3.1.96 \(\int \cos (c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [96]

3.1.96.1 Optimal result

Integrand size = 31, antiderivative size = 112 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a^2 A x+\frac {a^2 (2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {a^2 (2 A-3 C) \tan (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{2 d} \]

output

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

3.1.96.2 Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.83 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a^2 A x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {3 a^2 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 A \sin (c+d x)}{d}+\frac {2 a^2 C \tan (c+d x)}{d}+\frac {a^2 C \sec (c+d x) \tan (c+d x)}{2 d} \]

input

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

output

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

3.1.96.3 Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {3042, 4575, 3042, 4405, 3042, 4402, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4575

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4405

\(\displaystyle \frac {\frac {1}{2} \int (\sec (c+d x) a+a) \left (4 a^2 A-a^2 (2 A-3 C) \sec (c+d x)\right )dx-\frac {(2 A-C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4402

\(\displaystyle \frac {\frac {1}{2} \left (-a^3 (2 A-3 C) \int \sec ^2(c+d x)dx+a^3 (2 A+3 C) \int \sec (c+d x)dx+4 a^3 A x\right )-\frac {(2 A-C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{2} \left (\frac {a^3 (2 A+3 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (2 A-3 C) \tan (c+d x)}{d}+4 a^3 A x\right )-\frac {(2 A-C) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^2}{d}\)

input

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

output

(A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/d + (-1/2*((2*A - C)*(a^3 + a^3*Se 
c[c + d*x])*Tan[c + d*x])/d + (4*a^3*A*x + (a^3*(2*A + 3*C)*ArcTanh[Sin[c 
+ d*x]])/d - (a^3*(2*A - 3*C)*Tan[c + d*x])/d)/2)/a
 

3.1.96.3.1 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 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]
 

rule 4402

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + 
 (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d   Int[Csc[e + f*x]^2, x], x 
] + Simp[(b*c + a*d)   Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f 
}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
 

rule 4405

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + 
 (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 
*m]
 

rule 4575

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co 
t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( 
b*d*n)   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b 
*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 
 C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || 
 EqQ[m + n + 1, 0])
 

3.1.96.4 Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.02

method	result	size
derivativedivides	\(\frac {a^{2} A \sin \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (d x +c \right )+2 C \,a^{2} \tan \left (d x +c \right )+a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(114\)
default	\(\frac {a^{2} A \sin \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (d x +c \right )+2 C \,a^{2} \tan \left (d x +c \right )+a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(114\)
parallelrisch	\(\frac {a^{2} \left (-2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 d x A \cos \left (2 d x +2 c \right )+4 \sin \left (2 d x +2 c \right ) C +A \sin \left (3 d x +3 c \right )+\sin \left (d x +c \right ) \left (A +2 C \right )+4 d x A \right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(137\)
risch	\(2 a^{2} A x -\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{2} C \left ({\mathrm e}^{3 i \left (d x +c \right )}-4 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-4\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\)	\(190\)
norman	\(\frac {\frac {a^{2} \left (2 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-2 a^{2} A x +4 a^{2} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-4 a^{2} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 a^{2} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\frac {3 a^{2} \left (2 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {a^{2} \left (2 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{2} \left (6 A -5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a^{2} \left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(248\)

input

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

output

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

3.1.96.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.15 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, A a^{2} d x \cos \left (d x + c\right )^{2} + {\left (2 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 4 \, C a^{2} \cos \left (d x + c\right ) + C a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

input

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

output

1/4*(8*A*a^2*d*x*cos(d*x + c)^2 + (2*A + 3*C)*a^2*cos(d*x + c)^2*log(sin(d 
*x + c) + 1) - (2*A + 3*C)*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*( 
2*A*a^2*cos(d*x + c)^2 + 4*C*a^2*cos(d*x + c) + C*a^2)*sin(d*x + c))/(d*co 
s(d*x + c)^2)
 

3.1.96.6 Sympy [F]

\[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

input

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

output

a**2*(Integral(A*cos(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x) 
, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(C*cos(c + d* 
x)*sec(c + d*x)**2, x) + Integral(2*C*cos(c + d*x)*sec(c + d*x)**3, x) + I 
ntegral(C*cos(c + d*x)*sec(c + d*x)**4, x))
 

3.1.96.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.27 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} A a^{2} - C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]

input

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

output

1/4*(8*(d*x + c)*A*a^2 - C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log( 
sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^2*(log(sin(d*x + c) + 1 
) - log(sin(d*x + c) - 1)) + 2*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x 
+ c) - 1)) + 4*A*a^2*sin(d*x + c) + 8*C*a^2*tan(d*x + c))/d
 

3.1.96.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.36 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{2} + \frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + {\left (2 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

input

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

output

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

3.1.96.9 Mupad [B] (verification not implemented)

Time = 15.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.38 \[ \int \cos (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

input

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

output

(A*a^2*sin(c + d*x))/d + (4*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 
2)))/d + (2*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*C*a 
^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*sin(c + d*x) 
)/(d*cos(c + d*x)) + (C*a^2*sin(c + d*x))/(2*d*cos(c + d*x)^2)