3.91.1 \(\int \frac {e^{\frac {2 x}{\log (x)}} (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} (-2+2 \log (x)-2 \log ^2(x) \log (2 x))+e^{-15+x-x \log (2 x)} (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)))}{5 \log ^2(x)} \, dx\)

Optimal. Leaf size=31 \[ \frac {1}{5} e^{\frac {2 x}{\log (x)}} \left (-e^{-15+x-x \log (2 x)}+x\right )^2 \]

________________________________________________________________________________________

Rubi [F]  time = 6.61, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{5 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x] - 2*
Log[x]^2*Log[2*x]) + E^(-15 + x - x*Log[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/(5*Log
[x]^2),x]

[Out]

(E^((2*x)/Log[x])*x*(x - x*Log[x]))/(5*(Log[x]^(-2) - Log[x]^(-1))*Log[x]^2) - Defer[Int][(2^(1 - x)*E^(-15 +
x + (2*x)/Log[x]))/x^x, x]/5 + Defer[Int][(2^(2 - x)*E^(-15 + x + (2*x)/Log[x])*x^(1 - x))/Log[x]^2, x]/5 - De
fer[Int][(2^(1 - 2*x)*E^(-30 + 2*x + (2*x)/Log[x]))/(x^(2*x)*Log[x]^2), x]/5 - Defer[Int][(2^(2 - x)*E^(-15 +
x + (2*x)/Log[x])*x^(1 - x))/Log[x], x]/5 + Defer[Int][(2^(1 - 2*x)*E^(-30 + 2*x + (2*x)/Log[x]))/(x^(2*x)*Log
[x]), x]/5 + Defer[Int][2^(1 - x)*E^(-15 + x + (2*x)/Log[x])*x^(1 - x)*Log[2*x], x]/5 - Defer[Int][(2^(1 - 2*x
)*E^(-30 + 2*x + (2*x)/Log[x])*Log[2*x])/x^(2*x), x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{\log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (\frac {2 e^{\frac {2 x}{\log (x)}} x \left (-x+x \log (x)+\log ^2(x)\right )}{\log ^2(x)}-\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \left (1-\log (x)+\log ^2(x) \log (2 x)\right )}{\log ^2(x)}+\frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)+x \log ^2(x) \log (2 x)\right )}{\log ^2(x)}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \left (1-\log (x)+\log ^2(x) \log (2 x)\right )}{\log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)+x \log ^2(x) \log (2 x)\right )}{\log ^2(x)} \, dx+\frac {2}{5} \int \frac {e^{\frac {2 x}{\log (x)}} x \left (-x+x \log (x)+\log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}+\frac {1}{5} \int \left (\frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)\right )}{\log ^2(x)}+2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x)\right ) \, dx-\frac {1}{5} \int \left (\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} (1-\log (x))}{\log ^2(x)}+2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x)\right ) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}-\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} (1-\log (x))}{\log ^2(x)} \, dx+\frac {1}{5} \int \frac {2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \left (2 x-2 x \log (x)-\log ^2(x)\right )}{\log ^2(x)} \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}+\frac {1}{5} \int \left (-2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x}+\frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log ^2(x)}-\frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log (x)}\right ) \, dx-\frac {1}{5} \int \left (\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log ^2(x)}-\frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log (x)}\right ) \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ &=\frac {e^{\frac {2 x}{\log (x)}} x (x-x \log (x))}{5 \left (\frac {1}{\log ^2(x)}-\frac {1}{\log (x)}\right ) \log ^2(x)}-\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{-x} \, dx+\frac {1}{5} \int \frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log ^2(x)} \, dx-\frac {1}{5} \int \frac {2^{2-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x}}{\log (x)} \, dx+\frac {1}{5} \int \frac {2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x}}{\log (x)} \, dx+\frac {1}{5} \int 2^{1-x} e^{-15+x+\frac {2 x}{\log (x)}} x^{1-x} \log (2 x) \, dx-\frac {1}{5} \int 2^{1-2 x} e^{-30+2 x+\frac {2 x}{\log (x)}} x^{-2 x} \log (2 x) \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [F]  time = 0.73, size = 110, normalized size = 3.55 \begin {gather*} \frac {1}{5} \int \frac {e^{\frac {2 x}{\log (x)}} \left (-2 x^2+2 x^2 \log (x)+2 x \log ^2(x)+e^{-30+2 x-2 x \log (2 x)} \left (-2+2 \log (x)-2 \log ^2(x) \log (2 x)\right )+e^{-15+x-x \log (2 x)} \left (4 x-4 x \log (x)-2 \log ^2(x)+2 x \log ^2(x) \log (2 x)\right )\right )}{\log ^2(x)} \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x
] - 2*Log[x]^2*Log[2*x]) + E^(-15 + x - x*Log[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/
(5*Log[x]^2),x]

[Out]

Integrate[(E^((2*x)/Log[x])*(-2*x^2 + 2*x^2*Log[x] + 2*x*Log[x]^2 + E^(-30 + 2*x - 2*x*Log[2*x])*(-2 + 2*Log[x
] - 2*Log[x]^2*Log[2*x]) + E^(-15 + x - x*Log[2*x])*(4*x - 4*x*Log[x] - 2*Log[x]^2 + 2*x*Log[x]^2*Log[2*x])))/
Log[x]^2, x]/5

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 47, normalized size = 1.52 \begin {gather*} \frac {1}{5} \, {\left (x^{2} - 2 \, x e^{\left (-x \log \relax (2) - x \log \relax (x) + x - 15\right )} + e^{\left (-2 \, x \log \relax (2) - 2 \, x \log \relax (x) + 2 \, x - 30\right )}\right )} e^{\left (\frac {2 \, x}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+(2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x
*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x*log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="fri
cas")

