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

Optimal. Leaf size=22 \[ -4+e^{5-\frac {\log (x) (3+\log (x))}{\log \left (-3+e^4\right )}} \]

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Rubi [A]  time = 0.18, antiderivative size = 31, normalized size of antiderivative = 1.41, number of steps used = 5, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {12, 2274, 2288} \begin {gather*} e^{5-\frac {\log ^2(x)}{\log \left (e^4-3\right )}} x^{-\frac {3}{\log \left (e^4-3\right )}} \end {gather*}

Antiderivative was successfully verified.

[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 2274

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 2288

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

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Mathematica [A]  time = 0.03, size = 31, normalized size = 1.41 \begin {gather*} e^{5-\frac {\log ^2(x)}{\log \left (-3+e^4\right )}} x^{-\frac {3}{\log \left (-3+e^4\right )}} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.56, size = 26, normalized size = 1.18 \begin {gather*} e^{\left (-\frac {\log \relax (x)^{2} + 3 \, \log \relax (x) - 5 \, \log \left (e^{4} - 3\right )}{\log \left (e^{4} - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.25, size = 27, normalized size = 1.23 \begin {gather*} e^{\left (-\frac {\log \relax (x)^{2}}{\log \left (e^{4} - 3\right )} - \frac {3 \, \log \relax (x)}{\log \left (e^{4} - 3\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.11, size = 27, normalized size = 1.23

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.37, size = 26, normalized size = 1.18 \begin {gather*} e^{\left (-\frac {\log \relax (x)^{2} + 3 \, \log \relax (x) - 5 \, \log \left (e^{4} - 3\right )}{\log \left (e^{4} - 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 5.18, size = 30, normalized size = 1.36 \begin {gather*} \frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {{\ln \relax (x)}^2}{\ln \left ({\mathrm {e}}^4-3\right )}}}{x^{\frac {3}{\ln \left ({\mathrm {e}}^4-3\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.47, size = 26, normalized size = 1.18 \begin {gather*} e^{\frac {- \log {\relax (x )}^{2} - 3 \log {\relax (x )} + 5 \log {\left (-3 + e^{4} \right )}}{\log {\left (-3 + e^{4} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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