\(\int \frac {-2 x+2 (i \pi +\log (\frac {21}{5}))-e^{\frac {2+x}{2}} x (i \pi +\log (\frac {21}{5}))}{2 x (i \pi +\log (\frac {21}{5}))} \, dx\) [5570]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 57, antiderivative size = 31 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=-e^{\frac {1}{4} (4+2 x)}-\frac {x}{i \pi +\log \left (\frac {21}{5}\right )}+\log (x) \]

[Out]

ln(x)-exp(1+1/2*x)-x/(ln(21/5)+I*Pi)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2225, 45} \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=-e^{\frac {x}{2}+1}-\frac {x}{\log \left (\frac {21}{5}\right )+i \pi }+\frac {\left (\log \left (\frac {441}{25}\right )+2 i \pi \right ) \log (x)}{2 \left (\log \left (\frac {21}{5}\right )+i \pi \right )} \]

[In]

Int[(-2*x + 2*(I*Pi + Log[21/5]) - E^((2 + x)/2)*x*(I*Pi + Log[21/5]))/(2*x*(I*Pi + Log[21/5])),x]

[Out]

-E^(1 + x/2) - x/(I*Pi + Log[21/5]) + (((2*I)*Pi + Log[441/25])*Log[x])/(2*(I*Pi + Log[21/5]))

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{x} \, dx}{2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \\ & = \frac {\int \left (-i e^{1+\frac {x}{2}} \left (\pi -i \log \left (\frac {21}{5}\right )\right )+\frac {i \left (2 \pi +2 i x-i \log \left (\frac {441}{25}\right )\right )}{x}\right ) \, dx}{2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \\ & = -\left (\frac {1}{2} \int e^{1+\frac {x}{2}} \, dx\right )+\frac {\int \frac {2 \pi +2 i x-i \log \left (\frac {441}{25}\right )}{x} \, dx}{2 \left (\pi -i \log \left (\frac {21}{5}\right )\right )} \\ & = -e^{1+\frac {x}{2}}+\frac {\int \left (2 i+\frac {2 \pi -i \log \left (\frac {441}{25}\right )}{x}\right ) \, dx}{2 \left (\pi -i \log \left (\frac {21}{5}\right )\right )} \\ & = -e^{1+\frac {x}{2}}-\frac {x}{i \pi +\log \left (\frac {21}{5}\right )}+\frac {\left (2 \pi -i \log \left (\frac {441}{25}\right )\right ) \log (x)}{2 \left (\pi -i \log \left (\frac {21}{5}\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=-e^{1+\frac {x}{2}}-\frac {x}{i \pi +\log \left (\frac {21}{5}\right )}+\log (x) \]

[In]

Integrate[(-2*x + 2*(I*Pi + Log[21/5]) - E^((2 + x)/2)*x*(I*Pi + Log[21/5]))/(2*x*(I*Pi + Log[21/5])),x]

[Out]

-E^(1 + x/2) - x/(I*Pi + Log[21/5]) + Log[x]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10

method result size
parts \(\frac {-x +\left (\ln \left (\frac {21}{5}\right )+i \pi \right ) \ln \left (x \right )}{\ln \left (\frac {21}{5}\right )+i \pi }-{\mathrm e}^{1+\frac {x}{2}}\) \(34\)
norman \(\frac {\left (i \pi -\ln \left (21\right )+\ln \left (5\right )\right ) x}{\ln \left (21\right )^{2}-2 \ln \left (21\right ) \ln \left (5\right )+\ln \left (5\right )^{2}+\pi ^{2}}-{\mathrm e}^{1+\frac {x}{2}}+\ln \left (x \right )\) \(45\)
parallelrisch \(\frac {2 i \ln \left (x \right ) \pi -2 i \pi \,{\mathrm e}^{1+\frac {x}{2}}+2 \ln \left (x \right ) \ln \left (\frac {21}{5}\right )-2 \ln \left (\frac {21}{5}\right ) {\mathrm e}^{1+\frac {x}{2}}-2 x}{2 \ln \left (\frac {21}{5}\right )+2 i \pi }\) \(48\)
risch \(-\frac {x}{\ln \left (3\right )+\ln \left (7\right )-\ln \left (5\right )+i \pi }+\ln \left (x \right )-\frac {i {\mathrm e}^{1+\frac {x}{2}} \pi }{\ln \left (3\right )+\ln \left (7\right )-\ln \left (5\right )+i \pi }-\frac {{\mathrm e}^{1+\frac {x}{2}} \ln \left (3\right )}{\ln \left (3\right )+\ln \left (7\right )-\ln \left (5\right )+i \pi }-\frac {{\mathrm e}^{1+\frac {x}{2}} \ln \left (7\right )}{\ln \left (3\right )+\ln \left (7\right )-\ln \left (5\right )+i \pi }+\frac {{\mathrm e}^{1+\frac {x}{2}} \ln \left (5\right )}{\ln \left (3\right )+\ln \left (7\right )-\ln \left (5\right )+i \pi }\) \(121\)
derivativedivides \(\frac {2 i \pi \ln \left (\frac {x}{2}\right )-2 i \pi \,{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )-2 i \pi \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )-4-2 x -2 \ln \left (5\right ) \ln \left (\frac {x}{2}\right )+2 \ln \left (21\right ) \ln \left (\frac {x}{2}\right )+2 \ln \left (5\right ) {\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )-2 \ln \left (21\right ) {\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )+2 \ln \left (5\right ) \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )-2 \ln \left (21\right ) \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )}{2 \ln \left (\frac {21}{5}\right )+2 i \pi }\) \(134\)
default \(\frac {2 i \pi \ln \left (\frac {x}{2}\right )-2 i \pi \,{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )-2 i \pi \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )-4-2 x -2 \ln \left (5\right ) \ln \left (\frac {x}{2}\right )+2 \ln \left (21\right ) \ln \left (\frac {x}{2}\right )+2 \ln \left (5\right ) {\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )-2 \ln \left (21\right ) {\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )+2 \ln \left (5\right ) \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )-2 \ln \left (21\right ) \left ({\mathrm e}^{1+\frac {x}{2}}-{\mathrm e} \,\operatorname {Ei}_{1}\left (-\frac {x}{2}\right )\right )}{2 \ln \left (\frac {21}{5}\right )+2 i \pi }\) \(134\)

