3.55.83 \(\int \frac {e^{-x+\frac {e^{-x} (-x+e^x (15 x-3 x^2) \log (x))}{\log (x)}} (1+(-1+x) \log (x)+e^x (15-6 x) \log ^2(x))}{\log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ e^{x \left (3 (5-x)-\frac {e^{-x}}{\log (x)}\right )} \]

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Rubi [F]  time = 3.69, 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 {\exp \left (-x+\frac {e^{-x} \left (-x+e^x \left (15 x-3 x^2\right ) \log (x)\right )}{\log (x)}\right ) \left (1+(-1+x) \log (x)+e^x (15-6 x) \log ^2(x)\right )}{\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-x + (-x + E^x*(15*x - 3*x^2)*Log[x])/(E^x*Log[x]))*(1 + (-1 + x)*Log[x] + E^x*(15 - 6*x)*Log[x]^2))/L
og[x]^2,x]

[Out]

15*Defer[Int][E^(15*x - 3*x^2 - x/(E^x*Log[x])), x] - 6*Defer[Int][E^(15*x - 3*x^2 - x/(E^x*Log[x]))*x, x] + D
efer[Int][E^(14*x - 3*x^2 - x/(E^x*Log[x]))/Log[x]^2, x] - Defer[Int][E^(14*x - 3*x^2 - x/(E^x*Log[x]))/Log[x]
, x] + Defer[Int][(E^(14*x - 3*x^2 - x/(E^x*Log[x]))*x)/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) \left (1+(-1+x) \log (x)+e^x (15-6 x) \log ^2(x)\right )}{\log ^2(x)} \, dx\\ &=\int \left (-3 \exp \left (x-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (-5+2 x)+\frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (1-\log (x)+x \log (x))}{\log ^2(x)}\right ) \, dx\\ &=-\left (3 \int \exp \left (x-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (-5+2 x) \, dx\right )+\int \frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (1-\log (x)+x \log (x))}{\log ^2(x)} \, dx\\ &=-\left (3 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} (-5+2 x) \, dx\right )+\int \left (\frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right )}{\log ^2(x)}+\frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (-1+x)}{\log (x)}\right ) \, dx\\ &=-\left (3 \int \left (-5 e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}}+2 e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x\right ) \, dx\right )+\int \frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right )}{\log ^2(x)} \, dx+\int \frac {\exp \left (-\frac {e^{-x} x \left (1-14 e^x \log (x)+3 e^x x \log (x)\right )}{\log (x)}\right ) (-1+x)}{\log (x)} \, dx\\ &=-\left (6 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x \, dx\right )+15 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} \, dx+\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}}}{\log ^2(x)} \, dx+\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}} (-1+x)}{\log (x)} \, dx\\ &=-\left (6 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x \, dx\right )+15 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} \, dx+\int \left (-\frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}}}{\log (x)}+\frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x}{\log (x)}\right ) \, dx+\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}}}{\log ^2(x)} \, dx\\ &=-\left (6 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x \, dx\right )+15 \int e^{15 x-3 x^2-\frac {e^{-x} x}{\log (x)}} \, dx+\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}}}{\log ^2(x)} \, dx-\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}}}{\log (x)} \, dx+\int \frac {e^{14 x-3 x^2-\frac {e^{-x} x}{\log (x)}} x}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.25, size = 20, normalized size = 0.87 \begin {gather*} e^{x \left (15-3 x-\frac {e^{-x}}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + (-x + E^x*(15*x - 3*x^2)*Log[x])/(E^x*Log[x]))*(1 + (-1 + x)*Log[x] + E^x*(15 - 6*x)*Log[x]
^2))/Log[x]^2,x]

[Out]

E^(x*(15 - 3*x - 1/(E^x*Log[x])))

