\(\int \frac {(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx\) [45]

Optimal result

Integrand size = 33, antiderivative size = 105 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(3 A+2 C) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {(2 A+C) \tan (c+d x)}{a d}+\frac {(3 A+2 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))} \] Output:

1/2*(3*A+2*C)*arctanh(sin(d*x+c))/a/d-(2*A+C)*tan(d*x+c)/a/d+1/2*(3*A+2*C) 
*sec(d*x+c)*tan(d*x+c)/a/d-(A+C)*sec(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(284\) vs. \(2(105)=210\).

Time = 4.29 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.70 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (-4 (A+C) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left (-2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 A \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{2 a d (1+\cos (c+d x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 3521, 3042, 3227, 3042, 4254, 24, 4255, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {a (3 A+2 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a (2 A+C) \tan (c+d x)}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}\)

Input:

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

Output:

-(((A + C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))) + (-((a*(2 
*A + C)*Tan[c + d*x])/d) + a*(3*A + 2*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (S 
ec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

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 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

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 = 0.43 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {-3 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (\cos \left (2 d x +2 c \right )+1\right ) \left (A +\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \left (\left (2 A +C \right ) \cos \left (2 d x +2 c \right )+A \cos \left (d x +c \right )+A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d \left (\cos \left (2 d x +2 c \right )+1\right )}\)	\(120\)
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\left (\frac {3 A}{2}+C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\left (-\frac {3 A}{2}-C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(135\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {3 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\left (\frac {3 A}{2}+C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {3 A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\left (-\frac {3 A}{2}-C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(135\)
norman	\(\frac {\frac {A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{a d}+\frac {\left (7 A +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {3 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}-\frac {\left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (3 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (3 A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(196\)
risch	\(-\frac {i \left (3 A \,{\mathrm e}^{4 i \left (d x +c \right )}+2 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3 A \,{\mathrm e}^{3 i \left (d x +c \right )}+5 A \,{\mathrm e}^{2 i \left (d x +c \right )}+4 C \,{\mathrm e}^{2 i \left (d x +c \right )}+A \,{\mathrm e}^{i \left (d x +c \right )}+4 A +2 C \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a d}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a d}\)	\(202\)

Input:

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

Output:

1/2*(-3*(cos(2*d*x+2*c)+1)*(A+2/3*C)*ln(tan(1/2*d*x+1/2*c)-1)+3*(cos(2*d*x 
+2*c)+1)*(A+2/3*C)*ln(tan(1/2*d*x+1/2*c)+1)-2*((2*A+C)*cos(2*d*x+2*c)+A*co 
s(d*x+c)+A+C)*tan(1/2*d*x+1/2*c))/a/d/(cos(2*d*x+2*c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.45 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left ({\left (3 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (3 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A + 2 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right ) - A\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 239 vs. \(2 (101) = 202\).

Time = 0.06 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.28 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {A {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 2 \, C {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.24 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (3 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (3 \, A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \] Input:

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

Output:

1/2*((3*A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - (3*A + 2*C)*log(ab 
s(tan(1/2*d*x + 1/2*c) - 1))/a - 2*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x 
 + 1/2*c))/a + 2*(3*A*tan(1/2*d*x + 1/2*c)^3 - A*tan(1/2*d*x + 1/2*c))/((t 
an(1/2*d*x + 1/2*c)^2 - 1)^2*a))/d
 

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A}{2}+C\right )}{a\,d}-\frac {A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-3\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 285, normalized size of antiderivative = 2.71 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {4 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -2 \cos \left (d x +c \right ) a -2 \cos \left (d x +c \right ) c -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) c +3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} c -3 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) c -3 \sin \left (d x +c \right )^{2} a -2 \sin \left (d x +c \right )^{2} c +2 a +2 c}{2 \sin \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

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

Output:

(4*cos(c + d*x)*sin(c + d*x)**2*a + 2*cos(c + d*x)*sin(c + d*x)**2*c - 2*c 
os(c + d*x)*a - 2*cos(c + d*x)*c - 3*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**3*a - 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*c + 3*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)*a + 2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)*c + 3 
*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*a + 2*log(tan((c + d*x)/2) + 1) 
*sin(c + d*x)**3*c - 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a - 2*log(ta 
n((c + d*x)/2) + 1)*sin(c + d*x)*c - 3*sin(c + d*x)**2*a - 2*sin(c + d*x)* 
*2*c + 2*a + 2*c)/(2*sin(c + d*x)*a*d*(sin(c + d*x)**2 - 1))