3.8.67 \(\int \sec (c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [767]

3.8.67.1 Optimal result
3.8.67.2 Mathematica [A] (verified)
3.8.67.3 Rubi [A] (verified)
3.8.67.4 Maple [A] (verified)
3.8.67.5 Fricas [A] (verification not implemented)
3.8.67.6 Sympy [F]
3.8.67.7 Maxima [A] (verification not implemented)
3.8.67.8 Giac [B] (verification not implemented)
3.8.67.9 Mupad [B] (verification not implemented)

3.8.67.1 Optimal result

Integrand size = 36, antiderivative size = 93 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(b B+a C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(3 a B+2 b C) \tan (c+d x)}{3 d}+\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

output
1/2*(B*b+C*a)*arctanh(sin(d*x+c))/d+1/3*(3*B*a+2*C*b)*tan(d*x+c)/d+1/2*(B* 
b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*b*C*sec(d*x+c)^2*tan(d*x+c)/d
 
3.8.67.2 Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.72 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (b B+a C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 a B+6 b C+3 (b B+a C) \sec (c+d x)+2 b C \tan ^2(c+d x)\right )}{6 d} \]

input
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d* 
x]^2),x]
 
output
(3*(b*B + a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*a*B + 6*b*C + 3*(b* 
B + a*C)*Sec[c + d*x] + 2*b*C*Tan[c + d*x]^2))/(6*d)
 
3.8.67.3 Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.98, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4560

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4485

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4255

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

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

3.8.67.3.1 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]
 
3.8.67.4 Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87

method result size
parts \(\frac {\left (B b +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a B \tan \left (d x +c \right )}{d}-\frac {C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(81\)
derivativedivides \(\frac {B \tan \left (d x +c \right ) a +C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(105\)
default \(\frac {B \tan \left (d x +c \right ) a +C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(105\)
norman \(\frac {\frac {4 \left (3 a B +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (2 a B -B b -C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (2 a B +B b +C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) \(153\)
parallelrisch \(\frac {-9 \left (B b +C a \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (B b +C a \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (6 B b +6 C a \right ) \sin \left (2 d x +2 c \right )+\left (6 a B +4 C b \right ) \sin \left (3 d x +3 c \right )+6 \sin \left (d x +c \right ) \left (a B +2 C b \right )}{6 d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) \(159\)
risch \(-\frac {i \left (3 B b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 a B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{i \left (d x +c \right )}-3 C a \,{\mathrm e}^{i \left (d x +c \right )}-6 a B -4 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) \(201\)

input
int(sec(d*x+c)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RE 
TURNVERBOSE)
 
output
(B*b+C*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a*B/ 
d*tan(d*x+c)-C*b/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
 
3.8.67.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, B a + 2 \, C b\right )} \cos \left (d x + c\right )^{2} + 2 \, C b + 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, alg 
orithm="fricas")
 
output
1/12*(3*(C*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(C*a + B*b)*c 
os(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*B*a + 2*C*b)*cos(d*x + c)^2 
 + 2*C*b + 3*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)
 
3.8.67.6 Sympy [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 
output
Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**2, x)
 
3.8.67.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.37 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \]

input
integrate(sec(d*x+c)*(a+b*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*b - 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)) - 3*B*b* 
(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
 
3.8.67.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (85) = 170\).

Time = 0.31 (sec) , antiderivative size = 210, normalized size of antiderivative = 2.26 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, alg 
orithm="giac")
 
output
1/6*(3*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(C*a + B*b)*log( 
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*C*a*t 
an(1/2*d*x + 1/2*c)^5 - 3*B*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x + 
 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*b*tan(1/2*d*x + 1/2*c)^3 + 
 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a*tan(1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d 
*x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/ 
d
 
3.8.67.9 Mupad [B] (verification not implemented)

Time = 19.23 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.56 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B\,b+C\,a\right )}{d}-\frac {\left (2\,B\,a-B\,b-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,B\,a-\frac {4\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a+B\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input
int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)))/cos(c + d*x 
),x)
 
output
(atanh(tan(c/2 + (d*x)/2))*(B*b + C*a))/d - (tan(c/2 + (d*x)/2)*(2*B*a + B 
*b + C*a + 2*C*b) - tan(c/2 + (d*x)/2)^3*(4*B*a + (4*C*b)/3) + tan(c/2 + ( 
d*x)/2)^5*(2*B*a - B*b - C*a + 2*C*b))/(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))