Integrand size = 16, antiderivative size = 42 \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=(2-2 i) e^{(1+i) x} \operatorname {Hypergeometric2F1}\left (1-i,2,2-i,-e^{i x}\right )-\frac {e^x \sin (x)}{1+\cos (x)} \] Output:
(2-2*I)*exp((1+I)*x)*hypergeom([2, 1-I],[2-I],-exp(I*x))-exp(x)*sin(x)/(1+ cos(x))
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(87\) vs. \(2(42)=84\).
Time = 0.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 2.07 \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=-\frac {2 e^x \cos \left (\frac {x}{2}\right ) \left (2 i \cos \left (\frac {x}{2}\right ) \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )-(1+i) e^{i x} \cos \left (\frac {x}{2}\right ) \operatorname {Hypergeometric2F1}\left (1,1-i,2-i,-e^{i x}\right )-\sin \left (\frac {x}{2}\right )\right )}{1+\cos (x)} \] Input:
Integrate[(E^x*(1 - Sin[x]))/(1 + Cos[x]),x]
Output:
(-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]*Hypergeometric2F1[1, 1 - I, 2 - I, -E^(I*x)] - Sin[x/2]))/(1 + Cos[x])
Time = 0.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {4966, 2726, 4963, 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.
\(\displaystyle \int \frac {e^x (1-\sin (x))}{\cos (x)+1} \, dx\) |
\(\Big \downarrow \) 4966 |
\(\displaystyle \int \frac {e^x (\sin (x)+1)}{\cos (x)+1}dx-2 \int \frac {e^x \sin (x)}{\cos (x)+1}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^x \sin (x)}{\cos (x)+1}-2 \int \frac {e^x \sin (x)}{\cos (x)+1}dx\) |
\(\Big \downarrow \) 4963 |
\(\displaystyle \frac {e^x \sin (x)}{\cos (x)+1}-2 \int e^x \tan \left (\frac {x}{2}\right )dx\) |
\(\Big \downarrow \) 4942 |
\(\displaystyle \frac {e^x \sin (x)}{\cos (x)+1}-2 i \int \left (\frac {2 e^x}{1+e^{i x}}-e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^x \sin (x)}{\cos (x)+1}-2 i \left (-e^x+2 e^x \operatorname {Hypergeometric2F1}\left (-i,1,1-i,-e^{i x}\right )\right )\) |
Input:
Int[(E^x*(1 - Sin[x]))/(1 + Cos[x]),x]
Output:
(-2*I)*(-E^x + 2*E^x*Hypergeometric2F1[-I, 1, 1 - I, -E^(I*x)]) + (E^x*Sin [x])/(1 + Cos[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_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.) *(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Simp[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]
Int[((F_)^((c_.)*((a_.) + (b_.)*(x_)))*((h_) + (i_.)*Sin[(d_.) + (e_.)*(x_) ]))/(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_)), x_Symbol] :> Simp[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]
\[\int \frac {{\mathrm e}^{x} \left (-\sin \left (x \right )+1\right )}{1+\cos \left (x \right )}d x\]
Input:
int(exp(x)*(-sin(x)+1)/(1+cos(x)),x)
Output:
int(exp(x)*(-sin(x)+1)/(1+cos(x)),x)
\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \] Input:
integrate(exp(x)*(1-sin(x))/(1+cos(x)),x, algorithm="fricas")
Output:
integral(-(e^x*sin(x) - e^x)/(cos(x) + 1), x)
\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=- \int \left (- \frac {e^{x}}{\cos {\left (x \right )} + 1}\right )\, dx - \int \frac {e^{x} \sin {\left (x \right )}}{\cos {\left (x \right )} + 1}\, dx \] Input:
integrate(exp(x)*(1-sin(x))/(1+cos(x)),x)
Output:
-Integral(-exp(x)/(cos(x) + 1), x) - Integral(exp(x)*sin(x)/(cos(x) + 1), x)
\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \] Input:
integrate(exp(x)*(1-sin(x))/(1+cos(x)),x, algorithm="maxima")
Output:
-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)
\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int { -\frac {{\left (\sin \left (x\right ) - 1\right )} e^{x}}{\cos \left (x\right ) + 1} \,d x } \] Input:
integrate(exp(x)*(1-sin(x))/(1+cos(x)),x, algorithm="giac")
Output:
integrate(-(sin(x) - 1)*e^x/(cos(x) + 1), x)
Timed out. \[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\sin \left (x\right )-1\right )}{\cos \left (x\right )+1} \,d x \] Input:
int(-(exp(x)*(sin(x) - 1))/(cos(x) + 1),x)
Output:
-int((exp(x)*(sin(x) - 1))/(cos(x) + 1), x)
\[ \int \frac {e^x (1-\sin (x))}{1+\cos (x)} \, dx=\int \frac {e^{x}}{\cos \left (x \right )+1}d x -\left (\int \frac {e^{x} \sin \left (x \right )}{\cos \left (x \right )+1}d x \right ) \] Input:
int(exp(x)*(1-sin(x))/(1+cos(x)),x)
Output:
int(e**x/(cos(x) + 1),x) - int((e**x*sin(x))/(cos(x) + 1),x)