3.28.52 \(\int \frac {e^{-2+\frac {-27-135 e^2+\frac {1}{3} e^x (1+5 e^2) x}{e^2 x^2}} (54+270 e^2+\frac {1}{3} e^x x (-1+x+e^2 (-5+5 x)))}{x^3} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {\left (5+\frac {1}{e^2}\right ) \left (-27+\frac {e^x x}{3}\right )}{x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 1.48, antiderivative size = 35, normalized size of antiderivative = 1.67, number of steps used = 1, number of rules used = 1, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6706} \begin {gather*} \exp \left (-\frac {81 \left (1+5 e^2\right )-\left (1+5 e^2\right ) e^x x}{3 e^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-2 + (-27 - 135*E^2 + (E^x*(1 + 5*E^2)*x)/3)/(E^2*x^2))*(54 + 270*E^2 + (E^x*x*(-1 + x + E^2*(-5 + 5*x
)))/3))/x^3,x]

[Out]

E^(-1/3*(81*(1 + 5*E^2) - E^x*(1 + 5*E^2)*x)/(E^2*x^2))

Rule 6706

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 {gather*} \begin {aligned} \text {integral} &=\exp \left (-\frac {81 \left (1+5 e^2\right )-e^x \left (1+5 e^2\right ) x}{3 e^2 x^2}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.66, size = 26, normalized size = 1.24 \begin {gather*} e^{\frac {\left (1+5 e^2\right ) \left (-81+e^x x\right )}{3 e^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-2 + (-27 - 135*E^2 + (E^x*(1 + 5*E^2)*x)/3)/(E^2*x^2))*(54 + 270*E^2 + (E^x*x*(-1 + x + E^2*(-5
+ 5*x)))/3))/x^3,x]

[Out]

E^(((1 + 5*E^2)*(-81 + E^x*x))/(3*E^2*x^2))

________________________________________________________________________________________

fricas [B]  time = 0.69, size = 37, normalized size = 1.76 \begin {gather*} e^{\left (-\frac {{\left ({\left (2 \, x^{2} + 135\right )} e^{2} - {\left (5 \, e^{2} + 1\right )} e^{\left (x + \log \left (\frac {1}{3} \, x\right )\right )} + 27\right )} e^{\left (-2\right )}}{x^{2}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x-5)*exp(2)+x-1)*exp(x+log(1/3*x))+270*exp(2)+54)*exp(((5*exp(2)+1)*exp(x+log(1/3*x))-135*exp(2
)-27)/x^2/exp(2))/x^3/exp(2),x, algorithm="fricas")

[Out]

e^(-((2*x^2 + 135)*e^2 - (5*e^2 + 1)*e^(x + log(1/3*x)) + 27)*e^(-2)/x^2 + 2)

________________________________________________________________________________________

giac [B]  time = 0.33, size = 38, normalized size = 1.81 \begin {gather*} e^{\left (-\frac {27 \, e^{\left (-2\right )}}{x^{2}} + \frac {5 \, e^{\left (x + \log \left (\frac {1}{3} \, x\right )\right )}}{x^{2}} + \frac {e^{\left (x + \log \left (\frac {1}{3} \, x\right ) - 2\right )}}{x^{2}} - \frac {135}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x-5)*exp(2)+x-1)*exp(x+log(1/3*x))+270*exp(2)+54)*exp(((5*exp(2)+1)*exp(x+log(1/3*x))-135*exp(2
)-27)/x^2/exp(2))/x^3/exp(2),x, algorithm="giac")

[Out]

e^(-27*e^(-2)/x^2 + 5*e^(x + log(1/3*x))/x^2 + e^(x + log(1/3*x) - 2)/x^2 - 135/x^2)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 21, normalized size = 1.00

method	result	size
risch	\({\mathrm e}^{\frac {\left ({\mathrm e}^{x} x -81\right ) \left (5 \,{\mathrm e}^{2}+1\right ) {\mathrm e}^{-2}}{3 x^{2}}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((5*x-5)*exp(2)+x-1)*exp(x+ln(1/3*x))+270*exp(2)+54)*exp(((5*exp(2)+1)*exp(x+ln(1/3*x))-135*exp(2)-27)/x^
2/exp(2))/x^3/exp(2),x,method=_RETURNVERBOSE)

[Out]

exp(1/3*(exp(x)*x-81)*(5*exp(2)+1)*exp(-2)/x^2)

________________________________________________________________________________________

maxima [A]  time = 0.62, size = 30, normalized size = 1.43 \begin {gather*} e^{\left (\frac {e^{\left (x - 2\right )}}{3 \, x} + \frac {5 \, e^{x}}{3 \, x} - \frac {27 \, e^{\left (-2\right )}}{x^{2}} - \frac {135}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x-5)*exp(2)+x-1)*exp(x+log(1/3*x))+270*exp(2)+54)*exp(((5*exp(2)+1)*exp(x+log(1/3*x))-135*exp(2
)-27)/x^2/exp(2))/x^3/exp(2),x, algorithm="maxima")

[Out]

e^(1/3*e^(x - 2)/x + 5/3*e^x/x - 27*e^(-2)/x^2 - 135/x^2)

________________________________________________________________________________________

mupad [B]  time = 2.15, size = 33, normalized size = 1.57 \begin {gather*} {\mathrm {e}}^{-\frac {27\,{\mathrm {e}}^{-2}}{x^2}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^x}{3\,x}}\,{\mathrm {e}}^{-\frac {135}{x^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-2}\,{\mathrm {e}}^x}{3\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(exp(-2)*(135*exp(2) - exp(x + log(x/3))*(5*exp(2) + 1) + 27))/x^2)*exp(-2)*(270*exp(2) + exp(x + lo
g(x/3))*(x + exp(2)*(5*x - 5) - 1) + 54))/x^3,x)

[Out]

exp(-(27*exp(-2))/x^2)*exp((5*exp(x))/(3*x))*exp(-135/x^2)*exp((exp(-2)*exp(x))/(3*x))

________________________________________________________________________________________

sympy [A]  time = 0.33, size = 27, normalized size = 1.29 \begin {gather*} e^{\frac {\frac {x \left (1 + 5 e^{2}\right ) e^{x}}{3} - 135 e^{2} - 27}{x^{2} e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((5*x-5)*exp(2)+x-1)*exp(x+ln(1/3*x))+270*exp(2)+54)*exp(((5*exp(2)+1)*exp(x+ln(1/3*x))-135*exp(2)-
27)/x**2/exp(2))/x**3/exp(2),x)

[Out]

exp((x*(1 + 5*exp(2))*exp(x)/3 - 135*exp(2) - 27)*exp(-2)/x**2)

________________________________________________________________________________________