3.1.61 \(\int \frac {(A+C \cos ^2(c+d x)) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [61]

3.1.61.1 Optimal result

Integrand size = 31, antiderivative size = 115 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {A \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(7 A-3 C) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(22 A-3 C) \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \]

output

A*arctanh(sin(d*x+c))/a^3/d-1/5*(A+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^3-1/15 
*(7*A-3*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-1/15*(22*A-3*C)*sin(d*x+c)/d/ 
(a^3+a^3*cos(d*x+c))
 

3.1.61.2 Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.77 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {-240 A \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-5 (29 A-3 C) \sin \left (\frac {d x}{2}\right )+15 (5 A-C) \sin \left (c+\frac {d x}{2}\right )-95 A \sin \left (c+\frac {3 d x}{2}\right )+15 C \sin \left (c+\frac {3 d x}{2}\right )+15 A \sin \left (2 c+\frac {3 d x}{2}\right )-22 A \sin \left (2 c+\frac {5 d x}{2}\right )+3 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{30 a^3 d (1+\cos (c+d x))^3} \]

input

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^3,x]
 

output

(-240*A*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log 
[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2]*Sec[c/2]*(-5*(29 
*A - 3*C)*Sin[(d*x)/2] + 15*(5*A - C)*Sin[c + (d*x)/2] - 95*A*Sin[c + (3*d 
*x)/2] + 15*C*Sin[c + (3*d*x)/2] + 15*A*Sin[2*c + (3*d*x)/2] - 22*A*Sin[2* 
c + (5*d*x)/2] + 3*C*Sin[2*c + (5*d*x)/2]))/(30*a^3*d*(1 + Cos[c + d*x])^3 
)
 

3.1.61.3 Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3521, 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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

\(\displaystyle \frac {\int \frac {(5 a A-a (2 A-3 C) \cos (c+d x)) \sec (c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 a A-a (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {\left (15 a^2 A-a^2 (7 A-3 C) \cos (c+d x)\right ) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {15 a^2 A-a^2 (7 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\frac {\int 15 a^3 A \sec (c+d x)dx}{a^2}-\frac {a^2 (22 A-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {15 a A \int \sec (c+d x)dx-\frac {a^2 (22 A-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {15 a A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^2 (22 A-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {\frac {15 a A \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (22 A-3 C) \sin (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (7 A-3 C) \sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\)

input

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(a + a*Cos[c + d*x])^3,x]
 

output

-1/5*((A + C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/3*(a*(7*A - 3 
*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((15*a*A*ArcTanh[Sin[c + d* 
x]])/d - (a^2*(22*A - 3*C)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2) 
)/(5*a^2)
 

3.1.61.3.1 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 3521

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

rule 4257

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

3.1.61.4 Maple [A] (verified)

Time = 2.56 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.76

method	result	size
parallelrisch	\(\frac {-60 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {20 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+35 A -5 C \right )}{60 a^{3} d}\)	\(87\)
derivativedivides	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
default	\(\frac {-\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{5}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3}-7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
risch	\(-\frac {2 i \left (15 A \,{\mathrm e}^{4 i \left (d x +c \right )}+75 A \,{\mathrm e}^{3 i \left (d x +c \right )}-15 C \,{\mathrm e}^{3 i \left (d x +c \right )}+145 A \,{\mathrm e}^{2 i \left (d x +c \right )}-15 C \,{\mathrm e}^{2 i \left (d x +c \right )}+95 A \,{\mathrm e}^{i \left (d x +c \right )}-15 C \,{\mathrm e}^{i \left (d x +c \right )}+22 A -3 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\)	\(158\)
norman	\(\frac {-\frac {\left (A +C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}-\frac {\left (7 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (13 A +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {\left (23 A -3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {\left (37 A -3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2}}+\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\)	\(187\)

input

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+cos(d*x+c)*a)^3,x,method=_RETURNVERBO 
SE)
 

output

1/60*(-60*A*ln(tan(1/2*d*x+1/2*c)-1)+60*A*ln(tan(1/2*d*x+1/2*c)+1)-3*tan(1 
/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^4+20/3*A*tan(1/2*d*x+1/2*c)^2+35*A 
-5*C))/a^3/d
 

3.1.61.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.60 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (A \cos \left (d x + c\right )^{3} + 3 \, A \cos \left (d x + c\right )^{2} + 3 \, A \cos \left (d x + c\right ) + A\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left ({\left (22 \, A - 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (17 \, A - 3 \, C\right )} \cos \left (d x + c\right ) + 32 \, A - 3 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

input

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="f 
ricas")
 

output

1/30*(15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*lo 
g(sin(d*x + c) + 1) - 15*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos( 
d*x + c) + A)*log(-sin(d*x + c) + 1) - 2*((22*A - 3*C)*cos(d*x + c)^2 + 3* 
(17*A - 3*C)*cos(d*x + c) + 32*A - 3*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^ 
3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

3.1.61.6 Sympy [F]

\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {A \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]

input

integrate((A+C*cos(d*x+c)**2)*sec(d*x+c)/(a+a*cos(d*x+c))**3,x)
 

output

(Integral(A*sec(c + d*x)/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + 
d*x) + 1), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)/(cos(c + d*x)**3 + 
 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3
 

3.1.61.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.45 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {A {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {3 \, C {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

input

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="m 
axima")
 

output

-1/60*(A*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d* 
x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d* 
x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1 
) - 1)/a^3) - 3*C*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos 
(d*x + c) + 1)^5)/a^3)/d
 

3.1.61.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.14 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, A \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

input

integrate((A+C*cos(d*x+c)^2)*sec(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="g 
iac")
 

output

1/60*(60*A*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*A*log(abs(tan(1/2*d 
*x + 1/2*c) - 1))/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/ 
2*d*x + 1/2*c)^5 + 20*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*A*a^12*tan(1/2*d 
*x + 1/2*c) - 15*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
 

3.1.61.9 Mupad [B] (verification not implemented)

Time = 1.21 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.99 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {2\,A\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{12\,a^3}+\frac {3\,A-C}{12\,a^3}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{4\,a^3}+\frac {3\,A-C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

input

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)*(a + a*cos(c + d*x))^3),x)
 

output

(2*A*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - (tan(c/2 + (d*x)/2)^3*((A + C)/( 
12*a^3) + (3*A - C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)*((A + C)/(4*a^3) + 
(3*A - C)/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)