3.241 \(\int \cos (a+b x) \sec (c+b x) \, dx\)
Optimal. Leaf size=26 \[ \frac{\sin (a-c) \log (\cos (b x+c))}{b}+x \cos (a-c) \]
[Out]
x*Cos[a - c] + (Log[Cos[c + b*x]]*Sin[a - c])/b
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Rubi [A] time = 0.0132681, antiderivative size = 26, normalized size of antiderivative = 1.,
number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used =
{4583, 3475, 8} \[ \frac{\sin (a-c) \log (\cos (b x+c))}{b}+x \cos (a-c) \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]*Sec[c + b*x],x]
[Out]
x*Cos[a - c] + (Log[Cos[c + b*x]]*Sin[a - c])/b
Rule 4583
Int[Cos[v_]*Sec[w_]^(n_.), x_Symbol] :> -Dist[Sin[v - w], Int[Tan[w]*Sec[w]^(n - 1), x], x] + Dist[Cos[v - w],
Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]
Rule 3475
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rubi steps
\begin{align*} \int \cos (a+b x) \sec (c+b x) \, dx &=\cos (a-c) \int 1 \, dx-\sin (a-c) \int \tan (c+b x) \, dx\\ &=x \cos (a-c)+\frac{\log (\cos (c+b x)) \sin (a-c)}{b}\\ \end{align*}
Mathematica [A] time = 0.121403, size = 26, normalized size = 1. \[ \frac{\sin (a-c) \log (\cos (b x+c))}{b}+x \cos (a-c) \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]*Sec[c + b*x],x]
[Out]
x*Cos[a - c] + (Log[Cos[c + b*x]]*Sin[a - c])/b
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Maple [B] time = 0.199, size = 461, normalized size = 17.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(b*x+a)*sec(b*x+c),x)
[Out]
1/b/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(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))*cos(a)^2*sin(c)^2-2/b/(cos(a)
^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(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))*cos(a)*cos(c)*sin(a)*sin(c)+1/b/(cos(a)^
2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(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))*cos(c)^2*sin(a)^2+1/2/b/(cos(c)^2+sin(c)^
2)/(cos(a)^2+sin(a)^2)*ln(1+tan(b*x+a)^2)*cos(a)*sin(c)-1/2/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*ln(1+tan
(b*x+a)^2)*sin(a)*cos(c)+1/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*cos(a)*cos(c)*arctan(tan(b*x+a))+1/b/(cos
(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*sin(a)*sin(c)*arctan(tan(b*x+a))
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Maxima [B] time = 1.18199, size = 100, normalized size = 3.85 \begin{align*} \frac{2 \, b x \cos \left (-a + c\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 ) \sin \left (-a + c\right )}{2 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c),x, algorithm="maxima")
[Out]
1/2*(2*b*x*cos(-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)*sin(-a + c))/b
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Fricas [A] time = 0.533184, size = 73, normalized size = 2.81 \begin{align*} \frac{b x \cos \left (-a + c\right ) - \log \left (-\cos \left (b x + c\right )\right ) \sin \left (-a + c\right )}{b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c),x, algorithm="fricas")
[Out]
(b*x*cos(-a + c) - log(-cos(b*x + c))*sin(-a + c))/b
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Sympy [B] time = 109.34, size = 435, normalized size = 16.73 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c),x)
[Out]
-Piecewise((0, Eq(b, 0)), (-x, Eq(c, pi/2)), (x, Eq(c, -pi/2)), (-2*b*x*tan(c/2)/(b*tan(c/2)**2 + b) - log(tan
(b*x/2)**2 + 1)*tan(c/2)**2/(b*tan(c/2)**2 + b) + log(tan(b*x/2)**2 + 1)/(b*tan(c/2)**2 + b) + log(tan(b*x/2)
- tan(c/2)/(tan(c/2) - 1) - 1/(tan(c/2) - 1))*tan(c/2)**2/(b*tan(c/2)**2 + b) - log(tan(b*x/2) - tan(c/2)/(tan
(c/2) - 1) - 1/(tan(c/2) - 1))/(b*tan(c/2)**2 + b) + log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) +
1))*tan(c/2)**2/(b*tan(c/2)**2 + b) - log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) + 1))/(b*tan(c/2)
**2 + b), True))*sin(a) + Piecewise((x/cos(c), Eq(b, 0)), (-log(sin(b*x))/b, Eq(c, pi/2)), (log(sin(b*x))/b, E
q(c, -pi/2)), (-b*x*tan(c/2)**2/(b*tan(c/2)**2 + b) + b*x/(b*tan(c/2)**2 + b) + 2*log(tan(b*x/2)**2 + 1)*tan(c
/2)/(b*tan(c/2)**2 + b) - 2*log(tan(b*x/2) - tan(c/2)/(tan(c/2) - 1) - 1/(tan(c/2) - 1))*tan(c/2)/(b*tan(c/2)*
*2 + b) - 2*log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) + 1))*tan(c/2)/(b*tan(c/2)**2 + b), True))*
cos(a)
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Giac [B] time = 1.21496, size = 594, normalized size = 22.85 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c),x, algorithm="giac")
[Out]
((tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)*(b*x + a)/(tan(1/2*a)
^2*tan(1/2*c)^2 + tan(1/2*a)^2 + 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))*log(tan(b*x + a)^2 + 1)/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) + 2
*(tan(1/2*a)^4*tan(1/2*c)^2 - 2*tan(1/2*a)^3*tan(1/2*c)^3 + tan(1/2*a)^2*tan(1/2*c)^4 + 2*tan(1/2*a)^3*tan(1/2
*c) - 4*tan(1/2*a)^2*tan(1/2*c)^2 + 2*tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)^2 - 2*tan(1/2*a)*tan(1/2*c) + tan(1
/2*c)^2)*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)^4*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c)^4 + tan(1/2*a)^4*tan(1/2*c) - tan(
1/2*a)*tan(1/2*c)^4 + tan(1/2*a)^3 - tan(1/2*c)^3 + tan(1/2*a) - tan(1/2*c)))/b