3.4.7 \(\int \sec (c+d x) (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [307]

3.4.7.1 Optimal result

Integrand size = 36, antiderivative size = 86 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (B+C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a (3 B+2 C) \tan (c+d x)}{3 d}+\frac {a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

output

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

3.4.7.2 Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(181\) vs. \(2(86)=172\).

Time = 0.80 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.10 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a \sec ^3(c+d x) \left (9 (B+C) \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 (B+C) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-4 (3 B+4 C+3 (B+C) \cos (c+d x)+(3 B+2 C) \cos (2 (c+d x))) \sin (c+d x)\right )}{24 d} \]

input

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

output

-1/24*(a*Sec[c + d*x]^3*(9*(B + C)*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - Si 
n[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 3*(B + C)*Co 
s[3*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d* 
x)/2] + Sin[(c + d*x)/2]]) - 4*(3*B + 4*C + 3*(B + C)*Cos[c + d*x] + (3*B 
+ 2*C)*Cos[2*(c + d*x)])*Sin[c + d*x]))/d
 

3.4.7.3 Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4560, 3042, 4485, 3042, 4274, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

\(\displaystyle \int \sec ^2(c+d x) (a \sec (c+d x)+a) (B+C \sec (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4485

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

input

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

output

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

3.4.7.3.1 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 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]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4560

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

3.4.7.4 Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.94

method	result	size
parts	\(\frac {\left (a B +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a B \tan \left (d x +c \right )}{d}\)	\(81\)
derivativedivides	\(\frac {a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a B \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(105\)
default	\(\frac {a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a B \tan \left (d x +c \right )+C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(105\)
norman	\(\frac {-\frac {3 a \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a \left (B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}+\frac {4 a \left (3 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a \left (B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a \left (B +C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(121\)
parallelrisch	\(\frac {a \left (-\frac {3 \left (B +C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (B +C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (B +C \right ) \sin \left (2 d x +2 c \right )+\left (B +\frac {2 C}{3}\right ) \sin \left (3 d x +3 c \right )+\sin \left (d x +c \right ) \left (B +2 C \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(137\)
risch	\(-\frac {i a \left (3 B \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-6 B -4 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(193\)

input

int(sec(d*x+c)*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RE 
TURNVERBOSE)
 

output

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

3.4.7.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.22 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

input

integrate(sec(d*x+c)*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, alg 
orithm="fricas")
 

output

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

3.4.7.6 Sympy [F]

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

input

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

output

a*(Integral(B*sec(c + d*x)**2, x) + Integral(B*sec(c + d*x)**3, x) + Integ 
ral(C*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x))
 

3.4.7.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.48 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \]

input

integrate(sec(d*x+c)*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, alg 
orithm="maxima")
 

output

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

3.4.7.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.79 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (B a + C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input

integrate(sec(d*x+c)*(a+a*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, alg 
orithm="giac")
 

output

1/6*(3*(B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a + C*a)*log( 
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*t 
an(1/2*d*x + 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*tan(1/2*d*x 
+ 1/2*c)^3 + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan 
(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 

3.4.7.9 Mupad [B] (verification not implemented)

Time = 17.79 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.47 \[ \int \sec (c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B+C\right )}{d}-\frac {\left (B\,a+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,B\,a-\frac {4\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input

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

output

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