[Out]

1/5*(x^2 - 2*x*e^(-x*log(2) - x*log(x) + x - 15) + e^(-2*x*log(2) - 2*x*log(x) + 2*x - 30))*e^(2*x/log(x))

________________________________________________________________________________________

giac [B]  time = 0.26, size = 83, normalized size = 2.68 \begin {gather*} \frac {1}{5} \, x^{2} e^{\left (\frac {2 \, x}{\log \relax (x)}\right )} - \frac {2}{5} \, x e^{\left (-\frac {x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} - x \log \relax (x) - 2 \, x + 15 \, \log \relax (x)}{\log \relax (x)}\right )} + \frac {1}{5} \, e^{\left (-\frac {2 \, {\left (x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} - x \log \relax (x) - x + 15 \, \log \relax (x)\right )}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+(2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x
*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x*log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="gia
c")

[Out]

1/5*x^2*e^(2*x/log(x)) - 2/5*x*e^(-(x*log(2)*log(x) + x*log(x)^2 - x*log(x) - 2*x + 15*log(x))/log(x)) + 1/5*e
^(-2*(x*log(2)*log(x) + x*log(x)^2 - x*log(x) - x + 15*log(x))/log(x))

________________________________________________________________________________________

maple [B]  time = 0.11, size = 75, normalized size = 2.42




method result size



risch \(\frac {x^{2} {\mathrm e}^{\frac {2 x}{\ln \relax (x )}}}{5}-\frac {2 \left (\frac {1}{2}\right )^{x} x^{-x} x \,{\mathrm e}^{\frac {x \ln \relax (x )-15 \ln \relax (x )+2 x}{\ln \relax (x )}}}{5}+\frac {2^{-2 x} x^{-2 x} {\mathrm e}^{\frac {2 x \ln \relax (x )-30 \ln \relax (x )+2 x}{\ln \relax (x )}}}{5}\) \(75\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-2*ln(x)^2*ln(2*x)+2*ln(x)-2)*exp(-x*ln(2*x)+x-15)^2+(2*x*ln(x)^2*ln(2*x)-2*ln(x)^2-4*x*ln(x)+4*x)*e
xp(-x*ln(2*x)+x-15)+2*x*ln(x)^2+2*x^2*ln(x)-2*x^2)*exp(x/ln(x))^2/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*x^2*exp(2*x/ln(x))-2/5*(1/2)^x*x^(-x)*x*exp((x*ln(x)-15*ln(x)+2*x)/ln(x))+1/5*((1/2)^x)^2*(x^(-x))^2*exp(2
*(x*ln(x)-15*ln(x)+x)/ln(x))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*log(x)^2*log(2*x)+2*log(x)-2)*exp(-x*log(2*x)+x-15)^2+(2*x*log(x)^2*log(2*x)-2*log(x)^2-4*x
*log(x)+4*x)*exp(-x*log(2*x)+x-15)+2*x*log(x)^2+2*x^2*log(x)-2*x^2)*exp(x/log(x))^2/log(x)^2,x, algorithm="max
ima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
 of the expected type LIST

________________________________________________________________________________________

mupad [B]  time = 8.49, size = 54, normalized size = 1.74 \begin {gather*} {\mathrm {e}}^{\frac {2\,x}{\ln \relax (x)}}\,\left (\frac {x^2}{5}+\frac {{\mathrm {e}}^{2\,x-30}}{5\,2^{2\,x}\,x^{2\,x}}-\frac {2\,x\,{\mathrm {e}}^{x-15}}{5\,2^x\,x^x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x)/log(x))*(2*x*log(x)^2 + 2*x^2*log(x) - exp(2*x - 2*x*log(2*x) - 30)*(2*log(2*x)*log(x)^2 - 2*lo
g(x) + 2) + exp(x - x*log(2*x) - 15)*(4*x - 2*log(x)^2 - 4*x*log(x) + 2*x*log(2*x)*log(x)^2) - 2*x^2))/(5*log(
x)^2),x)

[Out]

exp((2*x)/log(x))*(x^2/5 + exp(2*x - 30)/(5*2^(2*x)*x^(2*x)) - (2*x*exp(x - 15))/(5*2^x*x^x))

________________________________________________________________________________________

sympy [B]  time = 33.25, size = 66, normalized size = 2.13 \begin {gather*} \frac {x^{2} e^{\frac {2 x}{\log {\relax (x )}}}}{5} - \frac {2 x e^{\frac {2 x}{\log {\relax (x )}}} e^{- x \left (\log {\relax (x )} + \log {\relax (2 )}\right ) + x - 15}}{5} + \frac {e^{\frac {2 x}{\log {\relax (x )}}} e^{- 2 x \left (\log {\relax (x )} + \log {\relax (2 )}\right ) + 2 x - 30}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-2*ln(x)**2*ln(2*x)+2*ln(x)-2)*exp(-x*ln(2*x)+x-15)**2+(2*x*ln(x)**2*ln(2*x)-2*ln(x)**2-4*x*ln
(x)+4*x)*exp(-x*ln(2*x)+x-15)+2*x*ln(x)**2+2*x**2*ln(x)-2*x**2)*exp(x/ln(x))**2/ln(x)**2,x)

[Out]

x**2*exp(2*x/log(x))/5 - 2*x*exp(2*x/log(x))*exp(-x*(log(x) + log(2)) + x - 15)/5 + exp(2*x/log(x))*exp(-2*x*(
log(x) + log(2)) + 2*x - 30)/5

________________________________________________________________________________________