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(b*x+a)/b-arctanh(sin(b*x+c))*sin(a-c)/b

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Rubi [A]  time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.05, size = 93, normalized size = 3.10 \[ \frac {2 i \sin (a-c) \tan ^{-1}\left (\frac {(\sin (c)+i \cos (c)) \left (\sin (c) \cos \left (\frac {b x}{2}\right )+\cos (c) \sin \left (\frac {b x}{2}\right )\right )}{\cos (c) \cos \left (\frac {b x}{2}\right )-i \sin (c) \cos \left (\frac {b x}{2}\right )}\right )}{b}+\frac {\sin (a) \sin (b x)}{b}-\frac {\cos (a) \cos (b x)}{b} \]

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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fricas [B]  time = 0.47, size = 196, normalized size = 6.53 \[ \frac {\frac {\sqrt {2} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right ) \sin \left (-2 \, a + 2 \, c\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - 4 \, \cos \left (b x + a\right )}{4 \, b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (b x + a\right ) \tan \left (b x + c\right )\,{d x} \]

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)

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maple [C]  time = 0.61, size = 97, normalized size = 3.23 \[ -\frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b} \]

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))-ln(exp(I*(b*x+a))+I*exp(I*(a-c)))/b*sin(a-c)+ln(exp(I*(b*x+a))-I*e
xp(I*(a-c)))/b*sin(a-c)

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maxima [B]  time = 0.51, size = 131, normalized size = 4.37 \[ -\frac {\log \left (\frac {\cos \left (b x + 2 \, c\right )^{2} + \cos \relax (c)^{2} - 2 \, \cos \relax (c) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} + 2 \, \cos \left (b x + 2 \, c\right ) \sin \relax (c) + \sin \relax (c)^{2}}{\cos \left (b x + 2 \, c\right )^{2} + \cos \relax (c)^{2} + 2 \, \cos \relax (c) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} - 2 \, \cos \left (b x + 2 \, c\right ) \sin \relax (c) + \sin \relax (c)^{2}}\right ) \sin \left (-a + c\right ) + 2 \, \cos \left (b x + a\right )}{2 \, b} \]

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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mupad [B]  time = 4.73, size = 237, normalized size = 7.90 \[ -\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2\,b}-\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2\,b}+\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}}-\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*tan(c + b*x),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\left (a + b x \right )} \tan {\left (b x + c \right )}\, dx \]

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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