3.79 \(\int \sin (x) \tan (n x) \, dx\)

Optimal. Leaf size=105 \[ -i e^{-i x} \text{Hypergeometric2F1}\left (1,-\frac{1}{2 n},1-\frac{1}{2 n},-e^{2 i n x}\right )-i e^{i x} \text{Hypergeometric2F1}\left (1,\frac{1}{2 n},\frac{1}{2} \left (\frac{1}{n}+2\right ),-e^{2 i n x}\right )+\frac{1}{2} i e^{-i x}+\frac{1}{2} i e^{i x} \]

[Out]

(I/2)/E^(I*x) + (I/2)*E^(I*x) - (I*Hypergeometric2F1[1, -1/(2*n), 1 - 1/(2*n), -E^((2*I)*n*x)])/E^(I*x) - I*E^
(I*x)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -E^((2*I)*n*x)]

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Rubi [A]  time = 0.0773561, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4557, 2194, 2251} \[ -i e^{-i x} \, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 i n x}\right )-i e^{i x} \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-e^{2 i n x}\right )+\frac{1}{2} i e^{-i x}+\frac{1}{2} i e^{i x} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]*Tan[n*x],x]

[Out]

(I/2)/E^(I*x) + (I/2)*E^(I*x) - (I*Hypergeometric2F1[1, -1/(2*n), 1 - 1/(2*n), -E^((2*I)*n*x)])/E^(I*x) - I*E^
(I*x)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -E^((2*I)*n*x)]

Rule 4557

Int[Sin[(a_.) + (b_.)*(x_)]*Tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Int[1/(E^(I*(a + b*x))*2) - E^(I*(a + b*x))/
2 - 1/(E^(I*(a + b*x))*(1 + E^(2*I*(c + d*x)))) + 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]

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 \sin (x) \tan (n x) \, dx &=\int \left (\frac{e^{-i x}}{2}-\frac{e^{i x}}{2}-\frac{e^{-i x}}{1+e^{2 i n x}}+\frac{e^{i x}}{1+e^{2 i n x}}\right ) \, dx\\ &=\frac{1}{2} \int e^{-i x} \, dx-\frac{1}{2} \int e^{i x} \, dx-\int \frac{e^{-i x}}{1+e^{2 i n x}} \, dx+\int \frac{e^{i x}}{1+e^{2 i n x}} \, dx\\ &=\frac{1}{2} i e^{-i x}+\frac{1}{2} i e^{i x}-i e^{-i x} \, _2F_1\left (1,-\frac{1}{2 n};1-\frac{1}{2 n};-e^{2 i n x}\right )-i e^{i x} \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-e^{2 i n x}\right )\\ \end{align*}

Mathematica [A]  time = 0.164693, size = 200, normalized size = 1.9 \[ -\frac{i e^{-2 i x} \left ((2 n+1) e^{i (2 n x+x)} \text{Hypergeometric2F1}\left (1,1-\frac{1}{2 n},2-\frac{1}{2 n},-e^{2 i n x}\right )+(2 n-1) \left ((2 n+1) e^{i x} \left (\text{Hypergeometric2F1}\left (1,-\frac{1}{2 n},1-\frac{1}{2 n},-e^{2 i n x}\right )+e^{2 i x} \text{Hypergeometric2F1}\left (1,\frac{1}{2 n},\frac{1}{2 n}+1,-e^{2 i n x}\right )\right )-e^{i (2 n+3) x} \text{Hypergeometric2F1}\left (1,\frac{1}{2 n}+1,\frac{1}{2 n}+2,-e^{2 i n x}\right )\right )\right )}{2 \left (4 n^2-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]*Tan[n*x],x]

[Out]

((-I/2)*(E^(I*(x + 2*n*x))*(1 + 2*n)*Hypergeometric2F1[1, 1 - 1/(2*n), 2 - 1/(2*n), -E^((2*I)*n*x)] + (-1 + 2*
n)*(-(E^(I*(3 + 2*n)*x)*Hypergeometric2F1[1, 1 + 1/(2*n), 2 + 1/(2*n), -E^((2*I)*n*x)]) + E^(I*x)*(1 + 2*n)*(H
ypergeometric2F1[1, -1/(2*n), 1 - 1/(2*n), -E^((2*I)*n*x)] + E^((2*I)*x)*Hypergeometric2F1[1, 1/(2*n), 1 + 1/(
2*n), -E^((2*I)*n*x)]))))/(E^((2*I)*x)*(-1 + 4*n^2))

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Maple [F]  time = 0.174, size = 0, normalized size = 0. \begin{align*} \int \sin \left ( x \right ) \tan \left ( nx \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)*tan(n*x),x)

[Out]

int(sin(x)*tan(n*x),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (x\right ) \tan \left (n x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(n*x),x, algorithm="maxima")

[Out]

integrate(sin(x)*tan(n*x), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sin \left (x\right ) \tan \left (n x\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(n*x),x, algorithm="fricas")

[Out]

integral(sin(x)*tan(n*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin{\left (x \right )} \tan{\left (n x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(n*x),x)

[Out]

Integral(sin(x)*tan(n*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sin \left (x\right ) \tan \left (n x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(n*x),x, algorithm="giac")

[Out]

integrate(sin(x)*tan(n*x), x)