3.16.38 \(\int \frac {e^{-22-e^4+18 x-2 x^2} (-2+18 x-4 x^2) \log ^2(5)}{x^3} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{10-e^4-2 (4-x)^2+2 x} \log ^2(5)}{x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 41, normalized size of antiderivative = 1.41, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 2288} \begin {gather*} \frac {e^{-2 x^2+18 x-e^4-22} \left (9 x-2 x^2\right ) \log ^2(5)}{(9-2 x) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-22 - E^4 + 18*x - 2*x^2)*(-2 + 18*x - 4*x^2)*Log[5]^2)/x^3,x]

[Out]

(E^(-22 - E^4 + 18*x - 2*x^2)*(9*x - 2*x^2)*Log[5]^2)/((9 - 2*x)*x^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 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} &=\log ^2(5) \int \frac {e^{-22-e^4+18 x-2 x^2} \left (-2+18 x-4 x^2\right )}{x^3} \, dx\\ &=\frac {e^{-22-e^4+18 x-2 x^2} \left (9 x-2 x^2\right ) \log ^2(5)}{(9-2 x) x^3}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 25, normalized size = 0.86 \begin {gather*} \frac {e^{-22-e^4+18 x-2 x^2} \log ^2(5)}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-22 - E^4 + 18*x - 2*x^2)*(-2 + 18*x - 4*x^2)*Log[5]^2)/x^3,x]

[Out]

(E^(-22 - E^4 + 18*x - 2*x^2)*Log[5]^2)/x^2

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 23, normalized size = 0.79 \begin {gather*} \frac {e^{\left (-2 \, x^{2} + 18 \, x - e^{4} - 22\right )} \log \relax (5)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+18*x-2)*exp(5)^2*log(5)^2*exp(-x^2+9*x-16)^2/x^3/exp(exp(4)),x, algorithm="fricas")

[Out]

e^(-2*x^2 + 18*x - e^4 - 22)*log(5)^2/x^2

________________________________________________________________________________________

giac [A]  time = 0.22, size = 23, normalized size = 0.79 \begin {gather*} \frac {e^{\left (-2 \, x^{2} + 18 \, x - e^{4} - 22\right )} \log \relax (5)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+18*x-2)*exp(5)^2*log(5)^2*exp(-x^2+9*x-16)^2/x^3/exp(exp(4)),x, algorithm="giac")

[Out]

e^(-2*x^2 + 18*x - e^4 - 22)*log(5)^2/x^2

________________________________________________________________________________________

maple [A]  time = 0.08, size = 24, normalized size = 0.83

method	result	size
risch	\(\frac {\ln \relax (5)^{2} {\mathrm e}^{-22-{\mathrm e}^{4}-2 x^{2}+18 x}}{x^{2}}\)	\(24\)
gosper	\(\frac {{\mathrm e}^{10} \ln \relax (5)^{2} {\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}^{-2 x^{2}+18 x -32}}{x^{2}}\)	\(31\)
norman	\(\frac {{\mathrm e}^{10} \ln \relax (5)^{2} {\mathrm e}^{-{\mathrm e}^{4}} {\mathrm e}^{-2 x^{2}+18 x -32}}{x^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^2+18*x-2)*exp(5)^2*ln(5)^2*exp(-x^2+9*x-16)^2/x^3/exp(exp(4)),x,method=_RETURNVERBOSE)

[Out]

1/x^2*ln(5)^2*exp(-22-exp(4)-2*x^2+18*x)

________________________________________________________________________________________

maxima [A]  time = 0.54, size = 23, normalized size = 0.79 \begin {gather*} \frac {e^{\left (-2 \, x^{2} + 18 \, x - e^{4} - 22\right )} \log \relax (5)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^2+18*x-2)*exp(5)^2*log(5)^2*exp(-x^2+9*x-16)^2/x^3/exp(exp(4)),x, algorithm="maxima")

[Out]

e^(-2*x^2 + 18*x - e^4 - 22)*log(5)^2/x^2

________________________________________________________________________________________

mupad [B]  time = 0.16, size = 25, normalized size = 0.86 \begin {gather*} \frac {{\mathrm {e}}^{-{\mathrm {e}}^4}\,{\mathrm {e}}^{18\,x}\,{\mathrm {e}}^{-22}\,{\mathrm {e}}^{-2\,x^2}\,{\ln \relax (5)}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(4))*exp(10)*exp(18*x - 2*x^2 - 32)*log(5)^2*(4*x^2 - 18*x + 2))/x^3,x)

[Out]

(exp(-exp(4))*exp(18*x)*exp(-22)*exp(-2*x^2)*log(5)^2)/x^2

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 27, normalized size = 0.93 \begin {gather*} \frac {e^{10} e^{- 2 x^{2} + 18 x - 32} \log {\relax (5 )}^{2}}{x^{2} e^{e^{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**2+18*x-2)*exp(5)**2*ln(5)**2*exp(-x**2+9*x-16)**2/x**3/exp(exp(4)),x)

[Out]

exp(10)*exp(-2*x**2 + 18*x - 32)*exp(-exp(4))*log(5)**2/x**2

________________________________________________________________________________________