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

Optimal. Leaf size=25 \[ e^x-x+2 \left (e^3+\log ^2\left (2-x+\log \left (\frac {5}{2}\right )\right )\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 21, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6688, 2194, 2390, 12, 2301} \begin {gather*} e^x-x+2 \log ^2\left (-x+2+\log \left (\frac {5}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 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 2301

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 2390

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 21, normalized size = 0.84 \begin {gather*} e^x-x+2 \log ^2\left (2-x+\log \left (\frac {5}{2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 1.41, size = 18, normalized size = 0.72 \begin {gather*} 2 \, \log \left (-x + \log \left (\frac {5}{2}\right ) + 2\right )^{2} - x + e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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maple [A]  time = 0.17, size = 19, normalized size = 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	\(-x +2 \ln \left (\ln \left (\frac {5}{2}\right )+2-x \right )^{2}+{\mathrm e}^{x}\)	\(19\)
risch	\(2 \ln \left (-\ln \relax (2)+\ln \relax (5)+2-x \right )^{2}-x +{\mathrm e}^{x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {5}{2} \, e^{2} E_{1}\left (-x + \log \left (\frac {5}{2}\right ) + 2\right ) \log \left (\frac {5}{2}\right ) + 5 \, e^{2} E_{1}\left (-x + \log \left (\frac {5}{2}\right ) + 2\right ) + {\left (\log \relax (5) - \log \relax (2) + 2\right )} \int \frac {e^{x}}{x^{2} - 2 \, x {\left (\log \relax (5) - \log \relax (2) + 2\right )} + \log \relax (5)^{2} - 2 \, {\left (\log \relax (5) + 2\right )} \log \relax (2) + \log \relax (2)^{2} + 4 \, \log \relax (5) + 4}\,{d x} - {\left (\log \left (\frac {5}{2}\right ) + 2\right )} \log \left (x - \log \left (\frac {5}{2}\right ) - 2\right ) + \log \left (\frac {5}{2}\right ) \log \left (x - \log \left (\frac {5}{2}\right ) - 2\right ) + 2 \, \log \left (-x + \log \relax (5) - \log \relax (2) + 2\right )^{2} - x + \frac {x e^{x}}{x - \log \relax (5) + \log \relax (2) - 2} + 2 \, \log \left (x - \log \left (\frac {5}{2}\right ) - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.23, size = 18, normalized size = 0.72 \begin {gather*} 2\,{\ln \left (\ln \left (\frac {5}{2}\right )-x+2\right )}^2-x+{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.36, size = 17, normalized size = 0.68 \begin {gather*} - x + e^{x} + 2 \log {\left (- x + \log {\left (\frac {5}{2} \right )} + 2 \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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