3.242 \(\int \cos (a+b x) \sec ^2(c+b x) \, dx\)
Optimal. Leaf size=35 \[ \frac{\cos (a-c) \tanh ^{-1}(\sin (b x+c))}{b}-\frac{\sin (a-c) \sec (b x+c)}{b} \]
[Out]
(ArcTanh[Sin[c + b*x]]*Cos[a - c])/b - (Sec[c + b*x]*Sin[a - c])/b
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Rubi [A] time = 0.0287126, antiderivative size = 35, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used =
{4583, 2606, 8, 3770} \[ \frac{\cos (a-c) \tanh ^{-1}(\sin (b x+c))}{b}-\frac{\sin (a-c) \sec (b x+c)}{b} \]
Antiderivative was successfully verified.
[In]
Int[Cos[a + b*x]*Sec[c + b*x]^2,x]
[Out]
(ArcTanh[Sin[c + b*x]]*Cos[a - c])/b - (Sec[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 2606
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Rule 8
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \cos (a+b x) \sec ^2(c+b x) \, dx &=\cos (a-c) \int \sec (c+b x) \, dx-\sin (a-c) \int \sec (c+b x) \tan (c+b x) \, dx\\ &=\frac{\tanh ^{-1}(\sin (c+b x)) \cos (a-c)}{b}-\frac{\sin (a-c) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+b x))}{b}\\ &=\frac{\tanh ^{-1}(\sin (c+b x)) \cos (a-c)}{b}-\frac{\sec (c+b x) \sin (a-c)}{b}\\ \end{align*}
Mathematica [C] time = 0.089295, size = 89, normalized size = 2.54 \[ -\frac{\sin (a-c) \sec (b x+c)}{b}-\frac{2 i \cos (a-c) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\sin (c) \cos \left (\frac{b x}{2}\right )+\cos (c) \sin \left (\frac{b x}{2}\right )\right )}{\cos (c) \cos \left (\frac{b x}{2}\right )-i \sin (c) \cos \left (\frac{b x}{2}\right )}\right )}{b} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[a + b*x]*Sec[c + b*x]^2,x]
[Out]
((-2*I)*ArcTan[((I*Cos[c] + Sin[c])*(Cos[(b*x)/2]*Sin[c] + Cos[c]*Sin[(b*x)/2]))/(Cos[c]*Cos[(b*x)/2] - I*Cos[
(b*x)/2]*Sin[c])]*Cos[a - c])/b - (Sec[c + b*x]*Sin[a - c])/b
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Maple [B] time = 0.451, size = 1049, normalized size = 30. \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)^2,x)
[Out]
2/b/(cos(a)*cos(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-
2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*
sin(a)^2+sin(a)^2*sin(c)^2)/(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)*cos(a)^2*sin(c)^2-4/b/(cos(a)*cos
(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1
/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)
^2*sin(c)^2)/(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)*cos(a)*cos(c)*sin(a)*sin(c)+2/b/(cos(a)*cos(c)*t
an(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1/2*a)
*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*si
n(c)^2)/(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)*cos(c)^2*sin(a)^2-2/b/(cos(a)*cos(c)*tan(1/2*b*x+1/2*
a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-
cos(a)*cos(c)-sin(a)*sin(c))/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)*cos(a)*
sin(c)+2/b/(cos(a)*cos(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*
sin(c)-2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+co
s(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)*sin(a)*cos(c)-2/b/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+si
n(a)^2*sin(c)^2)/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2)*arctan(1/2*(
2*(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*sin(c)-2*sin(a)*cos(c))/(-cos(a)^2*cos(c)^2-cos(a)
^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*cos(a)*cos(c)-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)/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^
