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

Optimal. Leaf size=41 \[ \frac{e^x \sin (x)}{1-\cos (x)}-(2-2 i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;e^{i x}\right ) \]

[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 = 0.113198, antiderivative size = 45, normalized size of antiderivative = 1.1, 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, 4461, 4443, 2194, 2251, 2288} \[ -4 i e^x \text{Hypergeometric2F1}\left (-i,1,1-i,e^{i x}\right )+2 i e^x-\frac{e^x \sin (x)}{1-\cos (x)} \]

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 4461

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))*Cot[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 4443

Int[Cot[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[(-I)^n, Int[ExpandInteg
rand[(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 &=2 \int \frac{e^x \sin (x)}{1-\cos (x)} \, dx+\int \frac{e^x (1-\sin (x))}{1-\cos (x)} \, dx\\ &=-\frac{e^x \sin (x)}{1-\cos (x)}+2 \int e^x \cot \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 = 0.208887, size = 100, normalized size = 2.44 \[ \frac{2 e^x \sin \left (\frac{x}{2}\right ) (\sin (x)+1) \left (2 i \, _2F_1\left (-i,1;1-i;e^{i x}\right ) \sin \left (\frac{x}{2}\right )+(1+i) e^{i x} \, _2F_1\left (1,1-i;2-i;e^{i x}\right ) \sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )}{(\cos (x)-1) \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{\frac{{{\rm e}^{x}} \left ( 1+\sin \left ( x \right ) \right ) }{1-\cos \left ( x \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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