\(\int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx\) [9328]

   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 [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 52, antiderivative size = 39 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-e^5+e^{e^{4 (4+x)}}+\frac {x (3-x+\log (4))}{i \pi +\log (5-\log (3))} \]

[Out]

(2*ln(2)+3-x)*x/ln(ln(3)-5)+exp(exp(4*x+16))-exp(5)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {12, 2320, 2225} \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^2}{\log (5-\log (3))+i \pi }+e^{e^{4 x+16}}+\frac {x (3+\log (4))}{\log (5-\log (3))+i \pi } \]

[In]

Int[(3 - 2*x + Log[4] + 4*E^(16 + E^(16 + 4*x) + 4*x)*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]]),x]

[Out]

E^E^(16 + 4*x) - x^2/(I*Pi + Log[5 - Log[3]]) + (x*(3 + Log[4]))/(I*Pi + Log[5 - Log[3]])

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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \left (3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))\right ) \, dx}{i \pi +\log (5-\log (3))} \\ & = -\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+4 \int e^{16+e^{16+4 x}+4 x} \, dx \\ & = -\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))}+\text {Subst}\left (\int e^{16+e^{16} x} \, dx,x,e^{4 x}\right ) \\ & = e^{e^{16+4 x}}-\frac {x^2}{i \pi +\log (5-\log (3))}+\frac {x (3+\log (4))}{i \pi +\log (5-\log (3))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=\frac {3 x-x^2+x \log (4)+e^{e^{4 (4+x)}} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \]

[In]

Integrate[(3 - 2*x + Log[4] + 4*E^(16 + E^(16 + 4*x) + 4*x)*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]])
,x]

[Out]

(3*x - x^2 + x*Log[4] + E^E^(4*(4 + x))*(I*Pi + Log[5 - Log[3]]))/(I*Pi + Log[5 - Log[3]])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.79

method result size
parts \(\frac {2 x \ln \left (2\right )-x^{2}+3 x}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}\) \(31\)
default \(\frac {-x^{2}+3 x +\ln \left (\ln \left (3\right )-5\right ) {\mathrm e}^{{\mathrm e}^{4 x +16}}+2 x \ln \left (2\right )}{\ln \left (\ln \left (3\right )-5\right )}\) \(36\)
norman \(\frac {\left (2 \ln \left (2\right )+3\right ) x}{\ln \left (\ln \left (3\right )-5\right )}-\frac {x^{2}}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}\) \(36\)
parallelrisch \(\frac {-x^{2}+\ln \left (\ln \left (3\right )-5\right ) {\mathrm e}^{{\mathrm e}^{4 x +16}}+x \left (2 \ln \left (2\right )+3\right )}{\ln \left (\ln \left (3\right )-5\right )}\) \(36\)
risch \(-\frac {x^{2}}{\ln \left (\ln \left (3\right )-5\right )}+\frac {3 x}{\ln \left (\ln \left (3\right )-5\right )}+{\mathrm e}^{{\mathrm e}^{4 x +16}}+\frac {2 x \ln \left (2\right )}{\ln \left (\ln \left (3\right )-5\right )}\) \(43\)

[In]

int((4*exp(4*x+16)*ln(ln(3)-5)*exp(exp(4*x+16))+2*ln(2)+3-2*x)/ln(ln(3)-5),x,method=_RETURNVERBOSE)

[Out]

1/ln(ln(3)-5)*(2*x*ln(2)-x^2+3*x)+exp(exp(4*x+16))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {{\left ({\left (x^{2} - 2 \, x \log \left (2\right ) - 3 \, x\right )} e^{\left (4 \, x + 16\right )} - e^{\left (4 \, x + e^{\left (4 \, x + 16\right )} + 16\right )} \log \left (\log \left (3\right ) - 5\right )\right )} e^{\left (-4 \, x - 16\right )}}{\log \left (\log \left (3\right ) - 5\right )} \]

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="fricas")

[Out]

-((x^2 - 2*x*log(2) - 3*x)*e^(4*x + 16) - e^(4*x + e^(4*x + 16) + 16)*log(log(3) - 5))*e^(-4*x - 16)/log(log(3
) - 5)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.05 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=- \frac {x^{2}}{\log {\left (5 - \log {\left (3 \right )} \right )} + i \pi } + \frac {x \left (2 \log {\left (2 \right )} + 3\right )}{\log {\left (5 - \log {\left (3 \right )} \right )} + i \pi } + e^{e^{16} e^{4 x}} \]

[In]

integrate((4*exp(4*x+16)*ln(ln(3)-5)*exp(exp(4*x+16))+2*ln(2)+3-2*x)/ln(ln(3)-5),x)

[Out]

-x**2/(log(5 - log(3)) + I*pi) + x*(2*log(2) + 3)/(log(5 - log(3)) + I*pi) + exp(exp(16)*exp(4*x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^{2} - 2 \, x \log \left (2\right ) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \left (3\right ) - 5\right ) - 3 \, x}{\log \left (\log \left (3\right ) - 5\right )} \]

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="maxima")

[Out]

-(x^2 - 2*x*log(2) - e^(e^(4*x + 16))*log(log(3) - 5) - 3*x)/log(log(3) - 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=-\frac {x^{2} - 2 \, x \log \left (2\right ) - e^{\left (e^{\left (4 \, x + 16\right )}\right )} \log \left (\log \left (3\right ) - 5\right ) - 3 \, x}{\log \left (\log \left (3\right ) - 5\right )} \]

[In]

integrate((4*exp(4*x+16)*log(log(3)-5)*exp(exp(4*x+16))+2*log(2)+3-2*x)/log(log(3)-5),x, algorithm="giac")

[Out]

-(x^2 - 2*x*log(2) - e^(e^(4*x + 16))*log(log(3) - 5) - 3*x)/log(log(3) - 5)

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.90 \[ \int \frac {3-2 x+\log (4)+4 e^{16+e^{16+4 x}+4 x} (i \pi +\log (5-\log (3)))}{i \pi +\log (5-\log (3))} \, dx=\frac {3\,x+x\,\ln \left (4\right )-x^2+\ln \left (\ln \left (3\right )-5\right )\,{\mathrm {e}}^{{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{16}}}{\ln \left (\ln \left (3\right )-5\right )} \]

[In]

int((2*log(2) - 2*x + 4*log(log(3) - 5)*exp(exp(4*x + 16))*exp(4*x + 16) + 3)/log(log(3) - 5),x)

[Out]

(3*x + x*log(4) - x^2 + log(log(3) - 5)*exp(exp(4*x)*exp(16)))/log(log(3) - 5)