\(\int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx\) [7545]

Optimal result

Integrand size = 31, antiderivative size = 15 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{e^{5-4 e^3 \log (16 x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 2306, 6838} \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{16^{-4 e^3} e^5 x^{-4 e^3}} \]

[In]

Int[(-4*E^(8 + E^(5 - 4*E^3*Log[16*x]) - 4*E^3*Log[16*x]))/x,x]

[Out]

E^(E^5/(16^(4*E^3)*x^(4*E^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 2306

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = -\left (4 \int \frac {e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx\right ) \\ & = -\left (4 \int 16^{-4 e^3} e^{8+e^{5-4 e^3 \log (16 x)}} x^{-1-4 e^3} \, dx\right ) \\ & = -\left (4^{1-8 e^3} \int e^{8+e^{5-4 e^3 \log (16 x)}} x^{-1-4 e^3} \, dx\right ) \\ & = e^{16^{-4 e^3} e^5 x^{-4 e^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{16^{-4 e^3} e^5 x^{-4 e^3}} \]

[In]

Integrate[(-4*E^(8 + E^(5 - 4*E^3*Log[16*x]) - 4*E^3*Log[16*x]))/x,x]

[Out]

E^(E^5/(16^(4*E^3)*x^(4*E^3)))

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87

method	result	size
derivativedivides	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{3} \ln \left (16 x \right )+5}}\)	\(13\)
default	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{3} \ln \left (16 x \right )+5}}\)	\(13\)
norman	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{3} \ln \left (16 x \right )+5}}\)	\(13\)
risch	\({\mathrm e}^{\left (16 x \right )^{-4 \,{\mathrm e}^{3}} {\mathrm e}^{5}}\)	\(13\)
parallelrisch	\({\mathrm e}^{{\mathrm e}^{-4 \,{\mathrm e}^{3} \ln \left (16 x \right )+5}}\)	\(13\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{\left (e^{\left (-4 \, e^{3} \log \left (16 \, x\right ) + 5\right )}\right )} \]

[In]

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

[Out]

e^(e^(-4*e^3*log(16*x) + 5))

Sympy [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{\frac {e^{5}}{65536^{e^{3}} x^{4 e^{3}}}} \]

[In]

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

[Out]

exp(exp(5)/(65536**exp(3)*x**(4*exp(3))))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{\left (e^{\left (-4 \, e^{3} \log \left (16 \, x\right ) + 5\right )}\right )} \]

[In]

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

[Out]

e^(e^(-4*e^3*log(16*x) + 5))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx=e^{\left (e^{\left (-4 \, e^{3} \log \left (16 \, x\right ) + 5\right )}\right )} \]

[In]

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

[Out]

e^(e^(-4*e^3*log(16*x) + 5))

Mupad [B] (verification not implemented)

Time = 13.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int -\frac {4 e^{8+e^{5-4 e^3 \log (16 x)}-4 e^3 \log (16 x)}}{x} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^5}{2^{16\,{\mathrm {e}}^3}\,x^{4\,{\mathrm {e}}^3}}} \]

[In]

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

[Out]

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