3.555 \(\int \frac{e^x}{1-\sin (x)} \, dx\)

Optimal. Leaf size=30 \[ (1+i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;-i e^{i x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0340965, antiderivative size = 30, 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 = {4456, 4450} \[ (1+i) e^{(1+i) x} \text{Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[E^x/(1 - Sin[x]),x]

[Out]

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

Rule 4456

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Dist[2^n*f^n,
Int[F^(c*(a + b*x))*Cos[d/2 + (e*x)/2 - (f*Pi)/(4*g)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &&
EqQ[f^2 - g^2, 0] && ILtQ[n, 0]

Rule 4450

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + Pi*(k_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[(2^n*E^(I*k*
n*Pi)*E^(I*n*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[n, n/2 - (I*b*c*Log[F])/(2*e), 1 + n/2 - (I*b*c*Log[
F])/(2*e), -(E^(2*I*k*Pi)*E^(2*I*(d + e*x)))])/(I*e*n + b*c*Log[F]), x] /; FreeQ[{F, a, b, c, d, e}, x] && Int
egerQ[4*k] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{e^x}{1-\sin (x)} \, dx &=\frac{1}{2} \int e^x \sec ^2\left (\frac{\pi }{4}+\frac{x}{2}\right ) \, dx\\ &=(1+i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;-i e^{i x}\right )\\ \end{align*}

Mathematica [B]  time = 0.62303, size = 61, normalized size = 2.03 \[ \frac{2 e^x \sin \left (\frac{x}{2}\right )}{\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )}+(1+i) (\sinh (x)+\cosh (x)) (1-(1+i) \, _2F_1(-i,1;1-i;\sin (x)-i \cos (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x/(1 - Sin[x]),x]

[Out]

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

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{x}}}{1-\sin \left ( x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-e^x/(sin(x) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(exp(x)/(sin(x) - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(-e^x/(sin(x) - 1), x)