2)^(1/2)*arctan(1/2*(2*(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*sin(c)-2*sin(a)*cos(c))/(-cos
(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*sin(a)*sin(c)
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Maxima [B] time = 1.86843, size = 528, normalized size = 15.09 \begin{align*} -\frac{2 \,{\left (\sin \left (b x + 2 \, a\right ) - \sin \left (b x + 2 \, c\right )\right )} \cos \left (2 \, b x + a + 2 \, c\right ) +{\left (\cos \left (2 \, b x + a + 2 \, c\right )^{2} \cos \left (-a + c\right ) + 2 \, \cos \left (2 \, b x + a + 2 \, c\right ) \cos \left (a\right ) \cos \left (-a + c\right ) + \cos \left (-a + c\right ) \sin \left (2 \, b x + a + 2 \, c\right )^{2} + 2 \, \cos \left (-a + c\right ) \sin \left (2 \, b x + a + 2 \, c\right ) \sin \left (a\right ) +{\left (\cos \left (a\right )^{2} + \sin \left (a\right )^{2}\right )} \cos \left (-a + c\right )\right )} \log \left (\frac{\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} - 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} + 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}{\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} + 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} - 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}\right ) - 2 \,{\left (\cos \left (b x + 2 \, a\right ) - \cos \left (b x + 2 \, c\right )\right )} \sin \left (2 \, b x + a + 2 \, c\right ) + 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, a\right ) - 2 \, \cos \left (a\right ) \sin \left (b x + 2 \, c\right ) - 2 \, \cos \left (b x + 2 \, a\right ) \sin \left (a\right ) + 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (a\right )}{2 \,{\left (b \cos \left (2 \, b x + a + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, b x + a + 2 \, c\right ) \cos \left (a\right ) + b \sin \left (2 \, b x + a + 2 \, c\right )^{2} + 2 \, b \sin \left (2 \, b x + a + 2 \, c\right ) \sin \left (a\right ) +{\left (\cos \left (a\right )^{2} + \sin \left (a\right )^{2}\right )} b\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c)^2,x, algorithm="maxima")
[Out]
-1/2*(2*(sin(b*x + 2*a) - sin(b*x + 2*c))*cos(2*b*x + a + 2*c) + (cos(2*b*x + a + 2*c)^2*cos(-a + c) + 2*cos(2
*b*x + a + 2*c)*cos(a)*cos(-a + c) + cos(-a + c)*sin(2*b*x + a + 2*c)^2 + 2*cos(-a + c)*sin(2*b*x + a + 2*c)*s
in(a) + (cos(a)^2 + sin(a)^2)*cos(-a + c))*log((cos(b*x + 2*c)^2 + cos(c)^2 - 2*cos(c)*sin(b*x + 2*c) + sin(b*
x + 2*c)^2 + 2*cos(b*x + 2*c)*sin(c) + sin(c)^2)/(cos(b*x + 2*c)^2 + cos(c)^2 + 2*cos(c)*sin(b*x + 2*c) + sin(
b*x + 2*c)^2 - 2*cos(b*x + 2*c)*sin(c) + sin(c)^2)) - 2*(cos(b*x + 2*a) - cos(b*x + 2*c))*sin(2*b*x + a + 2*c)
+ 2*cos(a)*sin(b*x + 2*a) - 2*cos(a)*sin(b*x + 2*c) - 2*cos(b*x + 2*a)*sin(a) + 2*cos(b*x + 2*c)*sin(a))/(b*c
os(2*b*x + a + 2*c)^2 + 2*b*cos(2*b*x + a + 2*c)*cos(a) + b*sin(2*b*x + a + 2*c)^2 + 2*b*sin(2*b*x + a + 2*c)*
sin(a) + (cos(a)^2 + sin(a)^2)*b)
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Fricas [A] time = 0.535566, size = 185, normalized size = 5.29 \begin{align*} \frac{\cos \left (b x + c\right ) \cos \left (-a + c\right ) \log \left (\sin \left (b x + c\right ) + 1\right ) - \cos \left (b x + c\right ) \cos \left (-a + c\right ) \log \left (-\sin \left (b x + c\right ) + 1\right ) + 2 \, \sin \left (-a + c\right )}{2 \, b \cos \left (b x + c\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c)^2,x, algorithm="fricas")
[Out]
1/2*(cos(b*x + c)*cos(-a + c)*log(sin(b*x + c) + 1) - cos(b*x + c)*cos(-a + c)*log(-sin(b*x + c) + 1) + 2*sin(
-a + c))/(b*cos(b*x + c))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(b*x+a)*sec(b*x+c)**2,x)
