\(\int \frac {e^{-\frac {3-4 e^3}{4 x}} (e^x (-3+4 e^3-4 x^2)+(3 x-4 e^3 x+4 x^2) \log (4))}{4 x^2 \log (4)} \, dx\) [5115]

Optimal result

Integrand size = 62, antiderivative size = 29 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=e^{-\frac {3}{4 x}+\frac {e^3}{x}} \left (x-\frac {e^x}{\log (4)}\right ) \]

[Out]

(x-1/2*exp(x)/ln(2))/exp(3/4/x-exp(3)/x)

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 6874, 6838, 2326} \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=e^{-\frac {3-4 e^3}{4 x}} x-\frac {e^{x-\frac {3-4 e^3}{4 x}}}{\log (4)} \]

[In]

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

[Out]

x/E^((3 - 4*E^3)/(4*x)) - E^(-1/4*(3 - 4*E^3)/x + x)/Log[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 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]

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]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {e^{\frac {-3+4 e^3}{4 x}} \left (-4 e^x+4 x \log (4)\right )}{4 \log (4)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {\left (8 x \ln \left (2\right )-4 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {4 \,{\mathrm e}^{3}-3}{4 x}}}{8 \ln \left (2\right )}\)	\(29\)
norman	\(\frac {\left (x^{2}-\frac {x \,{\mathrm e}^{x}}{2 \ln \left (2\right )}\right ) {\mathrm e}^{-\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}{x}\)	\(32\)
parallelrisch	\(\frac {\left (32 x^{2} \ln \left (2\right )-16 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{\frac {4 \,{\mathrm e}^{3}-3}{4 x}}}{32 \ln \left (2\right ) x}\)	\(37\)
parts	\(\frac {3 x \,{\mathrm e}^{-\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}}}{4 \left (-{\mathrm e}^{3}+\frac {3}{4}\right )}+{\mathrm e}^{3} \left (-\frac {x \,{\mathrm e}^{-\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}}}{-{\mathrm e}^{3}+\frac {3}{4}}+\operatorname {Ei}_{1}\left (\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}\right )\right )-{\mathrm e}^{3} \operatorname {Ei}_{1}\left (\frac {-{\mathrm e}^{3}+\frac {3}{4}}{x}\right )-\frac {{\mathrm e}^{x} {\mathrm e}^{-\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}{2 \ln \left (2\right )}\)	\(104\)

[In]

int(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*ln(2))/x^2/ln(2)/exp(1/4*(-4*exp(3)+3)/x),x,metho
d=_RETURNVERBOSE)

[Out]

1/8/ln(2)*(8*x*ln(2)-4*exp(x))*exp(1/4*(4*exp(3)-3)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {{\left (2 \, x \log \left (2\right ) - e^{x}\right )} e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \]

[In]

integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2))/x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x)
,x, algorithm="fricas")

[Out]

1/2*(2*x*log(2) - e^x)*e^(1/4*(4*e^3 - 3)/x)/log(2)

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {\left (2 x \log {\left (2 \right )} - e^{x}\right ) e^{- \frac {\frac {3}{4} - e^{3}}{x}}}{2 \log {\left (2 \right )}} \]

[In]

integrate(1/8*((4*exp(3)-4*x**2-3)*exp(x)+2*(-4*x*exp(3)+4*x**2+3*x)*ln(2))/x**2/ln(2)/exp(1/4*(-4*exp(3)+3)/x
),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {{\left (2 \, x \log \left (2\right ) - e^{x}\right )} e^{\left (\frac {e^{3}}{x} - \frac {3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \]

[In]

integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2))/x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x)
,x, algorithm="maxima")

[Out]

1/2*(2*x*log(2) - e^x)*e^(e^3/x - 3/4/x)/log(2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=\frac {2 \, x e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )} \log \left (2\right ) - e^{\left (\frac {4 \, x^{2} + 4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \left (2\right )} \]

[In]

integrate(1/8*((4*exp(3)-4*x^2-3)*exp(x)+2*(-4*x*exp(3)+4*x^2+3*x)*log(2))/x^2/log(2)/exp(1/4*(-4*exp(3)+3)/x)
,x, algorithm="giac")

[Out]

1/2*(2*x*e^(1/4*(4*e^3 - 3)/x)*log(2) - e^(1/4*(4*x^2 + 4*e^3 - 3)/x))/log(2)

Mupad [B] (verification not implemented)

Time = 13.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {3-4 e^3}{4 x}} \left (e^x \left (-3+4 e^3-4 x^2\right )+\left (3 x-4 e^3 x+4 x^2\right ) \log (4)\right )}{4 x^2 \log (4)} \, dx=-\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^3}{x}-\frac {3}{4\,x}}\,\left (4\,{\mathrm {e}}^x-x\,\ln \left (256\right )\right )}{8\,\ln \left (2\right )} \]

[In]

int((exp((exp(3) - 3/4)/x)*((log(2)*(3*x - 4*x*exp(3) + 4*x^2))/4 - (exp(x)*(4*x^2 - 4*exp(3) + 3))/8))/(x^2*l
og(2)),x)

[Out]

-(exp(exp(3)/x - 3/(4*x))*(4*exp(x) - x*log(256)))/(8*log(2))