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

Optimal. Leaf size=43 \[ \frac{e^x \cos (x)}{\sin (x)+1}-(2+2 i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;i e^{i x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.127377, antiderivative size = 47, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {4462, 4459, 4442, 2194, 2251, 2288} \[ 4 i e^x \text{Hypergeometric2F1}\left (i,1,1+i,-i e^{-i x}\right )-2 i e^x-\frac{e^x \cos (x)}{\sin (x)+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4462

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_)))*(Cos[(d_.) + (e_.)*(x_)]*(i_.) + (h_)))/((f_) + (g_.)*Sin[(d_.) + (e_.)
*(x_)]), x_Symbol] :> Dist[2*i, Int[F^(c*(a + b*x))*(Cos[d + e*x]/(f + g*Sin[d + e*x])), x], x] + Int[F^(c*(a
+ b*x))*((h - i*Cos[d + e*x])/(f + g*Sin[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 4459

Int[Cos[(d_.) + (e_.)*(x_)]^(m_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_
.), x_Symbol] :> Dist[g^n, Int[F^(c*(a + b*x))*Tan[(f*Pi)/(4*g) - d/2 - (e*x)/2]^m, x], x] /; FreeQ[{F, a, b,
c, d, e, f, g}, x] && EqQ[f^2 - g^2, 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+\cos (x))}{1+\sin (x)} \, dx &=2 \int \frac{e^x \cos (x)}{1+\sin (x)} \, dx+\int \frac{e^x (1-\cos (x))}{1+\sin (x)} \, dx\\ &=-\frac{e^x \cos (x)}{1+\sin (x)}+2 \int e^x \tan \left (\frac{\pi }{4}-\frac{x}{2}\right ) \, dx\\ &=-\frac{e^x \cos (x)}{1+\sin (x)}+2 i \int \left (-e^x+\frac{2 e^x}{1+e^{2 i \left (\frac{\pi }{4}-\frac{x}{2}\right )}}\right ) \, dx\\ &=-\frac{e^x \cos (x)}{1+\sin (x)}-2 i \int e^x \, dx+4 i \int \frac{e^x}{1+e^{2 i \left (\frac{\pi }{4}-\frac{x}{2}\right )}} \, dx\\ &=-2 i e^x+4 i e^x \, _2F_1\left (i,1;1+i;-i e^{-i x}\right )-\frac{e^x \cos (x)}{1+\sin (x)}\\ \end{align*}

Mathematica [A]  time = 0.186466, size = 73, normalized size = 1.7 \[ \frac{1}{2} (\cos (x)+1) \sec ^2\left (\frac{x}{2}\right ) \left (\frac{e^x \left ((1+2 i) \tan \left (\frac{x}{2}\right )-(1-2 i)\right )}{\tan \left (\frac{x}{2}\right )+1}-4 i (\sinh (x)+\cosh (x)) \, _2F_1(-i,1;1-i;i \cos (x)-\sin (x))\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{x}} \left ( \cos \left ( x \right ) +1 \right ) }{1+\sin \left ( x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(x)*(cos(x)+1)/(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} - 2 \,{\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+cos(x))/(1+sin(x)),x, algorithm="maxima")

[Out]

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

[Out]

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

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

[Out]

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

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

[Out]

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