3.6.55 \(\int \frac {e^x}{1-\sin (x)} \, dx\) [555]

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)*exp((1+I)*x)*hypergeom([2, 1-I],[2-I],-I*exp(I*x))

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Rubi [A]
time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, 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 = {4541, 4535} \begin {gather*} (1+i) e^{(1+i) x} \text {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right ) \end {gather*}

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 4535

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))/(I*e*n + b*c*Log[F]))*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))], x] /; FreeQ[{F, a, b, c, d, e}, x] && Int
egerQ[4*k] && IntegerQ[n]

Rule 4541

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 - f*(Pi/(4*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] &&
EqQ[f^2 - g^2, 0] && ILtQ[n, 0]

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*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(61\) vs. \(2(30)=60\).
time = 0.36, size = 61, normalized size = 2.03 \begin {gather*} \frac {2 e^x \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}+(1+i) (1-(1+i) \, _2F_1(-i,1;1-i;-i \cos (x)+\sin (x))) (\cosh (x)+\sinh (x)) \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{x}}{1-\sin \left (x \right )}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(exp(x)/(1-sin(x)),x, algorithm="fricas")

[Out]

integral(-e^x/(sin(x) - 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {e^{x}}{\sin {\left (x \right )} - 1}\, dx \end {gather*}

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 = 0.00, 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(exp(x)/(1-sin(x)),x, algorithm="giac")

[Out]

integrate(-e^x/(sin(x) - 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {{\mathrm {e}}^x}{\sin \left (x\right )-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(x)/(sin(x) - 1),x)

[Out]

-int(exp(x)/(sin(x) - 1), x)

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