\(\int \sec (a+b x) \sec (c+b x) \, dx\) [318]

Optimal result

Integrand size = 13, antiderivative size = 36 \[ \int \sec (a+b x) \sec (c+b x) \, dx=-\frac {\csc (a-c) \log (\cos (a+b x))}{b}+\frac {\csc (a-c) \log (\cos (c+b x))}{b} \] Output:

-csc(a-c)*ln(cos(b*x+a))/b+csc(a-c)*ln(cos(b*x+c))/b
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \sec (a+b x) \sec (c+b x) \, dx=-\frac {\csc (a-c) (\log (\cos (a+b x))-\log (\cos (c+b x)))}{b} \] Input:

Integrate[Sec[a + b*x]*Sec[c + b*x],x]
 

Output:

-((Csc[a - c]*(Log[Cos[a + b*x]] - Log[Cos[c + b*x]]))/b)
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5121, 3042, 3956}

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.

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

\(\Big \downarrow \) 5121

\(\displaystyle \csc (a-c) \int \tan (a+b x)dx-\csc (a-c) \int \tan (c+b x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \csc (a-c) \int \tan (a+b x)dx-\csc (a-c) \int \tan (c+b x)dx\)

\(\Big \downarrow \) 3956

\(\displaystyle \frac {\csc (a-c) \log (\cos (b x+c))}{b}-\frac {\csc (a-c) \log (\cos (a+b x))}{b}\)

Input:

Int[Sec[a + b*x]*Sec[c + b*x],x]
 

Output:

-((Csc[a - c]*Log[Cos[a + b*x]])/b) + (Csc[a - c]*Log[Cos[c + b*x]])/b
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3956

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

rule 5121

Int[Sec[(a_.) + (b_.)*(x_)]*Sec[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[-Csc[ 
(b*c - a*d)/d]   Int[Tan[a + b*x], x], x] + Simp[Csc[(b*c - a*d)/b]   Int[T 
an[c + d*x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b^2 - d^2, 0] && NeQ[b 
*c - a*d, 0]
 

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {\ln \left (\tan \left (b x +a \right ) \sin \left (a \right ) \cos \left (c \right )-\tan \left (b x +a \right ) \cos \left (a \right ) \sin \left (c \right )+\cos \left (a \right ) \cos \left (c \right )+\sin \left (a \right ) \sin \left (c \right )\right )}{b \left (\sin \left (a \right ) \cos \left (c \right )-\cos \left (a \right ) \sin \left (c \right )\right )}\)	\(54\)
risch	\(-\frac {2 i \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) {\mathrm e}^{i \left (a +c \right )}}{\left ({\mathrm e}^{2 i a}-{\mathrm e}^{2 i c}\right ) b}+\frac {2 i \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+{\mathrm e}^{2 i \left (a -c \right )}\right ) {\mathrm e}^{i \left (a +c \right )}}{\left ({\mathrm e}^{2 i a}-{\mathrm e}^{2 i c}\right ) b}\)	\(90\)

Input:

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

Output:

1/b/(sin(a)*cos(c)-cos(a)*sin(c))*ln(tan(b*x+a)*sin(a)*cos(c)-tan(b*x+a)*c 
os(a)*sin(c)+cos(a)*cos(c)+sin(a)*sin(c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (36) = 72\).

Time = 0.09 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.97 \[ \int \sec (a+b x) \sec (c+b x) \, dx=-\frac {\log \left (\cos \left (b x + c\right )^{2}\right ) - \log \left (\frac {4 \, {\left (2 \, \cos \left (b x + c\right ) \cos \left (-a + c\right ) \sin \left (b x + c\right ) \sin \left (-a + c\right ) + {\left (2 \, \cos \left (-a + c\right )^{2} - 1\right )} \cos \left (b x + c\right )^{2} - \cos \left (-a + c\right )^{2} + 1\right )}}{\cos \left (-a + c\right )^{2} + 2 \, \cos \left (-a + c\right ) + 1}\right )}{2 \, b \sin \left (-a + c\right )} \] Input:

integrate(sec(b*x+a)*sec(b*x+c),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate(sec(b*x+a)*sec(b*x+c),x)
 

Output:

Integral(sec(a + b*x)*sec(b*x + c), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (36) = 72\).

Time = 0.05 (sec) , antiderivative size = 349, normalized size of antiderivative = 9.69 \[ \int \sec (a+b x) \sec (c+b x) \, dx=-\frac {2 \, {\left ({\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (a + c\right ) + {\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (a + c\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, a\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, a\right )\right ) - 2 \, {\left ({\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \cos \left (a + c\right ) + {\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \sin \left (a + c\right )\right )} \arctan \left (\sin \left (2 \, b x\right ) - \sin \left (2 \, c\right ), \cos \left (2 \, b x\right ) + \cos \left (2 \, c\right )\right ) - {\left ({\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \cos \left (a + c\right ) - {\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \sin \left (a + c\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right ) + {\left ({\left (\sin \left (2 \, a\right ) - \sin \left (2 \, c\right )\right )} \cos \left (a + c\right ) - {\left (\cos \left (2 \, a\right ) - \cos \left (2 \, c\right )\right )} \sin \left (a + c\right )\right )} \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, c\right ) + \cos \left (2 \, c\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, c\right ) + \sin \left (2 \, c\right )^{2}\right )}{2 \, b \cos \left (2 \, a\right ) \cos \left (2 \, c\right ) - b \cos \left (2 \, c\right )^{2} + 2 \, b \sin \left (2 \, a\right ) \sin \left (2 \, c\right ) - b \sin \left (2 \, c\right )^{2} - {\left (\cos \left (2 \, a\right )^{2} + \sin \left (2 \, a\right )^{2}\right )} b} \] Input:

integrate(sec(b*x+a)*sec(b*x+c),x, algorithm="maxima")
 

