3.559 \(\int \frac{e^x (1-\sin (x))}{1+\cos (x)} \, dx\)

Optimal. Leaf size=42 \[ -\frac{e^x \sin (x)}{\cos (x)+1}+(2-2 i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;-e^{i x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.111279, antiderivative size = 44, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4463, 4460, 4442, 2194, 2251, 2288} \[ -4 i e^x \text{Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )+2 i e^x+\frac{e^x \sin (x)}{\cos (x)+1} \]

Antiderivative was successfully verified.

[In]

Int[(E^x*(1 - Sin[x]))/(1 + Cos[x]),x]

[Out]

(2*I)*E^x - (4*I)*E^x*Hypergeometric2F1[-I, 1, 1 - I, -E^(I*x)] + (E^x*Sin[x])/(1 + Cos[x])

Rule 4463

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

Rule 4460

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

Rule 4442

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

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])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

Mathematica [B]  time = 0.213225, size = 87, normalized size = 2.07 \[ -\frac{2 e^x \cos \left (\frac{x}{2}\right ) \left (2 i \, _2F_1\left (-i,1;1-i;-e^{i x}\right ) \cos \left (\frac{x}{2}\right )-(1+i) e^{i x} \, _2F_1\left (1,1-i;2-i;-e^{i x}\right ) \cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )}{\cos (x)+1} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 - Sin[x]))/(1 + Cos[x]),x]

[Out]

(-2*E^x*Cos[x/2]*((2*I)*Cos[x/2]*Hypergeometric2F1[-I, 1, 1 - I, -E^(I*x)] - (1 + I)*E^(I*x)*Cos[x/2]*Hypergeo
metric2F1[1, 1 - I, 2 - I, -E^(I*x)] - Sin[x/2]))/(1 + Cos[x])

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Maple [F]  time = 0.063, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{x}} \left ( 1-\sin \left ( x \right ) \right ) }{\cos \left ( x \right ) +1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*(1-sin(x))/(cos(x)+1),x)

[Out]

int(exp(x)*(1-sin(x))/(cos(x)+1),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{2 \,{\left (2 \,{\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-sin(x))/(1+cos(x)),x, algorithm="maxima")

[Out]

-2*(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
*sin(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} \sin \left (x\right ) - 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-sin(x))/(1+cos(x)),x, algorithm="fricas")

[Out]

integral(-(e^x*sin(x) - 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 - \int \frac{e^{x} \sin{\left (x \right )}}{\cos{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*(1-sin(x))/(1+cos(x)),x)

[Out]

-Integral(-exp(x)/(cos(x) + 1), x) - Integral(exp(x)*sin(x)/(cos(x) + 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (\sin \left (x\right ) - 1\right )} 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-sin(x))/(1+cos(x)),x, algorithm="giac")

[Out]

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