3.57.55 \(\int \frac {e^{\frac {1}{3} (-4-15 x)} (-75 \log ^2(x)+e^{\frac {2}{3} (-1+e^{\frac {x}{\log (x)}})} (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=29 \[ 15 e^{-\frac {4}{3}-5 x} \left (1+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )}\right ) \]

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Rubi [F]  time = 1.61, 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 {1}{3} (-4-15 x)} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-4 - 15*x)/3)*(-75*Log[x]^2 + E^((2*(-1 + E^(x/Log[x])))/3)*(-75*Log[x]^2 + E^(x/Log[x])*(-10 + 10*Lo
g[x]))))/Log[x]^2,x]

[Out]

15*E^(-4/3 - 5*x) - 75*Defer[Int][E^((-6 + 2*E^(x/Log[x]) - 15*x)/3), x] - 10*Defer[Int][E^(-2 + (2*E^(x/Log[x
]))/3 - 5*x + x/Log[x])/Log[x]^2, x] + 10*Defer[Int][E^(-2 + (2*E^(x/Log[x]))/3 - 5*x + x/Log[x])/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {4}{3}-5 x} \left (-75 \log ^2(x)+e^{\frac {2}{3} \left (-1+e^{\frac {x}{\log (x)}}\right )} \left (-75 \log ^2(x)+e^{\frac {x}{\log (x)}} (-10+10 \log (x))\right )\right )}{\log ^2(x)} \, dx\\ &=\int \left (-75 e^{-2-5 x} \left (e^{2/3}+e^{\frac {2}{3} e^{\frac {x}{\log (x)}}}\right )+\frac {10 e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}} (-1+\log (x))}{\log ^2(x)}\right ) \, dx\\ &=10 \int \frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}} (-1+\log (x))}{\log ^2(x)} \, dx-75 \int e^{-2-5 x} \left (e^{2/3}+e^{\frac {2}{3} e^{\frac {x}{\log (x)}}}\right ) \, dx\\ &=10 \int \left (-\frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log ^2(x)}+\frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log (x)}\right ) \, dx-75 \int \left (e^{-\frac {4}{3}-5 x}+e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x}\right ) \, dx\\ &=-\left (10 \int \frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log ^2(x)} \, dx\right )+10 \int \frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log (x)} \, dx-75 \int e^{-\frac {4}{3}-5 x} \, dx-75 \int e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x} \, dx\\ &=15 e^{-\frac {4}{3}-5 x}-10 \int \frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log ^2(x)} \, dx+10 \int \frac {e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x+\frac {x}{\log (x)}}}{\log (x)} \, dx-75 \int e^{\frac {1}{3} \left (-6+2 e^{\frac {x}{\log (x)}}-15 x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.10, size = 33, normalized size = 1.14 \begin {gather*} 15 e^{-\frac {4}{3}-5 x}+15 e^{-2+\frac {2}{3} e^{\frac {x}{\log (x)}}-5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-4 - 15*x)/3)*(-75*Log[x]^2 + E^((2*(-1 + E^(x/Log[x])))/3)*(-75*Log[x]^2 + E^(x/Log[x])*(-10 +
10*Log[x]))))/Log[x]^2,x]

[Out]

15*E^(-4/3 - 5*x) + 15*E^(-2 + (2*E^(x/Log[x]))/3 - 5*x)

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fricas [A]  time = 0.68, size = 26, normalized size = 0.90 \begin {gather*} 15 \, e^{\left (-5 \, x + \frac {2}{3} \, e^{\frac {x}{\log \relax (x)}} - 2\right )} + 15 \, e^{\left (-5 \, x - \frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x))-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/l
og(x)^2,x, algorithm="fricas")

[Out]

15*e^(-5*x + 2/3*e^(x/log(x)) - 2) + 15*e^(-5*x - 4/3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {undef} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x))-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/l
og(x)^2,x, algorithm="giac")

[Out]

undef

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maple [A]  time = 0.07, size = 27, normalized size = 0.93

method	result	size
risch	\(15 \,{\mathrm e}^{-5 x -\frac {4}{3}}+15 \,{\mathrm e}^{-5 x -2+\frac {2 \,{\mathrm e}^{\frac {x}{\ln \relax (x )}}}{3}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*ln(x)-10)*exp(x/ln(x))-75*ln(x)^2)*exp(1/3*exp(x/ln(x))-1/3)^2-75*ln(x)^2)/exp(5*x+4/3)/ln(x)^2,x,me
thod=_RETURNVERBOSE)

[Out]

15*exp(-5*x-4/3)+15*exp(-5*x-2+2/3*exp(x/ln(x)))

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maxima [A]  time = 0.44, size = 26, normalized size = 0.90 \begin {gather*} 15 \, e^{\left (-5 \, x + \frac {2}{3} \, e^{\frac {x}{\log \relax (x)}} - 2\right )} + 15 \, e^{\left (-5 \, x - \frac {4}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*log(x)-10)*exp(x/log(x))-75*log(x)^2)*exp(1/3*exp(x/log(x))-1/3)^2-75*log(x)^2)/exp(5*x+4/3)/l
og(x)^2,x, algorithm="maxima")

[Out]

15*e^(-5*x + 2/3*e^(x/log(x)) - 2) + 15*e^(-5*x - 4/3)

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mupad [B]  time = 3.59, size = 21, normalized size = 0.72 \begin {gather*} 15\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^{-2}\,\left ({\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{\frac {x}{\ln \relax (x)}}}{3}}+{\mathrm {e}}^{2/3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- 5*x - 4/3)*(75*log(x)^2 + exp((2*exp(x/log(x)))/3 - 2/3)*(75*log(x)^2 - exp(x/log(x))*(10*log(x) -
10))))/log(x)^2,x)

[Out]

15*exp(-5*x)*exp(-2)*(exp((2*exp(x/log(x)))/3) + exp(2/3))

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sympy [A]  time = 68.78, size = 37, normalized size = 1.28 \begin {gather*} 15 e^{- 5 x - \frac {4}{3}} e^{\frac {2 e^{\frac {x}{\log {\relax (x )}}}}{3} - \frac {2}{3}} + 15 e^{- 5 x - \frac {4}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*ln(x)-10)*exp(x/ln(x))-75*ln(x)**2)*exp(1/3*exp(x/ln(x))-1/3)**2-75*ln(x)**2)/exp(5*x+4/3)/ln(
x)**2,x)

[Out]

15*exp(-5*x - 4/3)*exp(2*exp(x/log(x))/3 - 2/3) + 15*exp(-5*x - 4/3)

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