3.79.47 \(\int \frac {e^x (-3+3 x)+(-3 e^x x+3 e^x \log (x)) \log (-x+\log (x))+(5 x-2 e^x x+(-5+2 e^x) \log (x)) \log ^2(-x+\log (x))}{(-e^x x+e^x \log (x)) \log ^2(-x+\log (x))} \, dx\)

Optimal. Leaf size=24 \[ 3+5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \]

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Rubi [F]  time = 0.49, 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^x (-3+3 x)+\left (-3 e^x x+3 e^x \log (x)\right ) \log (-x+\log (x))+\left (5 x-2 e^x x+\left (-5+2 e^x\right ) \log (x)\right ) \log ^2(-x+\log (x))}{\left (-e^x x+e^x \log (x)\right ) \log ^2(-x+\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-3 + 3*x) + (-3*E^x*x + 3*E^x*Log[x])*Log[-x + Log[x]] + (5*x - 2*E^x*x + (-5 + 2*E^x)*Log[x])*Log[-
x + Log[x]]^2)/((-(E^x*x) + E^x*Log[x])*Log[-x + Log[x]]^2),x]

[Out]

5/E^x + 2*x + 3*Defer[Int][1/((x - Log[x])*Log[-x + Log[x]]^2), x] - 3*Defer[Int][x/((x - Log[x])*Log[-x + Log
[x]]^2), x] + 3*Defer[Int][Log[-x + Log[x]]^(-1), x]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 23, normalized size = 0.96 \begin {gather*} 5 e^{-x}+2 x+\frac {3 x}{\log (-x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-3 + 3*x) + (-3*E^x*x + 3*E^x*Log[x])*Log[-x + Log[x]] + (5*x - 2*E^x*x + (-5 + 2*E^x)*Log[x])
*Log[-x + Log[x]]^2)/((-(E^x*x) + E^x*Log[x])*Log[-x + Log[x]]^2),x]

[Out]

5/E^x + 2*x + (3*x)/Log[-x + Log[x]]

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fricas [A]  time = 0.68, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \relax (x)\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)*log(x)-3*exp(x)*x)*log(log(x)-x)+(3*
x-3)*exp(x))/(exp(x)*log(x)-exp(x)*x)/log(log(x)-x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.26, size = 37, normalized size = 1.54 \begin {gather*} \frac {2 \, x \log \left (-x + \log \relax (x)\right ) + 5 \, e^{\left (-x\right )} \log \left (-x + \log \relax (x)\right ) + 3 \, x}{\log \left (-x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)*log(x)-3*exp(x)*x)*log(log(x)-x)+(3*
x-3)*exp(x))/(exp(x)*log(x)-exp(x)*x)/log(log(x)-x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 26, normalized size = 1.08




method result size



risch \(\left (2 \,{\mathrm e}^{x} x +5\right ) {\mathrm e}^{-x}+\frac {3 x}{\ln \left (\ln \relax (x )-x \right )}\) \(26\)
default \(5 \,{\mathrm e}^{-x}+\frac {3 x +2 \ln \left (\ln \relax (x )-x \right ) x}{\ln \left (\ln \relax (x )-x \right )}\) \(32\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*exp(x)-5)*ln(x)-2*exp(x)*x+5*x)*ln(ln(x)-x)^2+(3*exp(x)*ln(x)-3*exp(x)*x)*ln(ln(x)-x)+(3*x-3)*exp(x))
/(exp(x)*ln(x)-exp(x)*x)/ln(ln(x)-x)^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.39, size = 35, normalized size = 1.46 \begin {gather*} \frac {{\left (3 \, x e^{x} + {\left (2 \, x e^{x} + 5\right )} \log \left (-x + \log \relax (x)\right )\right )} e^{\left (-x\right )}}{\log \left (-x + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(x)-5)*log(x)-2*exp(x)*x+5*x)*log(log(x)-x)^2+(3*exp(x)*log(x)-3*exp(x)*x)*log(log(x)-x)+(3*
x-3)*exp(x))/(exp(x)*log(x)-exp(x)*x)/log(log(x)-x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.87, size = 62, normalized size = 2.58 \begin {gather*} 5\,x+5\,{\mathrm {e}}^{-x}-3\,\ln \relax (x)+\frac {3}{x-1}-\frac {3\,x^2}{x-1}-\frac {3\,\ln \relax (x)}{x-1}+\frac {3\,x}{\ln \left (\ln \relax (x)-x\right )}+\frac {3\,x\,\ln \relax (x)}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*x + 5*exp(-x) - 3*log(x) + 3/(x - 1) - (3*x^2)/(x - 1) - (3*log(x))/(x - 1) + (3*x)/log(log(x) - x) + (3*x*l
og(x))/(x - 1)

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sympy [A]  time = 0.59, size = 17, normalized size = 0.71 \begin {gather*} 2 x + \frac {3 x}{\log {\left (- x + \log {\relax (x )} \right )}} + 5 e^{- x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*exp(x)-5)*ln(x)-2*exp(x)*x+5*x)*ln(ln(x)-x)**2+(3*exp(x)*ln(x)-3*exp(x)*x)*ln(ln(x)-x)+(3*x-3)*
exp(x))/(exp(x)*ln(x)-exp(x)*x)/ln(ln(x)-x)**2,x)

[Out]

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

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