\(\int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} (-25-64 x+128 x^2)}{16 x^2} \, dx\) [9861]

Optimal result

Integrand size = 41, antiderivative size = 19 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=\frac {4 e^{-5+\frac {25}{64 x}+2 x}}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.073, Rules used = {12, 6820, 2326} \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=-\frac {4 e^{2 x+\frac {25}{64 x}-5} \left (25-128 x^2\right )}{\left (128-\frac {25}{x^2}\right ) x^3} \]

[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 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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=\frac {4 e^{-5+\frac {25}{64 x}+2 x}}{x} \]

[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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=4 \, e^{\left (\frac {128 \, x^{2} - 64 \, x \log \left (x\right ) - 320 \, x + 25}{64 \, x}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=4 e^{\frac {2 x^{2} - x \log {\left (x \right )} - 5 x + \frac {25}{64}}{x}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=\frac {4 \, e^{\left (2 \, x + \frac {25}{64 \, x} - 5\right )}}{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=4 \, e^{\left (2 \, x + \frac {25}{64 \, x} - \log \left (x\right ) - 5\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 15.42 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {25-320 x+128 x^2-64 x \log (x)}{64 x}} \left (-25-64 x+128 x^2\right )}{16 x^2} \, dx=\frac {4\,{\mathrm {e}}^{2\,x+\frac {25}{64\,x}-5}}{x} \]

[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