3.26.1 \(\int \frac {e^{\frac {64+16 e^{40+2 x}+e^{20+x} (-64-128 \log (5))+256 \log (5)+256 \log ^2(5)}{x^2}} (-1280+e^{40+2 x} (-320+320 x)-5120 \log (5)-5120 \log ^2(5)+e^{20+x} (1280-640 x+(2560-1280 x) \log (5)))}{x^3} \, dx\)

Optimal. Leaf size=24 \[ 10 e^{\frac {16 \left (2-e^{20+x}+4 \log (5)\right )^2}{x^2}} \]

________________________________________________________________________________________

Rubi [A]  time = 3.97, antiderivative size = 47, normalized size of antiderivative = 1.96, number of steps used = 1, number of rules used = 1, integrand size = 87, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.011, Rules used = {6706} \begin {gather*} 2\ 5^{\frac {256}{x^2}+1} \exp \left (\frac {16 \left (e^{2 x+40}-4 e^{x+20} (1+\log (25))+4 \left (1+4 \log ^2(5)\right )\right )}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((64 + 16*E^(40 + 2*x) + E^(20 + x)*(-64 - 128*Log[5]) + 256*Log[5] + 256*Log[5]^2)/x^2)*(-1280 + E^(40
+ 2*x)*(-320 + 320*x) - 5120*Log[5] - 5120*Log[5]^2 + E^(20 + x)*(1280 - 640*x + (2560 - 1280*x)*Log[5])))/x^
3,x]

[Out]

2*5^(1 + 256/x^2)*E^((16*(E^(40 + 2*x) + 4*(1 + 4*Log[5]^2) - 4*E^(20 + x)*(1 + Log[25])))/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} &=2\ 5^{1+\frac {256}{x^2}} \exp \left (\frac {16 \left (e^{40+2 x}+4 \left (1+4 \log ^2(5)\right )-4 e^{20+x} (1+\log (25))\right )}{x^2}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [F]  time = 7.51, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {64+16 e^{40+2 x}+e^{20+x} (-64-128 \log (5))+256 \log (5)+256 \log ^2(5)}{x^2}} \left (-1280+e^{40+2 x} (-320+320 x)-5120 \log (5)-5120 \log ^2(5)+e^{20+x} (1280-640 x+(2560-1280 x) \log (5))\right )}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((64 + 16*E^(40 + 2*x) + E^(20 + x)*(-64 - 128*Log[5]) + 256*Log[5] + 256*Log[5]^2)/x^2)*(-1280 +
E^(40 + 2*x)*(-320 + 320*x) - 5120*Log[5] - 5120*Log[5]^2 + E^(20 + x)*(1280 - 640*x + (2560 - 1280*x)*Log[5]
)))/x^3,x]

[Out]

Integrate[(E^((64 + 16*E^(40 + 2*x) + E^(20 + x)*(-64 - 128*Log[5]) + 256*Log[5] + 256*Log[5]^2)/x^2)*(-1280 +
E^(40 + 2*x)*(-320 + 320*x) - 5120*Log[5] - 5120*Log[5]^2 + E^(20 + x)*(1280 - 640*x + (2560 - 1280*x)*Log[5]
)))/x^3, x]

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 40, normalized size = 1.67 \begin {gather*} 10 \, e^{\left (-\frac {16 \, {\left (4 \, {\left (2 \, \log \relax (5) + 1\right )} e^{\left (x + 20\right )} - 16 \, \log \relax (5)^{2} - e^{\left (2 \, x + 40\right )} - 16 \, \log \relax (5) - 4\right )}}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((320*x-320)*exp(20+x)^2+((-1280*x+2560)*log(5)-640*x+1280)*exp(20+x)-5120*log(5)^2-5120*log(5)-1280
)*exp((16*exp(20+x)^2+(-128*log(5)-64)*exp(20+x)+256*log(5)^2+256*log(5)+64)/x^2)/x^3,x, algorithm="fricas")

[Out]

10*e^(-16*(4*(2*log(5) + 1)*e^(x + 20) - 16*log(5)^2 - e^(2*x + 40) - 16*log(5) - 4)/x^2)

________________________________________________________________________________________

giac [B]  time = 0.40, size = 56, normalized size = 2.33 \begin {gather*} 10 \, e^{\left (-\frac {128 \, e^{\left (x + 20\right )} \log \relax (5)}{x^{2}} + \frac {256 \, \log \relax (5)^{2}}{x^{2}} + \frac {16 \, e^{\left (2 \, x + 40\right )}}{x^{2}} - \frac {64 \, e^{\left (x + 20\right )}}{x^{2}} + \frac {256 \, \log \relax (5)}{x^{2}} + \frac {64}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((320*x-320)*exp(20+x)^2+((-1280*x+2560)*log(5)-640*x+1280)*exp(20+x)-5120*log(5)^2-5120*log(5)-1280
)*exp((16*exp(20+x)^2+(-128*log(5)-64)*exp(20+x)+256*log(5)^2+256*log(5)+64)/x^2)/x^3,x, algorithm="giac")

