3.99.61 \(\int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} (-25-64 x+128 x^2)}{16 x^2} \, dx\)

Optimal. Leaf size=19 \[ \frac {4 e^{-5+\frac {25}{64 x}+2 x}}{x} \]

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Rubi [A]  time = 0.24, antiderivative size = 35, normalized size of antiderivative = 1.84, number of steps used = 3, number of rules used = 3, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 6688, 2288} \begin {gather*} -\frac {4 e^{2 x+\frac {25}{64 x}-5} \left (25-128 x^2\right )}{\left (128-\frac {25}{x^2}\right ) x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((25 - 320*x + 128*x^2 - 64*x*Log[x])/(64*x))*(-25 - 64*x + 128*x^2))/(16*x^2),x]

[Out]

(-4*E^(-5 + 25/(64*x) + 2*x)*(25 - 128*x^2))/((128 - 25/x^2)*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]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{16} \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{x^2} \, dx\\ &=\frac {1}{16} \int \frac {e^{-5+\frac {25}{64 x}+2 x} \left (-25-64 x+128 x^2\right )}{x^3} \, dx\\ &=-\frac {4 e^{-5+\frac {25}{64 x}+2 x} \left (25-128 x^2\right )}{\left (128-\frac {25}{x^2}\right ) x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.00 \begin {gather*} \frac {4 e^{-5+\frac {25}{64 x}+2 x}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((25 - 320*x + 128*x^2 - 64*x*Log[x])/(64*x))*(-25 - 64*x + 128*x^2))/(16*x^2),x]

[Out]

(4*E^(-5 + 25/(64*x) + 2*x))/x

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fricas [A]  time = 0.94, size = 23, normalized size = 1.21 \begin {gather*} 4 \, e^{\left (\frac {128 \, x^{2} - 64 \, x \log \relax (x) - 320 \, x + 25}{64 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(128*x^2-64*x-25)*exp(1/64*(-64*x*log(x)+128*x^2-320*x+25)/x)/x^2,x, algorithm="fricas")

[Out]

4*e^(1/64*(128*x^2 - 64*x*log(x) - 320*x + 25)/x)

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giac [A]  time = 0.15, size = 17, normalized size = 0.89 \begin {gather*} 4 \, e^{\left (2 \, x + \frac {25}{64 \, x} - \log \relax (x) - 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(128*x^2-64*x-25)*exp(1/64*(-64*x*log(x)+128*x^2-320*x+25)/x)/x^2,x, algorithm="giac")

[Out]

4*e^(2*x + 25/64/x - log(x) - 5)

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maple [A]  time = 0.09, size = 22, normalized size = 1.16

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{\frac {128 x^{2}-320 x +25}{64 x}}}{x}\)	\(22\)
gosper	\(4 \,{\mathrm e}^{-\frac {64 x \ln \relax (x )-128 x^{2}+320 x -25}{64 x}}\)	\(24\)
norman	\(4 \,{\mathrm e}^{\frac {-64 x \ln \relax (x )+128 x^{2}-320 x +25}{64 x}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/16*(128*x^2-64*x-25)*exp(1/64*(-64*x*ln(x)+128*x^2-320*x+25)/x)/x^2,x,method=_RETURNVERBOSE)

[Out]

4/x*exp(1/64*(128*x^2-320*x+25)/x)

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maxima [A]  time = 0.41, size = 16, normalized size = 0.84 \begin {gather*} \frac {4 \, e^{\left (2 \, x + \frac {25}{64 \, x} - 5\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(128*x^2-64*x-25)*exp(1/64*(-64*x*log(x)+128*x^2-320*x+25)/x)/x^2,x, algorithm="maxima")

[Out]

4*e^(2*x + 25/64/x - 5)/x

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mupad [B]  time = 5.63, size = 16, normalized size = 0.84 \begin {gather*} \frac {4\,{\mathrm {e}}^{2\,x+\frac {25}{64\,x}-5}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(5*x + x*log(x) - 2*x^2 - 25/64)/x)*(64*x - 128*x^2 + 25))/(16*x^2),x)

[Out]

(4*exp(2*x + 25/(64*x) - 5))/x

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sympy [A]  time = 0.34, size = 20, normalized size = 1.05 \begin {gather*} 4 e^{\frac {2 x^{2} - x \log {\relax (x )} - 5 x + \frac {25}{64}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/16*(128*x**2-64*x-25)*exp(1/64*(-64*x*ln(x)+128*x**2-320*x+25)/x)/x**2,x)

[Out]

4*exp((2*x**2 - x*log(x) - 5*x + 25/64)/x)

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