Output:

-(2*((cos(2*a) - cos(2*c))*cos(a + c) + (sin(2*a) - sin(2*c))*sin(a + c))* 
arctan2(sin(2*b*x) - sin(2*a), cos(2*b*x) + cos(2*a)) - 2*((cos(2*a) - cos 
(2*c))*cos(a + c) + (sin(2*a) - sin(2*c))*sin(a + c))*arctan2(sin(2*b*x) - 
 sin(2*c), cos(2*b*x) + cos(2*c)) - ((sin(2*a) - sin(2*c))*cos(a + c) - (c 
os(2*a) - cos(2*c))*sin(a + c))*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + 
 cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2) + ((sin(2 
*a) - sin(2*c))*cos(a + c) - (cos(2*a) - cos(2*c))*sin(a + c))*log(cos(2*b 
*x)^2 + 2*cos(2*b*x)*cos(2*c) + cos(2*c)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*s 
in(2*c) + sin(2*c)^2))/(2*b*cos(2*a)*cos(2*c) - b*cos(2*c)^2 + 2*b*sin(2*a 
)*sin(2*c) - b*sin(2*c)^2 - (cos(2*a)^2 + sin(2*a)^2)*b)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (36) = 72\).

Time = 0.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 4.75 \[ \int \sec (a+b x) \sec (c+b x) \, dx=\frac {{\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} + \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )} \log \left ({\left | 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right ) - \tan \left (\frac {1}{2} \, a\right )^{2} - 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) - \tan \left (\frac {1}{2} \, c\right )^{2} + 1 \right |}\right )}{2 \, {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right ) - \tan \left (\frac {1}{2} \, c\right )\right )} b} \] Input:

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

Output:

1/2*(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*log(abs( 
2*tan(b*x + a)*tan(1/2*a)^2*tan(1/2*c) - 2*tan(b*x + a)*tan(1/2*a)*tan(1/2 
*c)^2 + tan(1/2*a)^2*tan(1/2*c)^2 + 2*tan(b*x + a)*tan(1/2*a) - tan(1/2*a) 
^2 - 2*tan(b*x + a)*tan(1/2*c) + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 
1))/((tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan 
(1/2*c))*b)
 

Mupad [B] (verification not implemented)

Time = 25.03 (sec) , antiderivative size = 249, normalized size of antiderivative = 6.92 \[ \int \sec (a+b x) \sec (c+b x) \, dx=\frac {2\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}\,\left (\ln \left (-\frac {2\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}\,\left (4\,b\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}\right )}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )}+{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}\right )-\ln \left (-\frac {2\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}\,\left (4\,b\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{a\,4{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}\right )}{b-b\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}+{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,2{}\mathrm {i}}\,4{}\mathrm {i}\right )\right )}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )} \] Input:

int(1/(cos(a + b*x)*cos(c + b*x)),x)
 

Output:

(2*(-exp(a*2i - c*2i))^(1/2)*(log(exp(a*1i)*exp(a*2i)*exp(-c*1i)*exp(b*x*2 
i)*4i - (2*(-exp(a*2i)*exp(-c*2i))^(1/2)*(4*b*exp(a*2i)*exp(-c*2i) + 2*b*e 
xp(a*2i)*exp(b*x*2i) + 2*b*exp(a*4i)*exp(-c*2i)*exp(b*x*2i)))/(b*(exp(a*2i 
)*exp(-c*2i) - 1))) - log(exp(a*1i)*exp(a*2i)*exp(-c*1i)*exp(b*x*2i)*4i - 
(2*(-exp(a*2i)*exp(-c*2i))^(1/2)*(4*b*exp(a*2i)*exp(-c*2i) + 2*b*exp(a*2i) 
*exp(b*x*2i) + 2*b*exp(a*4i)*exp(-c*2i)*exp(b*x*2i)))/(b - b*exp(a*2i)*exp 
(-c*2i)))))/(b*(exp(a*2i - c*2i) - 1))
 

Reduce [F]

\[ \int \sec (a+b x) \sec (c+b x) \, dx=\frac {4 \left (\int \frac {1}{\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\tan \left (\frac {b x}{2}+\frac {c}{2}\right )^{2}-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+1}d x \right ) b +\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {c}{2}\right )-1\right )-\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {c}{2}\right )+1\right )+\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-b x}{b} \] Input:

int(sec(b*x+a)*sec(b*x+c),x)
 

Output:

(4*int(1/(tan((b*x + c)/2)**2*tan((a + b*x)/2)**2 - tan((b*x + c)/2)**2 - 
tan((a + b*x)/2)**2 + 1),x)*b + log(tan((b*x + c)/2) - 1) - log(tan((b*x + 
 c)/2) + 1) + log(tan((a + b*x)/2) - 1) - log(tan((a + b*x)/2) + 1) - b*x) 
/b