\(\int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx\) [52]

Optimal result

Integrand size = 29, antiderivative size = 79 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {(4 A-B) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {(A-B) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \] Output:

A*arctanh(sin(d*x+c))/a^2/d-1/3*(4*A-B)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/ 
3*(A-B)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(170\) vs. \(2(79)=158\).

Time = 1.02 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.15 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (6 A \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(A-B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+2 (4 A-B) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+(A-B) \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \] Input:

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

Output:

(-2*Cos[(c + d*x)/2]*(6*A*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[( 
c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + (A - B)*Sec[c/2 
]*Sin[(d*x)/2] + 2*(4*A - B)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + (A 
 - B)*Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3042, 3457, 3042, 3457, 27, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int 3 a^2 A \sec (c+d x)dx}{a^2}-\frac {(4 A-B) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 A \int \sec (c+d x)dx-\frac {(4 A-B) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {3 A \text {arctanh}(\sin (c+d x))}{d}-\frac {(4 A-B) \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {(A-B) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A - B)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((3*A*ArcTanh[Sin 
[c + d*x]])/d - ((4*A - B)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(3*a^2)
 

Defintions of rubi rules used

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3457

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

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

Maple [A] (verified)

Time = 1.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95

method	result	size
parallelrisch	\(\frac {-6 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (A -B \right )+9 A -3 B \right )}{6 a^{2} d}\)	\(75\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\)	\(91\)
risch	\(-\frac {2 i \left (3 A \,{\mathrm e}^{2 i \left (d x +c \right )}+9 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}+4 A -B \right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\)	\(110\)
norman	\(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}-\frac {\left (3 A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {\left (5 A -2 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{a \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\)	\(137\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))^2,x,method=_RETURNVERBOSE 
)
 

Output:

1/6*(-6*A*ln(tan(1/2*d*x+1/2*c)-1)+6*A*ln(tan(1/2*d*x+1/2*c)+1)-tan(1/2*d* 
x+1/2*c)*(tan(1/2*d*x+1/2*c)^2*(A-B)+9*A-3*B))/a^2/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.66 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A \cos \left (d x + c\right )^{2} + 2 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (4 \, A - B\right )} \cos \left (d x + c\right ) + 5 \, A - 2 \, B\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))^2,x, algorithm="fri 
cas")
 

Output:

1/6*(3*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + A)*log(sin(d*x + c) + 1) - 3 
*(A*cos(d*x + c)^2 + 2*A*cos(d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*((4* 
A - B)*cos(d*x + c) + 5*A - 2*B)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a 
^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))**2,x)
 

Output:

(Integral(A*sec(c + d*x)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x) + Inte 
gral(B*cos(c + d*x)*sec(c + d*x)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1), x 
))/a**2
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.84 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {A {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}\right )} - \frac {B {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, d} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))^2,x, algorithm="max 
ima")
 

Output:

-1/6*(A*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) 
 + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin( 
d*x + c)/(cos(d*x + c) + 1) - 1)/a^2) - B*(3*sin(d*x + c)/(cos(d*x + c) + 
1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2)/d
 

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)/(a+a*cos(d*x+c))^2,x, algorithm="gia 
c")
 

Output:

1/6*(6*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*A*log(abs(tan(1/2*d*x 
+ 1/2*c) - 1))/a^2 - (A*a^4*tan(1/2*d*x + 1/2*c)^3 - B*a^4*tan(1/2*d*x + 1 
/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) - 3*B*a^4*tan(1/2*d*x + 1/2*c))/a^6 
)/d
 

Mupad [B] (verification not implemented)

Time = 41.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^2\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{6\,a^2\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A}{a^2}+\frac {A-B}{2\,a^2}\right )}{d} \] Input:

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

Output:

(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^2*d) - (tan(c/2 + (d*x)/2)^3*(A - B))/( 
6*a^2*d) - (tan(c/2 + (d*x)/2)*(A/a^2 + (A - B)/(2*a^2)))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14 \[ \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a +\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{6 a^{2} d} \] Input:

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

Output:

( - 6*log(tan((c + d*x)/2) - 1)*a + 6*log(tan((c + d*x)/2) + 1)*a - tan((c 
 + d*x)/2)**3*a + tan((c + d*x)/2)**3*b - 9*tan((c + d*x)/2)*a + 3*tan((c 
+ d*x)/2)*b)/(6*a**2*d)