[Out]

10*e^(-128*e^(x + 20)*log(5)/x^2 + 256*log(5)^2/x^2 + 16*e^(2*x + 40)/x^2 - 64*e^(x + 20)/x^2 + 256*log(5)/x^2
+ 64/x^2)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 39, normalized size = 1.62

method	result	size
norman	\(10 \,{\mathrm e}^{\frac {16 \,{\mathrm e}^{40+2 x}+\left (-128 \ln \relax (5)-64\right ) {\mathrm e}^{20+x}+256 \ln \relax (5)^{2}+256 \ln \relax (5)+64}{x^{2}}}\)	\(39\)
risch	\(10 \,{\mathrm e}^{\frac {-128 \,{\mathrm e}^{20+x} \ln \relax (5)+256 \ln \relax (5)^{2}+16 \,{\mathrm e}^{40+2 x}-64 \,{\mathrm e}^{20+x}+256 \ln \relax (5)+64}{x^{2}}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((320*x-320)*exp(20+x)^2+((-1280*x+2560)*ln(5)-640*x+1280)*exp(20+x)-5120*ln(5)^2-5120*ln(5)-1280)*exp((16
*exp(20+x)^2+(-128*ln(5)-64)*exp(20+x)+256*ln(5)^2+256*ln(5)+64)/x^2)/x^3,x,method=_RETURNVERBOSE)

[Out]

10*exp((16*exp(20+x)^2+(-128*ln(5)-64)*exp(20+x)+256*ln(5)^2+256*ln(5)+64)/x^2)

________________________________________________________________________________________

maxima [B]  time = 0.75, size = 56, normalized size = 2.33 \begin {gather*} 10 \, e^{\left (-\frac {128 \, e^{\left (x + 20\right )} \log \relax (5)}{x^{2}} + \frac {256 \, \log \relax (5)^{2}}{x^{2}} + \frac {16 \, e^{\left (2 \, x + 40\right )}}{x^{2}} - \frac {64 \, e^{\left (x + 20\right )}}{x^{2}} + \frac {256 \, \log \relax (5)}{x^{2}} + \frac {64}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((320*x-320)*exp(20+x)^2+((-1280*x+2560)*log(5)-640*x+1280)*exp(20+x)-5120*log(5)^2-5120*log(5)-1280
)*exp((16*exp(20+x)^2+(-128*log(5)-64)*exp(20+x)+256*log(5)^2+256*log(5)+64)/x^2)/x^3,x, algorithm="maxima")

[Out]

10*e^(-128*e^(x + 20)*log(5)/x^2 + 256*log(5)^2/x^2 + 16*e^(2*x + 40)/x^2 - 64*e^(x + 20)/x^2 + 256*log(5)/x^2
+ 64/x^2)

________________________________________________________________________________________

mupad [B]  time = 1.70, size = 56, normalized size = 2.33 \begin {gather*} 2\,5^{1-\frac {128\,\left ({\mathrm {e}}^{20}\,{\mathrm {e}}^x-2\right )}{x^2}}\,{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{40}}{x^2}}\,{\mathrm {e}}^{\frac {256\,{\ln \relax (5)}^2}{x^2}}\,{\mathrm {e}}^{\frac {64}{x^2}}\,{\mathrm {e}}^{-\frac {64\,{\mathrm {e}}^{20}\,{\mathrm {e}}^x}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((256*log(5) + 16*exp(2*x + 40) - exp(x + 20)*(128*log(5) + 64) + 256*log(5)^2 + 64)/x^2)*(5120*log(5
) - exp(2*x + 40)*(320*x - 320) + 5120*log(5)^2 + exp(x + 20)*(640*x + log(5)*(1280*x - 2560) - 1280) + 1280))
/x^3,x)

[Out]

2*5^(1 - (128*(exp(20)*exp(x) - 2))/x^2)*exp((16*exp(2*x)*exp(40))/x^2)*exp((256*log(5)^2)/x^2)*exp(64/x^2)*ex
p(-(64*exp(20)*exp(x))/x^2)

________________________________________________________________________________________

sympy [B]  time = 0.40, size = 41, normalized size = 1.71 \begin {gather*} 10 e^{\frac {\left (- 128 \log {\relax (5 )} - 64\right ) e^{x + 20} + 16 e^{2 x + 40} + 64 + 256 \log {\relax (5 )} + 256 \log {\relax (5 )}^{2}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((320*x-320)*exp(20+x)**2+((-1280*x+2560)*ln(5)-640*x+1280)*exp(20+x)-5120*ln(5)**2-5120*ln(5)-1280)
*exp((16*exp(20+x)**2+(-128*ln(5)-64)*exp(20+x)+256*ln(5)**2+256*ln(5)+64)/x**2)/x**3,x)

[Out]

10*exp(((-128*log(5) - 64)*exp(x + 20) + 16*exp(2*x + 40) + 64 + 256*log(5) + 256*log(5)**2)/x**2)

________________________________________________________________________________________