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fricas [A]  time = 2.67, size = 29, normalized size = 1.26 \begin {gather*} e^{\left (x - \frac {{\left ({\left (3 \, x^{2} - 14 \, x\right )} e^{x} \log \relax (x) + x\right )} e^{\left (-x\right )}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+15)*exp(x)*log(x)^2+(x-1)*log(x)+1)*exp(((-3*x^2+15*x)*exp(x)*log(x)-x)/exp(x)/log(x))/exp(x)
/log(x)^2,x, algorithm="fricas")

[Out]

e^(x - ((3*x^2 - 14*x)*e^x*log(x) + x)*e^(-x)/log(x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (3 \, {\left (2 \, x - 5\right )} e^{x} \log \relax (x)^{2} - {\left (x - 1\right )} \log \relax (x) - 1\right )} e^{\left (-x - \frac {{\left (3 \, {\left (x^{2} - 5 \, x\right )} e^{x} \log \relax (x) + x\right )} e^{\left (-x\right )}}{\log \relax (x)}\right )}}{\log \relax (x)^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+15)*exp(x)*log(x)^2+(x-1)*log(x)+1)*exp(((-3*x^2+15*x)*exp(x)*log(x)-x)/exp(x)/log(x))/exp(x)
/log(x)^2,x, algorithm="giac")

[Out]

integrate(-(3*(2*x - 5)*e^x*log(x)^2 - (x - 1)*log(x) - 1)*e^(-x - (3*(x^2 - 5*x)*e^x*log(x) + x)*e^(-x)/log(x
))/log(x)^2, x)

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maple [A]  time = 0.09, size = 28, normalized size = 1.22

method	result	size
risch	\({\mathrm e}^{-\frac {x \left (3 x \,{\mathrm e}^{x} \ln \relax (x )-15 \,{\mathrm e}^{x} \ln \relax (x )+1\right ) {\mathrm e}^{-x}}{\ln \relax (x )}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x+15)*exp(x)*ln(x)^2+(x-1)*ln(x)+1)*exp(((-3*x^2+15*x)*exp(x)*ln(x)-x)/exp(x)/ln(x))/exp(x)/ln(x)^2,x
,method=_RETURNVERBOSE)

[Out]

exp(-x*(3*x*exp(x)*ln(x)-15*exp(x)*ln(x)+1)*exp(-x)/ln(x))

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maxima [A]  time = 0.52, size = 21, normalized size = 0.91 \begin {gather*} e^{\left (-3 \, x^{2} + 15 \, x - \frac {x e^{\left (-x\right )}}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+15)*exp(x)*log(x)^2+(x-1)*log(x)+1)*exp(((-3*x^2+15*x)*exp(x)*log(x)-x)/exp(x)/log(x))/exp(x)
/log(x)^2,x, algorithm="maxima")

[Out]

e^(-3*x^2 + 15*x - x*e^(-x)/log(x))

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mupad [B]  time = 3.62, size = 23, normalized size = 1.00 \begin {gather*} {\mathrm {e}}^{15\,x}\,{\mathrm {e}}^{-3\,x^2}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^{-x}}{\ln \relax (x)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp(-(exp(-x)*(x - exp(x)*log(x)*(15*x - 3*x^2)))/log(x))*(log(x)*(x - 1) - exp(x)*log(x)^2*(6*x
- 15) + 1))/log(x)^2,x)

[Out]

exp(15*x)*exp(-3*x^2)*exp(-(x*exp(-x))/log(x))

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sympy [A]  time = 0.51, size = 24, normalized size = 1.04 \begin {gather*} e^{\frac {\left (- x + \left (- 3 x^{2} + 15 x\right ) e^{x} \log {\relax (x )}\right ) e^{- x}}{\log {\relax (x )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x+15)*exp(x)*ln(x)**2+(x-1)*ln(x)+1)*exp(((-3*x**2+15*x)*exp(x)*ln(x)-x)/exp(x)/ln(x))/exp(x)/l
n(x)**2,x)

[Out]

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

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