2}}\, c}\right ) C c}{\sqrt {-a^{2}+c^{2}}\, \left (a +c \right ) \left (a -c \right ) e}\)	\(446\)

Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^2,x,method=_RETURNVERBO 
SE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 458, normalized size of antiderivative = 3.88 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\left [-\frac {2 \, B a^{4} - 4 \, B a^{2} c^{2} + 2 \, B c^{4} + {\left (A a^{2} c - C a c^{2} + {\left (A a c^{2} - C c^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}} \log \left (\frac {{\left (2 \, a^{2} - c^{2}\right )} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2} + 2 \, {\left (a \cos \left (e x + d\right ) \sin \left (e x + d\right ) + c \cos \left (e x + d\right )\right )} \sqrt {-a^{2} + c^{2}}}{c^{2} \cos \left (e x + d\right )^{2} - 2 \, a c \sin \left (e x + d\right ) - a^{2} - c^{2}}\right ) + 2 \, {\left (C a^{3} c - A a^{2} c^{2} - C a c^{3} + A c^{4}\right )} \cos \left (e x + d\right )}{2 \, {\left ({\left (a^{4} c^{2} - 2 \, a^{2} c^{4} + c^{6}\right )} e \sin \left (e x + d\right ) + {\left (a^{5} c - 2 \, a^{3} c^{3} + a c^{5}\right )} e\right )}}, -\frac {B a^{4} - 2 \, B a^{2} c^{2} + B c^{4} + {\left (A a^{2} c - C a c^{2} + {\left (A a c^{2} - C c^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt {a^{2} - c^{2}} \arctan \left (-\frac {a \sin \left (e x + d\right ) + c}{\sqrt {a^{2} - c^{2}} \cos \left (e x + d\right )}\right ) + {\left (C a^{3} c - A a^{2} c^{2} - C a c^{3} + A c^{4}\right )} \cos \left (e x + d\right )}{{\left (a^{4} c^{2} - 2 \, a^{2} c^{4} + c^{6}\right )} e \sin \left (e x + d\right ) + {\left (a^{5} c - 2 \, a^{3} c^{3} + a c^{5}\right )} e}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))**2,x)
 

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^2,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 [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.53 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + c}{\sqrt {a^{2} - c^{2}}}\right )\right )} {\left (A a - C c\right )}}{{\left (a^{2} - c^{2}\right )}^{\frac {3}{2}}} + \frac {B a^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - C a c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + A c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - B c^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - C a^{2} + A a c}{{\left (a^{3} - a c^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, c \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a\right )}}\right )}}{e} \] Input:

integrate((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^2,x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.24 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.92 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\frac {\frac {2\,\left (A\,c-C\,a\right )}{a^2-c^2}+\frac {2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (B\,a^2+A\,c^2-B\,c^2-C\,a\,c\right )}{a\,\left (a^2-c^2\right )}}{e\,\left (a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2+2\,c\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+a\right )}+\frac {2\,\mathrm {atan}\left (\frac {\left (a^2-c^2\right )\,\left (\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (A\,a-C\,c\right )}{{\left (a+c\right )}^{3/2}\,{\left (a-c\right )}^{3/2}}+\frac {2\,\left (a^2\,c-c^3\right )\,\left (A\,a-C\,c\right )}{{\left (a+c\right )}^{3/2}\,\left (a^2-c^2\right )\,{\left (a-c\right )}^{3/2}}\right )}{2\,\left (A\,a-C\,c\right )}\right )\,\left (A\,a-C\,c\right )}{e\,{\left (a+c\right )}^{3/2}\,{\left (a-c\right )}^{3/2}} \] Input:

int((A + B*cos(d + e*x) + C*sin(d + e*x))/(a + c*sin(d + e*x))^2,x)
 

Output:

((2*(A*c - C*a))/(a^2 - c^2) + (2*tan(d/2 + (e*x)/2)*(B*a^2 + A*c^2 - B*c^ 
2 - C*a*c))/(a*(a^2 - c^2)))/(e*(a + 2*c*tan(d/2 + (e*x)/2) + a*tan(d/2 + 
(e*x)/2)^2)) + (2*atan(((a^2 - c^2)*((2*a*tan(d/2 + (e*x)/2)*(A*a - C*c))/ 
((a + c)^(3/2)*(a - c)^(3/2)) + (2*(a^2*c - c^3)*(A*a - C*c))/((a + c)^(3/ 
2)*(a^2 - c^2)*(a - c)^(3/2))))/(2*(A*a - C*c)))*(A*a - C*c))/(e*(a + c)^( 
3/2)*(a - c)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.20 \[ \int \frac {A+B \cos (d+e x)+C \sin (d+e x)}{(a+c \sin (d+e x))^2} \, dx=\frac {2 \sqrt {a^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a +c}{\sqrt {a^{2}-c^{2}}}\right ) \sin \left (e x +d \right ) c^{2}+2 \sqrt {a^{2}-c^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) a +c}{\sqrt {a^{2}-c^{2}}}\right ) a c -a^{2} b +b \,c^{2}}{c e \left (\sin \left (e x +d \right ) a^{2} c -\sin \left (e x +d \right ) c^{3}+a^{3}-a \,c^{2}\right )} \] Input:

int((A+B*cos(e*x+d)+C*sin(e*x+d))/(a+c*sin(e*x+d))^2,x)
 

Output:

(2*sqrt(a**2 - c**2)*atan((tan((d + e*x)/2)*a + c)/sqrt(a**2 - c**2))*sin( 
d + e*x)*c**2 + 2*sqrt(a**2 - c**2)*atan((tan((d + e*x)/2)*a + c)/sqrt(a** 
2 - c**2))*a*c - a**2*b + b*c**2)/(c*e*(sin(d + e*x)*a**2*c - sin(d + e*x) 
*c**3 + a**3 - a*c**2))