\(\int \frac {-2+x-\log (\frac {5}{2})+e^x (2-x+\log (\frac {5}{2}))-4 \log (2-x+\log (\frac {5}{2}))}{2-x+\log (\frac {5}{2})} \, dx\) [4399]

Optimal result

Integrand size = 46, antiderivative size = 25 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=e^x-x+2 \left (e^3+\log ^2\left (2-x+\log \left (\frac {5}{2}\right )\right )\right ) \]

[Out]

exp(x)+2*ln(ln(5/2)+2-x)^2+2*exp(3)-x

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6820, 2225, 2437, 12, 2338} \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=e^x-x+2 \log ^2\left (-x+2+\log \left (\frac {5}{2}\right )\right ) \]

[In]

Int[(-2 + x - Log[5/2] + E^x*(2 - x + Log[5/2]) - 4*Log[2 - x + Log[5/2]])/(2 - x + Log[5/2]),x]

[Out]

E^x - x + 2*Log[2 - x + Log[5/2]]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \left (-1+e^x+\frac {4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{-2+x-\log \left (\frac {5}{2}\right )}\right ) \, dx \\ & = -x+4 \int \frac {\log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{-2+x-\log \left (\frac {5}{2}\right )} \, dx+\int e^x \, dx \\ & = e^x-x-4 \text {Subst}\left (\int \frac {\left (2+\log \left (\frac {5}{2}\right )\right ) \log (x)}{x \left (-2-\log \left (\frac {5}{2}\right )\right )} \, dx,x,2-x+\log \left (\frac {5}{2}\right )\right ) \\ & = e^x-x+4 \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,2-x+\log \left (\frac {5}{2}\right )\right ) \\ & = e^x-x+2 \log ^2\left (2-x+\log \left (\frac {5}{2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=e^x-x+2 \log ^2\left (2-x+\log \left (\frac {5}{2}\right )\right ) \]

[In]

Integrate[(-2 + x - Log[5/2] + E^x*(2 - x + Log[5/2]) - 4*Log[2 - x + Log[5/2]])/(2 - x + Log[5/2]),x]

[Out]

E^x - x + 2*Log[2 - x + Log[5/2]]^2

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76

method	result	size
default	\(-x +2 \ln \left (\ln \left (\frac {5}{2}\right )+2-x \right )^{2}+{\mathrm e}^{x}\)	\(19\)
norman	\(-x +2 \ln \left (\ln \left (\frac {5}{2}\right )+2-x \right )^{2}+{\mathrm e}^{x}\)	\(19\)
parts	\(-x +2 \ln \left (\ln \left (\frac {5}{2}\right )+2-x \right )^{2}+{\mathrm e}^{x}\)	\(19\)
risch	\(2 \ln \left (\ln \left (5\right )-\ln \left (2\right )+2-x \right )^{2}-x +{\mathrm e}^{x}\)	\(23\)
parallelrisch	\(-4+2 \ln \left (\ln \left (\frac {5}{2}\right )+2-x \right )^{2}-2 \ln \left (\frac {5}{2}\right )-x +{\mathrm e}^{x}\)	\(24\)

[In]

int((-4*ln(ln(5/2)+2-x)+(ln(5/2)+2-x)*exp(x)-ln(5/2)+x-2)/(ln(5/2)+2-x),x,method=_RETURNVERBOSE)

[Out]

-x+2*ln(ln(5/2)+2-x)^2+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=2 \, \log \left (-x + \log \left (\frac {5}{2}\right ) + 2\right )^{2} - x + e^{x} \]

[In]

integrate((-4*log(log(5/2)+2-x)+(log(5/2)+2-x)*exp(x)-log(5/2)+x-2)/(log(5/2)+2-x),x, algorithm="fricas")

[Out]

2*log(-x + log(5/2) + 2)^2 - x + e^x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=- x + e^{x} + 2 \log {\left (- x + \log {\left (\frac {5}{2} \right )} + 2 \right )}^{2} \]

[In]

integrate((-4*ln(ln(5/2)+2-x)+(ln(5/2)+2-x)*exp(x)-ln(5/2)+x-2)/(ln(5/2)+2-x),x)

[Out]

-x + exp(x) + 2*log(-x + log(5/2) + 2)**2

Maxima [F]

\[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=\int { \frac {{\left (x - \log \left (\frac {5}{2}\right ) - 2\right )} e^{x} - x + \log \left (\frac {5}{2}\right ) + 4 \, \log \left (-x + \log \left (\frac {5}{2}\right ) + 2\right ) + 2}{x - \log \left (\frac {5}{2}\right ) - 2} \,d x } \]

[In]

integrate((-4*log(log(5/2)+2-x)+(log(5/2)+2-x)*exp(x)-log(5/2)+x-2)/(log(5/2)+2-x),x, algorithm="maxima")

[Out]

5/2*e^2*exp_integral_e(1, -x + log(5/2) + 2)*log(5/2) + 5*e^2*exp_integral_e(1, -x + log(5/2) + 2) + (log(5) -
log(2) + 2)*integrate(e^x/(x^2 - 2*x*(log(5) - log(2) + 2) + log(5)^2 - 2*(log(5) + 2)*log(2) + log(2)^2 + 4*
log(5) + 4), x) - (log(5/2) + 2)*log(x - log(5/2) - 2) + log(5/2)*log(x - log(5/2) - 2) + 2*log(-x + log(5) -
log(2) + 2)^2 - x + x*e^x/(x - log(5) + log(2) - 2) + 2*log(x - log(5/2) - 2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=2 \, \log \left (-x + \log \left (\frac {5}{2}\right ) + 2\right )^{2} - x + e^{x} + \log \left (\frac {5}{2}\right ) + 2 \]

[In]

integrate((-4*log(log(5/2)+2-x)+(log(5/2)+2-x)*exp(x)-log(5/2)+x-2)/(log(5/2)+2-x),x, algorithm="giac")

[Out]

2*log(-x + log(5/2) + 2)^2 - x + e^x + log(5/2) + 2

Mupad [B] (verification not implemented)

Time = 10.82 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {-2+x-\log \left (\frac {5}{2}\right )+e^x \left (2-x+\log \left (\frac {5}{2}\right )\right )-4 \log \left (2-x+\log \left (\frac {5}{2}\right )\right )}{2-x+\log \left (\frac {5}{2}\right )} \, dx=2\,{\ln \left (\ln \left (\frac {5}{2}\right )-x+2\right )}^2-x+{\mathrm {e}}^x \]

[In]

int(-(4*log(log(5/2) - x + 2) - x + log(5/2) - exp(x)*(log(5/2) - x + 2) + 2)/(log(5/2) - x + 2),x)

[Out]

exp(x) - x + 2*log(log(5/2) - x + 2)^2