3.77.83 \(\int \frac {e^{-x} (-1+(1-x) \log (-\frac {x}{2})-\log ^2(-\frac {x}{2}))}{\log ^2(-\frac {x}{2})} \, dx\)

Optimal. Leaf size=23 \[ \frac {e^{-x} \left (x+\frac {x^2}{\log \left (-\frac {x}{2}\right )}\right )}{x} \]

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Rubi [F]  time = 0.48, 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} \left (-1+(1-x) \log \left (-\frac {x}{2}\right )-\log ^2\left (-\frac {x}{2}\right )\right )}{\log ^2\left (-\frac {x}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

E^(-x) - 4*Defer[Subst][Defer[Int][x/(E^(2*x)*Log[-x]), x], x, x/2] + 2*Defer[Subst][Defer[Int][E^(2*x)/Log[x]
^2, x], x, -1/2*x] - 2*Defer[Subst][Defer[Int][E^(2*x)/Log[x], x], x, -1/2*x]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 18, normalized size = 0.78 \begin {gather*} e^{-x} \left (1+\frac {x}{\log \left (-\frac {x}{2}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1 + x/Log[-1/2*x])/E^x

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fricas [A]  time = 0.56, size = 23, normalized size = 1.00 \begin {gather*} \frac {x e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (-\frac {1}{2} \, x\right )}{\log \left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 23, normalized size = 1.00 \begin {gather*} \frac {x e^{\left (-x\right )} + e^{\left (-x\right )} \log \left (-\frac {1}{2} \, x\right )}{\log \left (-\frac {1}{2} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 18, normalized size = 0.78

method	result	size
norman	\(\frac {\left (x +\ln \left (-\frac {x}{2}\right )\right ) {\mathrm e}^{-x}}{\ln \left (-\frac {x}{2}\right )}\)	\(18\)
risch	\({\mathrm e}^{-x}+\frac {x \,{\mathrm e}^{-x}}{\ln \left (-\frac {x}{2}\right )}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x+ln(-1/2*x))/exp(x)/ln(-1/2*x)

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maxima [A]  time = 0.49, size = 24, normalized size = 1.04 \begin {gather*} -\frac {x}{e^{x} \log \relax (2) - e^{x} \log \left (-x\right )} + e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.99, size = 17, normalized size = 0.74 \begin {gather*} {\mathrm {e}}^{-x}+\frac {x\,{\mathrm {e}}^{-x}}{\ln \left (-\frac {x}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(-x) + (x*exp(-x))/log(-x/2)

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sympy [A]  time = 0.27, size = 17, normalized size = 0.74 \begin {gather*} \frac {\left (x + \log {\left (- \frac {x}{2} \right )}\right ) e^{- x}}{\log {\left (- \frac {x}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x + log(-x/2))*exp(-x)/log(-x/2)

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