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

Optimal. Leaf size=29 \[ e^{-\frac {3}{4 x}+\frac {e^3}{x}} \left (x-\frac {e^x}{\log (4)}\right ) \]

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Rubi [A]  time = 0.64, antiderivative size = 43, normalized size of antiderivative = 1.48, number of steps used = 5, number of rules used = 4, integrand size = 62, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 6742, 6706, 2288} \begin {gather*} e^{-\frac {3-4 e^3}{4 x}} x-\frac {e^{x-\frac {3-4 e^3}{4 x}}}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[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 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]

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]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 35, normalized size = 1.21 \begin {gather*} \frac {e^{\frac {-3+4 e^3}{4 x}} \left (-4 e^x+4 x \log (4)\right )}{4 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.51, size = 28, normalized size = 0.97 \begin {gather*} \frac {{\left (2 \, x \log \relax (2) - e^{x}\right )} e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.16, size = 43, normalized size = 1.48 \begin {gather*} \frac {2 \, x e^{\left (\frac {4 \, e^{3} - 3}{4 \, x}\right )} \log \relax (2) - e^{\left (\frac {4 \, x^{2} + 4 \, e^{3} - 3}{4 \, x}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.12, size = 29, normalized size = 1.00

method	result	size
risch	\(\frac {\left (8 x \ln \relax (2)-4 \,{\mathrm e}^{x}\right ) {\mathrm e}^{\frac {4 \,{\mathrm e}^{3}-3}{4 x}}}{8 \ln \relax (2)}\)	\(29\)
norman	\(\frac {\left (x^{2}-\frac {x \,{\mathrm e}^{x}}{2 \ln \relax (2)}\right ) {\mathrm e}^{-\frac {-4 \,{\mathrm e}^{3}+3}{4 x}}}{x}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.50, size = 29, normalized size = 1.00 \begin {gather*} \frac {{\left (2 \, x \log \relax (2) - e^{x}\right )} e^{\left (\frac {e^{3}}{x} - \frac {3}{4 \, x}\right )}}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.33, size = 29, normalized size = 1.00 \begin {gather*} -\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 \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 1.93, size = 24, normalized size = 0.83 \begin {gather*} \frac {\left (2 x \log {\relax (2 )} - e^{x}\right ) e^{- \frac {\frac {3}{4} - e^{3}}{x}}}{2 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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