3.249 \(\int \cos (a+b x) \tan (c+b x) \, dx\)

Optimal. Leaf size=30 \[ -\frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}-\frac{\cos (a+b x)}{b} \]

[Out]

-(Cos[a + b*x]/b) - (ArcTanh[Sin[c + b*x]]*Sin[a - c])/b

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Rubi [A]  time = 0.0176683, antiderivative size = 30, 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 = {4579, 2638, 3770} \[ -\frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}-\frac{\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Tan[c + b*x],x]

[Out]

-(Cos[a + b*x]/b) - (ArcTanh[Sin[c + b*x]]*Sin[a - c])/b

Rule 4579

Int[Cos[v_]*Tan[w_]^(n_.), x_Symbol] :> Int[Sin[v]*Tan[w]^(n - 1), x] - Dist[Sin[v - w], Int[Sec[w]*Tan[w]^(n
- 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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) \tan (c+b x) \, dx &=-(\sin (a-c) \int \sec (c+b x) \, dx)+\int \sin (a+b x) \, dx\\ &=-\frac{\cos (a+b x)}{b}-\frac{\tanh ^{-1}(\sin (c+b x)) \sin (a-c)}{b}\\ \end{align*}

Mathematica [C]  time = 0.0595658, size = 93, normalized size = 3.1 \[ \frac{2 i \sin (a-c) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\sin (c) \cos \left (\frac{b x}{2}\right )+\cos (c) \sin \left (\frac{b x}{2}\right )\right )}{\cos (c) \cos \left (\frac{b x}{2}\right )-i \sin (c) \cos \left (\frac{b x}{2}\right )}\right )}{b}+\frac{\sin (a) \sin (b x)}{b}-\frac{\cos (a) \cos (b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Tan[c + b*x],x]

[Out]

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

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Maple [C]  time = 0.066, size = 97, normalized size = 3.2 \begin{align*} -{\frac{{{\rm e}^{i \left ( bx+a \right ) }}}{2\,b}}-{\frac{{{\rm e}^{-i \left ( bx+a \right ) }}}{2\,b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-i{{\rm e}^{i \left ( a-c \right ) }} \right ) \sin \left ( a-c \right ) }{b}}-{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+i{{\rm e}^{i \left ( a-c \right ) }} \right ) \sin \left ( a-c \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*tan(b*x+c),x)

[Out]

-1/2*exp(I*(b*x+a))/b-1/2/b*exp(-I*(b*x+a))+1/b*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))*sin(a-c)-1/b*ln(exp(I*(b*x+a
))+I*exp(I*(a-c)))*sin(a-c)

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Maxima [B]  time = 1.86161, size = 177, normalized size = 5.9 \begin{align*} -\frac{\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 ) \sin \left (-a + c\right ) + 2 \, \cos \left (b x + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.53063, size = 531, normalized size = 17.7 \begin{align*} \frac{\frac{\sqrt{2} \log \left (\frac{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac{2 \, \sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right ) \sin \left (-2 \, a + 2 \, c\right )}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}} - 4 \, \cos \left (b x + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*log((2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - 2*sqrt(2)*(
(cos(-2*a + 2*c) + 1)*sin(b*x + a) + cos(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c)
- 3)/(2*cos(b*x + a)^2*cos(-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) + 1))*
sin(-2*a + 2*c)/sqrt(cos(-2*a + 2*c) + 1) - 4*cos(b*x + a))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*tan(b*x+c),x)

[Out]

Integral(cos(a + b*x)*tan(b*x + c), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (b x + a\right ) \tan \left (b x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(b*x + a)*tan(b*x + c), x)