Integrand size = 18, antiderivative size = 46 \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=(2+2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-i e^{i x}\right )-\frac {e^x \cos (x)}{1-\sin (x)} \] Output:
(2+2*I)*exp((1+I)*x)*hypergeom([2, 1-I],[2-I],-I*exp(I*x))-exp(x)*cos(x)/( 1-sin(x))
Time = 0.67 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.57 \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\frac {1}{2} (-1+\cos (x)) \csc ^2\left (\frac {x}{2}\right ) \left (-\frac {e^x \left ((1-2 i)+(1+2 i) \cot \left (\frac {x}{2}\right )\right )}{-1+\cot \left (\frac {x}{2}\right )}+4 i \operatorname {Hypergeometric2F1}(-i,1,1-i,-i \cos (x)+\sin (x)) (\cosh (x)+\sinh (x))\right ) \] Input:
Integrate[(E^x*(1 - Cos[x]))/(1 - Sin[x]),x]
Output:
((-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
Time = 0.39 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4965, 2726, 4962, 25, 4942, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx\) |
| \(\Big \downarrow \) 4965 |
| \(\displaystyle \int \frac {e^x (\cos (x)+1)}{1-\sin (x)}dx-2 \int \frac {e^x \cos (x)}{1-\sin (x)}dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {e^x \cos (x)}{1-\sin (x)}-2 \int \frac {e^x \cos (x)}{1-\sin (x)}dx\) |
| \(\Big \downarrow \) 4962 |
| \(\displaystyle 2 \int -e^x \tan \left (\frac {x}{2}+\frac {\pi }{4}\right )dx+\frac {e^x \cos (x)}{1-\sin (x)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {e^x \cos (x)}{1-\sin (x)}-2 \int e^x \tan \left (\frac {x}{2}+\frac {\pi }{4}\right )dx\) |
| \(\Big \downarrow \) 4942 |
| \(\displaystyle \frac {e^x \cos (x)}{1-\sin (x)}-2 i \int \left (\frac {2 e^x}{1+e^{\frac {1}{2} i (2 x+\pi )}}-e^x\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^x \cos (x)}{1-\sin (x)}-2 i \left (-e^x+2 e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-i e^{i x}\right )\right )\) |
Input:
Int[(E^x*(1 - Cos[x]))/(1 - Sin[x]),x]
Output:
(-2*I)*(-E^x + 2*E^x*Hypergeometric2F1[-I, 1, 1 - I, (-I)*E^(I*x)]) + (E^x *Cos[x])/(1 - Sin[x])
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]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Tan[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[I^n Int[ExpandIntegrand[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]
Int[Cos[(d_.) + (e_.)*(x_)]^(m_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)])^(n_.), x_Symbol] :> Simp[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]
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_)))*(Cos[(d_.) + (e_.)*(x_)]*(i_.) + (h_ )))/((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)]), x_Symbol] :> Simp[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]
\[\int \frac {{\mathrm e}^{x} \left (1-\cos \left (x \right )\right )}{-\sin \left (x \right )+1}d x\]
Input:
int(exp(x)*(1-cos(x))/(-sin(x)+1),x)
Output:
int(exp(x)*(1-cos(x))/(-sin(x)+1),x)
\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \] Input:
integrate(exp(x)*(1-cos(x))/(1-sin(x)),x, algorithm="fricas")
Output:
integral((cos(x) - 1)*e^x/(sin(x) - 1), x)
\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int \frac {\left (\cos {\left (x \right )} - 1\right ) e^{x}}{\sin {\left (x \right )} - 1}\, dx \] Input:
integrate(exp(x)*(1-cos(x))/(1-sin(x)),x)
Output:
Integral((cos(x) - 1)*exp(x)/(sin(x) - 1), x)
\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \] Input:
integrate(exp(x)*(1-cos(x))/(1-sin(x)),x, algorithm="maxima")
Output:
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)
\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int { \frac {{\left (\cos \left (x\right ) - 1\right )} e^{x}}{\sin \left (x\right ) - 1} \,d x } \] Input:
integrate(exp(x)*(1-cos(x))/(1-sin(x)),x, algorithm="giac")
Output:
integrate((cos(x) - 1)*e^x/(sin(x) - 1), x)
Timed out. \[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (\cos \left (x\right )-1\right )}{\sin \left (x\right )-1} \,d x \] Input:
int((exp(x)*(cos(x) - 1))/(sin(x) - 1),x)
Output:
int((exp(x)*(cos(x) - 1))/(sin(x) - 1), x)
\[ \int \frac {e^x (1-\cos (x))}{1-\sin (x)} \, dx=e^{x}-\left (\int \frac {e^{x}}{\sin \left (x \right )-1}d x \right )+2 \left (\int \frac {e^{x}}{\tan \left (\frac {x}{2}\right )-1}d x \right ) \] Input:
int(exp(x)*(1-cos(x))/(1-sin(x)),x)
Output:
e**x - int(e**x/(sin(x) - 1),x) + 2*int(e**x/(tan(x/2) - 1),x)