3.553 \(\int \frac{e^x}{1-\cos (x)} \, dx\)

Optimal. Leaf size=26 \[ (-1+i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;e^{i x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0321949, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4458, 4453} \[ (-1+i) e^{(1+i) x} \text{Hypergeometric2F1}\left (1-i,2,2-i,e^{i x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(1 - Cos[x]),x]

[Out]

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

Rule 4458

Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[2^n*f^n,
Int[F^(c*(a + b*x))*Sin[d/2 + (e*x)/2]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f + g, 0] &
& ILtQ[n, 0]

Rule 4453

Int[Csc[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(-2*I)^n*E^(I*n*(d + e*
x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 - (I*b*c*Log[F])/(2*e), 1 + n/2 - (I*b*c*L
og[F])/(2*e), E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps

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

Mathematica [B]  time = 0.07335, size = 84, normalized size = 3.23 \[ \frac{(1+i) e^x \sin \left (\frac{x}{2}\right ) \left ((1+i) \, _2F_1\left (-i,1;1-i;e^{i x}\right ) \sin \left (\frac{x}{2}\right )+e^{i x} \, _2F_1\left (1,1-i;2-i;e^{i x}\right ) \sin \left (\frac{x}{2}\right )+(1-i) \cos \left (\frac{x}{2}\right )\right )}{\cos (x)-1} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(1 - Cos[x]),x]

[Out]

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

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Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{x}}}{1-\cos \left ( x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)/(1-cos(x)),x)

[Out]

int(exp(x)/(1-cos(x)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left ({\left (\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1\right )} \int \frac{e^{x} \sin \left (x\right )}{\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1}\,{d x} - e^{x} \sin \left (x\right )\right )}}{\cos \left (x\right )^{2} + \sin \left (x\right )^{2} - 2 \, \cos \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(1-cos(x)),x, algorithm="maxima")

[Out]

2*((cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)*integrate(e^x*sin(x)/(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1), x) - e^x*si
n(x))/(cos(x)^2 + sin(x)^2 - 2*cos(x) + 1)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{e^{x}}{\cos \left (x\right ) - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(1-cos(x)),x, algorithm="fricas")

[Out]

integral(-e^x/(cos(x) - 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{e^{x}}{\cos{\left (x \right )} - 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(1-cos(x)),x)

[Out]

-Integral(exp(x)/(cos(x) - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{e^{x}}{\cos \left (x\right ) - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)/(1-cos(x)),x, algorithm="giac")

[Out]

integrate(-e^x/(cos(x) - 1), x)