3.6.28 \(\int (a+b \cos (c+d x)) (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [528]

3.6.28.1 Optimal result

Integrand size = 31, antiderivative size = 86 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {b (A+2 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (2 A+3 C) \tan (c+d x)}{3 d}+\frac {A b \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a A \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

output

1/2*b*(A+2*C)*arctanh(sin(d*x+c))/d+1/3*a*(2*A+3*C)*tan(d*x+c)/d+1/2*A*b*s 
ec(d*x+c)*tan(d*x+c)/d+1/3*a*A*sec(d*x+c)^2*tan(d*x+c)/d
 

3.6.28.2 Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 b (A+2 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 a (A+C)+3 A b \sec (c+d x)+2 a A \tan ^2(c+d x)\right )}{6 d} \]

input

Integrate[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

output

(3*b*(A + 2*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*a*(A + C) + 3*A*b*S 
ec[c + d*x] + 2*a*A*Tan[c + d*x]^2))/(6*d)
 

3.6.28.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 3511, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.

\(\displaystyle \int \sec ^4(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 3511

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3500

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {2 a (2 A+3 C) \tan (c+d x)}{d}+\frac {3 b (A+2 C) \text {arctanh}(\sin (c+d x))}{d}\right )+\frac {3 A b \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a A \tan (c+d x) \sec ^2(c+d x)}{3 d}\)

input

Int[(a + b*Cos[c + d*x])*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

output

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

3.6.28.3.1 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 3500

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

rule 3511

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ 
(-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( 
b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2))   Int[(a 
 + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d 
)) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b 
*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e 
, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -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 4257

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

3.6.28.4 Maple [A] (verified)

Time = 7.42 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a C \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(88\)
default	\(\frac {-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a C \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(88\)
parts	\(-\frac {a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(96\)
parallelrisch	\(\frac {-9 b \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 b \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 \left (A +\frac {3 C}{2}\right ) a \sin \left (3 d x +3 c \right )+6 b \sin \left (2 d x +2 c \right ) A +12 \left (A +\frac {C}{2}\right ) a \sin \left (d x +c \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(147\)
risch	\(-\frac {i \left (3 A b \,{\mathrm e}^{5 i \left (d x +c \right )}-6 C a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 a C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A b \,{\mathrm e}^{i \left (d x +c \right )}-4 a A -6 a C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {A b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(174\)
norman	\(\frac {-\frac {2 \left (2 a A -A b -2 a C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a A -A b +2 a C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (2 a A +A b -2 a C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 a A +A b +2 a C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (14 a A -9 A b +6 a C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (14 a A +9 A b +6 a C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {b \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(250\)

input

int((a+cos(d*x+c)*b)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVERBO 
SE)
 

output

1/d*(-a*A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*C*tan(d*x+c)+A*b*(1/2*sec(d 
*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*b*ln(sec(d*x+c)+tan(d*x+ 
c)))
 

3.6.28.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 2 \, C\right )} b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, A b \cos \left (d x + c\right ) + 2 \, A a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

input

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="f 
ricas")
 

output

1/12*(3*(A + 2*C)*b*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + 2*C)*b*c 
os(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(2*A + 3*C)*a*cos(d*x + c)^2 + 
 3*A*b*cos(d*x + c) + 2*A*a)*sin(d*x + c))/(d*cos(d*x + c)^3)
 

3.6.28.6 Sympy [F]

\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

input

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
 

output

Integral((A + C*cos(c + d*x)**2)*(a + b*cos(c + d*x))*sec(c + d*x)**4, x)
 

3.6.28.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a - 3 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a \tan \left (d x + c\right )}{12 \, d} \]

input

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="m 
axima")
 

output

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a - 3*A*b*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*b* 
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*tan(d*x + c))/d
 

3.6.28.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (78) = 156\).

Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.14 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A b + 2 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input

integrate((a+b*cos(d*x+c))*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="g 
iac")
 

output

1/6*(3*(A*b + 2*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*b + 2*C*b)* 
log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 + 6*C 
*a*tan(1/2*d*x + 1/2*c)^5 - 3*A*b*tan(1/2*d*x + 1/2*c)^5 - 4*A*a*tan(1/2*d 
*x + 1/2*c)^3 - 12*C*a*tan(1/2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*x + 1/2*c) 
 + 6*C*a*tan(1/2*d*x + 1/2*c) + 3*A*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 
 1/2*c)^2 - 1)^3)/d
 

3.6.28.9 Mupad [B] (verification not implemented)

Time = 3.55 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.59 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+2\,C\right )}{d}-\frac {\left (2\,A\,a-A\,b+2\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A\,a}{3}-4\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+2\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x)))/cos(c + d*x)^4,x)
 

output

(b*atanh(tan(c/2 + (d*x)/2))*(A + 2*C))/d - (tan(c/2 + (d*x)/2)*(2*A*a + A 
*b + 2*C*a) - tan(c/2 + (d*x)/2)^3*((4*A*a)/3 + 4*C*a) + tan(c/2 + (d*x)/2 
)^5*(2*A*a - A*b + 2*C*a))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/ 
2)^4 + tan(c/2 + (d*x)/2)^6 - 1))