3.6.60 \(\int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx\) [560]

Optimal. Leaf size=46 \[ (2+2 i) e^{(1+i) x} \, _2F_1\left (1-i,2;2-i;-i e^{i x}\right )-\frac {e^x \cos (x)}{1-\sin (x)} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 48, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4550, 4547, 4527, 2225, 2283, 2326} \begin {gather*} -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)}{1-\sin (x)} \end {gather*}

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 2225

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 2283

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_.) + (g_.)*(x_))), x_Symbol] :> Simp
[a^p*(G^(h*(f + g*x))/(g*h*Log[G]))*Hypergeometric2F1[-p, g*h*(Log[G]/(d*e*Log[F])), g*h*(Log[G]/(d*e*Log[F]))
+ 1, Simplify[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, p}, x] && (ILtQ[p, 0] || G
tQ[a, 0])

Rule 2326

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]

Rule 4527

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 4547

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 4550

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]

Rubi steps

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

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Mathematica [A]
time = 0.44, size = 72, normalized size = 1.57 \begin {gather*} \frac {1}{2} (-1+\cos (x)) \csc ^2\left (\frac {x}{2}\right ) \left (-\frac {e^x \left ((1-2 i)+(1+2 i) \cot \left (\frac {x}{2}\right )\right )}{-1+\cot \left (\frac {x}{2}\right )}+4 i \, _2F_1(-i,1;1-i;-i \cos (x)+\sin (x)) (\cosh (x)+\sinh (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {{\mathrm e}^{x} \left (1-\cos \left (x \right )\right )}{1-\sin \left (x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(x)*(1-cos(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-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 = 0.00, 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-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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\cos {\left (x \right )} - 1\right ) 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-cos(x))/(1-sin(x)),x)

[Out]

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

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Giac [F]
time = 0.00, 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-cos(x))/(1-sin(x)),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {e}}^x\,\left (\cos \left (x\right )-1\right )}{\sin \left (x\right )-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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