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

Optimal. Leaf size=36 \[ \frac{\csc (a-c) \log (\cos (b x+c))}{b}-\frac{\csc (a-c) \log (\cos (a+b x))}{b} \]

[Out]

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

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Rubi [A]  time = 0.019024, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4610, 3475} \[ \frac{\csc (a-c) \log (\cos (b x+c))}{b}-\frac{\csc (a-c) \log (\cos (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]*Sec[c + b*x],x]

[Out]

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

Rule 4610

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

Rule 3475

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

Rubi steps

\begin{align*} \int \sec (a+b x) \sec (c+b x) \, dx &=\csc (a-c) \int \tan (a+b x) \, dx-\csc (a-c) \int \tan (c+b x) \, dx\\ &=-\frac{\csc (a-c) \log (\cos (a+b x))}{b}+\frac{\csc (a-c) \log (\cos (c+b x))}{b}\\ \end{align*}

Mathematica [A]  time = 0.224254, size = 28, normalized size = 0.78 \[ -\frac{\csc (a-c) (\log (\cos (a+b x))-\log (\cos (b x+c)))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.231, size = 55, normalized size = 1.5 \begin{align*} -{\frac{\ln \left ( -\tan \left ( bx+a \right ) \cos \left ( a \right ) \sin \left ( c \right ) +\tan \left ( bx+a \right ) \sin \left ( a \right ) \cos \left ( c \right ) +\cos \left ( a \right ) \cos \left ( c \right ) +\sin \left ( a \right ) \sin \left ( c \right ) \right ) }{b \left ( \cos \left ( a \right ) \sin \left ( c \right ) -\sin \left ( a \right ) \cos \left ( c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.05778, size = 471, normalized size = 13.08 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(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) - (cos(2*a) - cos(2*c))*sin(a + c))*lo
g(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) + ((s
in(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)*sin(2*c) + sin(2*c)^2))/(2*b*cos(2*a)*cos(2*c) - b*cos(2*c)^2 + 2*b*s
in(2*a)*sin(2*c) - b*sin(2*c)^2 - (cos(2*a)^2 + sin(2*a)^2)*b)

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Fricas [B]  time = 2.60332, size = 273, normalized size = 7.58 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sec{\left (a + b x \right )} \sec{\left (b x + c \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.24713, size = 231, normalized size = 6.42 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) - ta
n(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*b)