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

Optimal. Leaf size=134 \[ -\frac {e^{-i (a+b x)} \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )}{b}+\frac {e^{i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;-e^{2 i (c+d x)}\right )}{b}+\frac {e^{-i (a+b x)}}{2 b}-\frac {e^{i (a+b x)}}{2 b} \]

[Out]

1/2/b/exp(I*(b*x+a))-1/2*exp(I*(b*x+a))/b-hypergeom([1, -1/2*b/d],[1-1/2*b/d],-exp(2*I*(d*x+c)))/b/exp(I*(b*x+
a))+exp(I*(b*x+a))*hypergeom([1, 1/2*b/d],[1+1/2*b/d],-exp(2*I*(d*x+c)))/b

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Rubi [A]  time = 0.12, antiderivative size = 134, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 1.66, size = 114, normalized size = 0.85 \[ \frac {e^{-i (a+b x)} \left (2 e^{2 i (a+b x)} \, _2F_1\left (1,\frac {b}{2 d};\frac {b}{2 d}+1;-e^{2 i (c+d x)}\right )-e^{2 i (a+b x)}-2 \, _2F_1\left (1,-\frac {b}{2 d};1-\frac {b}{2 d};-e^{2 i (c+d x)}\right )+1\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cos \left (b x + a\right ) \tan \left (d x + c\right ), x\right ) \]

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

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)

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maple [F]  time = 1.59, size = 0, normalized size = 0.00 \[ \int \cos \left (b x +a \right ) \tan \left (d x +c \right )\, dx \]

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

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,\mathrm {tan}\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*tan(c + d*x),x)

[Out]

int(cos(a + b*x)*tan(c + d*x), x)

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

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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