\(\int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\) [279]

Optimal result

Integrand size = 27, antiderivative size = 61 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {(2 a A+b B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d} \] Output:

1/2*(2*A*a+B*b)*arctanh(sin(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/2*b*B*sec(d 
*x+c)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {a A \coth ^{-1}(\sin (c+d x))}{d}+\frac {b B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A b \tan (c+d x)}{d}+\frac {a B \tan (c+d x)}{d}+\frac {b B \sec (c+d x) \tan (c+d x)}{2 d} \] Input:

Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
 

Output:

(a*A*ArcCoth[Sin[c + d*x]])/d + (b*B*ArcTanh[Sin[c + d*x]])/(2*d) + (A*b*T 
an[c + d*x])/d + (a*B*Tan[c + d*x])/d + (b*B*Sec[c + d*x]*Tan[c + d*x])/(2 
*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {3042, 4485, 3042, 4274, 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.

\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4485

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4257

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

Input:

Int[Sec[c + d*x]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]
 

Output:

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

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23

method	result	size
derivativedivides	\(\frac {A a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
default	\(\frac {A a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B a \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(75\)
parts	\(\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{d}+\frac {A a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(76\)
parallelrisch	\(\frac {-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A a +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (1+\cos \left (2 d x +2 c \right )\right ) \left (A a +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (A b +B a \right ) \sin \left (2 d x +2 c \right )+B b \sin \left (d x +c \right )}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(110\)
norman	\(\frac {\frac {\left (2 A b +2 B a +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (2 A b +2 B a -B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}-\frac {\left (2 A a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A a +B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(123\)
risch	\(-\frac {i \left (B b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-B b \,{\mathrm e}^{i \left (d x +c \right )}-2 A b -2 B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A a}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A a}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}\)	\(160\)

Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/d*(A*a*ln(sec(d*x+c)+tan(d*x+c))+B*a*tan(d*x+c)+A*b*tan(d*x+c)+B*b*(1/2* 
sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.57 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, A a + B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A a + B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B b + 2 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="frica 
s")
 

Output:

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

Sympy [F]

\[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

Output:

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.44 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=-\frac {B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, B a \tan \left (d x + c\right ) - 4 \, A b \tan \left (d x + c\right )}{4 \, d} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxim 
a")
 

Output:

-1/4*(B*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + l 
og(sin(d*x + c) - 1)) - 4*A*a*log(sec(d*x + c) + tan(d*x + c)) - 4*B*a*tan 
(d*x + c) - 4*A*b*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (57) = 114\).

Time = 0.14 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.51 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {{\left (2 \, A a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac" 
)
 

Output:

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

Mupad [B] (verification not implemented)

Time = 13.18 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.70 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,b+2\,B\,a+B\,b\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,b+2\,B\,a-B\,b\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A\,a+B\,b\right )}{d} \] Input:

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x)))/cos(c + d*x),x)
 

Output:

(tan(c/2 + (d*x)/2)*(2*A*b + 2*B*a + B*b) - tan(c/2 + (d*x)/2)^3*(2*A*b + 
2*B*a - B*b))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1)) + (a 
tanh(tan(c/2 + (d*x)/2))*(2*A*a + B*b))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 211, normalized size of antiderivative = 3.46 \[ \int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx=\frac {-4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{2}+2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{2}-2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{2}-\sin \left (d x +c \right ) b^{2}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)
 

Output:

( - 4*cos(c + d*x)*sin(c + d*x)*a*b - 2*log(tan((c + d*x)/2) - 1)*sin(c + 
d*x)**2*a**2 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**2 + 2*log(tan( 
(c + d*x)/2) - 1)*a**2 + log(tan((c + d*x)/2) - 1)*b**2 + 2*log(tan((c + d 
*x)/2) + 1)*sin(c + d*x)**2*a**2 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)* 
*2*b**2 - 2*log(tan((c + d*x)/2) + 1)*a**2 - log(tan((c + d*x)/2) + 1)*b** 
2 - sin(c + d*x)*b**2)/(2*d*(sin(c + d*x)**2 - 1))