3.3.53 \(\int \cos (a+b x) \tan (c+d x) \, dx\) [253]

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

[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 = 0.08, 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 = {4656, 2225, 2283} \begin {gather*} -\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} \end {gather*}

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, -1/2*b/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 2225

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 2283

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

Rule 4656

Int[Cos[(a_.) + (b_.)*(x_)]*Tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[(-I)*(1/(E^(I*(a + b*x))*2)) - I*(E^(I*(
a + b*x))/2) + I*(1/(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] /; FreeQ[{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 = 1.71, size = 197, normalized size = 1.47 \begin {gather*} \frac {e^{-i (a-2 c+b x)} \left (b (b+2 d) e^{2 i d x} \left (1+e^{2 i c}\right ) \, _2F_1\left (1,1-\frac {b}{2 d};2-\frac {b}{2 d};-e^{2 i (c+d x)}\right )+(b-2 d) \left ((b+2 d) \left (-1+e^{2 i (a+b x)}\right )-b e^{2 i (a+(b+d) x)} \left (1+e^{2 i c}\right ) \, _2F_1\left (1,1+\frac {b}{2 d};2+\frac {b}{2 d};-e^{2 i (c+d x)}\right )\right )\right )}{\left (b^3-4 b d^2\right ) \left (1+e^{2 i c}\right )}+\frac {\sin (a+b x) \tan (c)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(b + 2*d)*E^((2*I)*d*x)*(1 + E^((2*I)*c))*Hypergeometric2F1[1, 1 - b/(2*d), 2 - b/(2*d), -E^((2*I)*(c + d*x
))] + (b - 2*d)*((b + 2*d)*(-1 + E^((2*I)*(a + b*x))) - b*E^((2*I)*(a + (b + d)*x))*(1 + E^((2*I)*c))*Hypergeo
metric2F1[1, 1 + b/(2*d), 2 + b/(2*d), -E^((2*I)*(c + d*x))]))/((b^3 - 4*b*d^2)*E^(I*(a - 2*c + b*x))*(1 + E^(
(2*I)*c))) + (Sin[a + b*x]*Tan[c])/b

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Maple [F]
time = 0.04, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \cos {\left (a + b x \right )} \tan {\left (c + d x \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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