[In]

int(1/2*(-x*(ln(21/5)+I*Pi)*exp(1+1/2*x)+2*ln(21/5)+2*I*Pi-2*x)/x/(ln(21/5)+I*Pi),x,method=_RETURNVERBOSE)

[Out]

1/(ln(21/5)+I*Pi)*(-x+(ln(21/5)+I*Pi)*ln(x))-exp(1+1/2*x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.19 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=\frac {{\left (-i \, \pi - \log \left (\frac {21}{5}\right )\right )} e^{\left (\frac {1}{2} \, x + 1\right )} + {\left (i \, \pi + \log \left (\frac {21}{5}\right )\right )} \log \left (x\right ) - x}{i \, \pi + \log \left (\frac {21}{5}\right )} \]

[In]

integrate(1/2*(-x*(log(21/5)+I*pi)*exp(1+1/2*x)+2*log(21/5)+2*I*pi-2*x)/x/(log(21/5)+I*pi),x, algorithm="frica
s")

[Out]

((-I*pi - log(21/5))*e^(1/2*x + 1) + (I*pi + log(21/5))*log(x) - x)/(I*pi + log(21/5))

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=\frac {- x - \left (- \log {\left (21 \right )} + \log {\left (5 \right )} - i \pi \right ) \log {\left (x \right )}}{- \log {\left (5 \right )} + \log {\left (21 \right )} + i \pi } - e e^{\frac {x}{2}} \]

[In]

integrate(1/2*(-x*(ln(21/5)+I*pi)*exp(1+1/2*x)+2*ln(21/5)+2*I*pi-2*x)/x/(ln(21/5)+I*pi),x)

[Out]

(-x - (-log(21) + log(5) - I*pi)*log(x))/(-log(5) + log(21) + I*pi) - E*exp(x/2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=\frac {-i \, \pi e^{\left (\frac {1}{2} \, x + 1\right )} - e^{\left (\frac {1}{2} \, x + 1\right )} \log \left (\frac {21}{5}\right ) + i \, \pi \log \left (x\right ) + \log \left (\frac {21}{5}\right ) \log \left (x\right ) - x}{i \, \pi + \log \left (\frac {21}{5}\right )} \]

[In]

integrate(1/2*(-x*(log(21/5)+I*pi)*exp(1+1/2*x)+2*log(21/5)+2*I*pi-2*x)/x/(log(21/5)+I*pi),x, algorithm="maxim
a")

[Out]

(-I*pi*e^(1/2*x + 1) - e^(1/2*x + 1)*log(21/5) + I*pi*log(x) + log(21/5)*log(x) - x)/(I*pi + log(21/5))

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.45 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=-\frac {i \, \pi e^{\left (\frac {1}{2} \, x + 1\right )} + e^{\left (\frac {1}{2} \, x + 1\right )} \log \left (\frac {21}{5}\right ) - i \, \pi \log \left (\frac {1}{2} \, x\right ) - \log \left (\frac {21}{5}\right ) \log \left (\frac {1}{2} \, x\right ) + x}{i \, \pi + \log \left (\frac {21}{5}\right )} \]

[In]

integrate(1/2*(-x*(log(21/5)+I*pi)*exp(1+1/2*x)+2*log(21/5)+2*I*pi-2*x)/x/(log(21/5)+I*pi),x, algorithm="giac"
)

[Out]

-(I*pi*e^(1/2*x + 1) + e^(1/2*x + 1)*log(21/5) - I*pi*log(1/2*x) - log(21/5)*log(1/2*x) + x)/(I*pi + log(21/5)
)

Mupad [B] (verification not implemented)

Time = 11.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {-2 x+2 \left (i \pi +\log \left (\frac {21}{5}\right )\right )-e^{\frac {2+x}{2}} x \left (i \pi +\log \left (\frac {21}{5}\right )\right )}{2 x \left (i \pi +\log \left (\frac {21}{5}\right )\right )} \, dx=\ln \left (x\right )-\frac {{\mathrm {e}}^{\frac {x}{2}+1}\,\left (2\,\Pi ^2+2\,{\ln \left (\frac {21}{5}\right )}^2\right )-x\,\left (-2\,\ln \left (\frac {21}{5}\right )+\Pi \,2{}\mathrm {i}\right )}{2\,\Pi ^2+2\,{\ln \left (\frac {21}{5}\right )}^2} \]

[In]

int((Pi*1i - x + log(21/5) - (x*exp(x/2 + 1)*(Pi*1i + log(21/5)))/2)/(x*(Pi*1i + log(21/5))),x)

[Out]

log(x) - (exp(x/2 + 1)*(2*Pi^2 + 2*log(21/5)^2) - x*(Pi*2i - 2*log(21/5)))/(2*Pi^2 + 2*log(21/5)^2)