[Out]
Timed out
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Giac [B] time = 1.37998, size = 1810, normalized size = 51.71 \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)^2,x, algorithm="giac")
[Out]
-((tan(1/2*a)^3*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c)^2 + tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c)
+ 5*tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)^3 - 5*tan(1/2*a)^2*tan(1/2*c) + 5*tan(1/
2*a)*tan(1/2*c)^2 - tan(1/2*c)^3 - tan(1/2*a)^2 + 5*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 - tan(1/2*a) + tan(1/
2*c) + 1)*log(abs(-tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c) + tan(1/2*b*x + 1/2*a)*tan(1/2*a) - tan(1/2*b*x
+ 1/2*a)*tan(1/2*c) + tan(1/2*a)*tan(1/2*c) - tan(1/2*b*x + 1/2*a) + tan(1/2*a) - tan(1/2*c) + 1))/(tan(1/2*a)
^3*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c)^2 + tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c) + tan(1/2*a)
^2*tan(1/2*c)^2 + tan(1/2*a)*tan(1/2*c)^3 - tan(1/2*a)^3 + tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 +
tan(1/2*c)^3 + tan(1/2*a)^2 + tan(1/2*a)*tan(1/2*c) + tan(1/2*c)^2 - tan(1/2*a) + tan(1/2*c) + 1) - (tan(1/2*
a)^3*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c)^2 - tan(1/2*a)^2*tan(1/2*c)^3 - tan(1/2*a)^3*tan(1/2*c) + 5*tan(1/
2*a)^2*tan(1/2*c)^2 - tan(1/2*a)*tan(1/2*c)^3 - tan(1/2*a)^3 + 5*tan(1/2*a)^2*tan(1/2*c) - 5*tan(1/2*a)*tan(1/
2*c)^2 + tan(1/2*c)^3 - tan(1/2*a)^2 + 5*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c) + 1)*l
og(abs(-tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c) - tan(1/2*b*x + 1/2*a)*tan(1/2*a) + tan(1/2*b*x + 1/2*a)*ta
n(1/2*c) - tan(1/2*a)*tan(1/2*c) - tan(1/2*b*x + 1/2*a) + tan(1/2*a) - tan(1/2*c) - 1))/(tan(1/2*a)^3*tan(1/2*
c)^3 + tan(1/2*a)^3*tan(1/2*c)^2 - tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)^3*tan(1/2*c) + tan(1/2*a)^2*tan(1/2*
c)^2 + tan(1/2*a)*tan(1/2*c)^3 + tan(1/2*a)^3 - tan(1/2*a)^2*tan(1/2*c) + tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*c)
^3 + tan(1/2*a)^2 + tan(1/2*a)*tan(1/2*c) + tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c) + 1) - 4*(2*tan(1/2*b*x + 1
/2*a)*tan(1/2*a)^4*tan(1/2*c)^2 - 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a)^3*tan(1/2*c)^3 + tan(1/2*a)^4*tan(1/2*c)^3
+ 2*tan(1/2*b*x + 1/2*a)*tan(1/2*a)^2*tan(1/2*c)^4 - tan(1/2*a)^3*tan(1/2*c)^4 + 4*tan(1/2*b*x + 1/2*a)*tan(1
/2*a)^3*tan(1/2*c) - tan(1/2*a)^4*tan(1/2*c) - 8*tan(1/2*b*x + 1/2*a)*tan(1/2*a)^2*tan(1/2*c)^2 + 6*tan(1/2*a)
^3*tan(1/2*c)^2 + 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c)^3 - 6*tan(1/2*a)^2*tan(1/2*c)^3 + tan(1/2*a)*ta
n(1/2*c)^4 + 2*tan(1/2*b*x + 1/2*a)*tan(1/2*a)^2 - tan(1/2*a)^3 - 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c)
+ 6*tan(1/2*a)^2*tan(1/2*c) + 2*tan(1/2*b*x + 1/2*a)*tan(1/2*c)^2 - 6*tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*c)^3
+ tan(1/2*a) - tan(1/2*c))/((tan(1/2*b*x + 1/2*a)^2*tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*b*x + 1/2*a)^2*tan(1/2
*a)^2 + 4*tan(1/2*b*x + 1/2*a)^2*tan(1/2*a)*tan(1/2*c) - 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a)^2*tan(1/2*c) - tan(
1/2*b*x + 1/2*a)^2*tan(1/2*c)^2 + 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a)*tan(1/2*c)^2 - tan(1/2*a)^2*tan(1/2*c)^2 +
tan(1/2*b*x + 1/2*a)^2 - 4*tan(1/2*b*x + 1/2*a)*tan(1/2*a) + tan(1/2*a)^2 + 4*tan(1/2*b*x + 1/2*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)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/
2*c) - tan(1/2*c)^2 + 1)))/b