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

Optimal. Leaf size=134 \[ -\frac{e^{-i (a+b x)} \text{Hypergeometric2F1}\left (1,-\frac{b}{2 d},1-\frac{b}{2 d},-e^{2 i (c+d x)}\right )}{b}+\frac{e^{i (a+b x)} \text{Hypergeometric2F1}\left (1,\frac{b}{2 d},\frac{b}{2 d}+1,-e^{2 i (c+d x)}\right )}{b}+\frac{e^{-i (a+b x)}}{2 b}-\frac{e^{i (a+b x)}}{2 b} \]

[Out]

1/(2*b*E^(I*(a + b*x))) - E^(I*(a + b*x))/(2*b) - Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), -E^((2*I)*(c + d
*x))]/(b*E^(I*(a + b*x))) + (E^(I*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), -E^((2*I)*(c + d*x))])
/b

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Rubi [A]  time = 0.119283, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4560, 2194, 2251} \[ -\frac{e^{-i (a+b x)} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 i (c+d x)}\right )}{b}+\frac{e^{i (a+b x)} \, _2F_1\left (1,\frac{b}{2 d};\frac{b}{2 d}+1;-e^{2 i (c+d x)}\right )}{b}+\frac{e^{-i (a+b x)}}{2 b}-\frac{e^{i (a+b x)}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Tan[c + d*x],x]

[Out]

1/(2*b*E^(I*(a + b*x))) - E^(I*(a + b*x))/(2*b) - Hypergeometric2F1[1, -b/(2*d), 1 - b/(2*d), -E^((2*I)*(c + d
*x))]/(b*E^(I*(a + b*x))) + (E^(I*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), -E^((2*I)*(c + d*x))])
/b

Rule 4560

Int[Cos[(a_.) + (b_.)*(x_)]*Tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[-(I/(E^(I*(a + b*x))*2)) - (I*E^(I*(a +
b*x)))/2 + I/(E^(I*(a + b*x))*(1 + E^(2*I*(c + d*x)))) + (I*E^(I*(a + b*x)))/(1 + E^(2*I*(c + d*x))), x] /; Fr
eeQ[{a, b, c, d}, x] && NeQ[b^2 - d^2, 0]

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2251

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[(a^p*G^(h*(f + g*x))*Hypergeometric2F1[-p, (g*h*Log[G])/(d*e*Log[F]), (g*h*Log[G])/(d*e*Log[F]) + 1, Simplify
[-((b*F^(e*(c + d*x)))/a)]])/(g*h*Log[G]), x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] ||
GtQ[a, 0])

Rubi steps

\begin{align*} \int \cos (a+b x) \tan (c+d x) \, dx &=\int \left (-\frac{1}{2} i e^{-i (a+b x)}-\frac{1}{2} i e^{i (a+b x)}+\frac{i e^{-i (a+b x)}}{1+e^{2 i (c+d x)}}+\frac{i e^{i (a+b x)}}{1+e^{2 i (c+d x)}}\right ) \, dx\\ &=-\left (\frac{1}{2} i \int e^{-i (a+b x)} \, dx\right )-\frac{1}{2} i \int e^{i (a+b x)} \, dx+i \int \frac{e^{-i (a+b x)}}{1+e^{2 i (c+d x)}} \, dx+i \int \frac{e^{i (a+b x)}}{1+e^{2 i (c+d x)}} \, dx\\ &=\frac{e^{-i (a+b x)}}{2 b}-\frac{e^{i (a+b x)}}{2 b}-\frac{e^{-i (a+b x)} \, _2F_1\left (1,-\frac{b}{2 d};1-\frac{b}{2 d};-e^{2 i (c+d x)}\right )}{b}+\frac{e^{i (a+b x)} \, _2F_1\left (1,\frac{b}{2 d};1+\frac{b}{2 d};-e^{2 i (c+d x)}\right )}{b}\\ \end{align*}

Mathematica [A]  time = 1.8844, size = 142, normalized size = 1.06 \[ \frac{e^{-i (a+b x)} \left ((b-2 d) \left (2 e^{2 i (a+b x)} \text{Hypergeometric2F1}\left (1,\frac{b}{2 d},\frac{b}{2 d}+1,-e^{2 i (c+d x)}\right )-e^{2 i (a+b x)}-1\right )+2 b e^{2 i (c+d x)} \text{Hypergeometric2F1}\left (1,1-\frac{b}{2 d},2-\frac{b}{2 d},-e^{2 i (c+d x)}\right )\right )}{2 b (b-2 d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*b*E^((2*I)*(c + d*x))*Hypergeometric2F1[1, 1 - b/(2*d), 2 - b/(2*d), -E^((2*I)*(c + d*x))] + (b - 2*d)*(-1
- E^((2*I)*(a + b*x)) + 2*E^((2*I)*(a + b*x))*Hypergeometric2F1[1, b/(2*d), 1 + b/(2*d), -E^((2*I)*(c + d*x))]
))/(2*b*(b - 2*d)*E^(I*(a + b*x)))

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Maple [F]  time = 0.175, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( bx+a \right ) \tan \left ( dx+c \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(cos(b*x + a)*tan(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cos(a + b*x)*tan(c + d*x), 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 (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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