\(\int \frac {e^{\frac {5 \log (-3+e^4)-3 \log (x)-\log ^2(x)}{\log (-3+e^4)}} (-3-2 \log (x))}{x \log (-3+e^4)} \, dx\) [7855]

Optimal result

Integrand size = 48, antiderivative size = 22 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=-4+e^{5-\frac {\log (x) (3+\log (x))}{\log \left (-3+e^4\right )}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 2306, 2326} \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{5-\frac {\log ^2(x)}{\log \left (e^4-3\right )}} x^{-\frac {3}{\log \left (e^4-3\right )}} \]

[In]

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

[Out]

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

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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\exp \left (\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}\right ) (-3-2 \log (x))}{x} \, dx}{\log \left (-3+e^4\right )} \\ & = \frac {\int \frac {e^{\frac {-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} \left (-3+e^4\right )^{\frac {5}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x} \, dx}{\log \left (-3+e^4\right )} \\ & = \frac {e^5 \int \frac {e^{\frac {-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x} \, dx}{\log \left (-3+e^4\right )} \\ & = \frac {e^5 \int e^{-\frac {\log ^2(x)}{\log \left (-3+e^4\right )}} x^{-1-\frac {3}{\log \left (-3+e^4\right )}} (-3-2 \log (x)) \, dx}{\log \left (-3+e^4\right )} \\ & = e^{5-\frac {\log ^2(x)}{\log \left (-3+e^4\right )}} x^{-\frac {3}{\log \left (-3+e^4\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{5-\frac {\log ^2(x)}{\log \left (-3+e^4\right )}} x^{-\frac {3}{\log \left (-3+e^4\right )}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23

method	result	size
risch	\({\mathrm e}^{-\frac {\ln \left (x \right )^{2}+3 \ln \left (x \right )-5 \ln \left ({\mathrm e}^{4}-3\right )}{\ln \left ({\mathrm e}^{4}-3\right )}}\)	\(27\)
parallelrisch	\({\mathrm e}^{-\frac {\ln \left (x \right )^{2}+3 \ln \left (x \right )-5 \ln \left ({\mathrm e}^{4}-3\right )}{\ln \left ({\mathrm e}^{4}-3\right )}}\)	\(27\)
derivativedivides	\({\mathrm e}^{\frac {5 \ln \left ({\mathrm e}^{4}-3\right )-\ln \left (x \right )^{2}-3 \ln \left (x \right )}{\ln \left ({\mathrm e}^{4}-3\right )}}\)	\(28\)
default	\({\mathrm e}^{\frac {5 \ln \left ({\mathrm e}^{4}-3\right )-\ln \left (x \right )^{2}-3 \ln \left (x \right )}{\ln \left ({\mathrm e}^{4}-3\right )}}\)	\(28\)
norman	\({\mathrm e}^{\frac {5 \ln \left ({\mathrm e}^{4}-3\right )-\ln \left (x \right )^{2}-3 \ln \left (x \right )}{\ln \left ({\mathrm e}^{4}-3\right )}}\)	\(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{\left (-\frac {\log \left (x\right )^{2} + 3 \, \log \left (x\right ) - 5 \, \log \left (e^{4} - 3\right )}{\log \left (e^{4} - 3\right )}\right )} \]

[In]

integrate((-2*log(x)-3)*exp((5*log(exp(4)-3)-log(x)^2-3*log(x))/log(exp(4)-3))/x/log(exp(4)-3),x, algorithm="f
ricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{\frac {- \log {\left (x \right )}^{2} - 3 \log {\left (x \right )} + 5 \log {\left (-3 + e^{4} \right )}}{\log {\left (-3 + e^{4} \right )}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{\left (-\frac {\log \left (x\right )^{2} + 3 \, \log \left (x\right ) - 5 \, \log \left (e^{4} - 3\right )}{\log \left (e^{4} - 3\right )}\right )} \]

[In]

integrate((-2*log(x)-3)*exp((5*log(exp(4)-3)-log(x)^2-3*log(x))/log(exp(4)-3))/x/log(exp(4)-3),x, algorithm="m
axima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=e^{\left (-\frac {\log \left (x\right )^{2}}{\log \left (e^{4} - 3\right )} - \frac {3 \, \log \left (x\right )}{\log \left (e^{4} - 3\right )} + 5\right )} \]

[In]

integrate((-2*log(x)-3)*exp((5*log(exp(4)-3)-log(x)^2-3*log(x))/log(exp(4)-3))/x/log(exp(4)-3),x, algorithm="g
iac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36 \[ \int \frac {e^{\frac {5 \log \left (-3+e^4\right )-3 \log (x)-\log ^2(x)}{\log \left (-3+e^4\right )}} (-3-2 \log (x))}{x \log \left (-3+e^4\right )} \, dx=\frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {{\ln \left (x\right )}^2}{\ln \left ({\mathrm {e}}^4-3\right )}}}{x^{\frac {3}{\ln \left ({\mathrm {e}}^4-3\right )}}} \]

[In]

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

[Out]

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