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

Optimal. Leaf size=19 \[ 2-\frac {2}{5 \log \left (\left (-e^x+x\right ) \log (3)\right )} \]

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Rubi [A]  time = 0.09, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6686} \begin {gather*} -\frac {2}{5 \log \left (x \log (3)-e^x \log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(5*Log[-(E^x*Log[3]) + x*Log[3]])

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.89 \begin {gather*} -\frac {2}{5 \log \left (\left (-e^x+x\right ) \log (3)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-2/(5*Log[(-E^x + x)*Log[3]])

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fricas [A]  time = 0.50, size = 16, normalized size = 0.84 \begin {gather*} -\frac {2}{5 \, \log \left (x \log \relax (3) - e^{x} \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/log(x*log(3) - e^x*log(3))

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giac [A]  time = 0.14, size = 16, normalized size = 0.84 \begin {gather*} -\frac {2}{5 \, \log \left (x \log \relax (3) - e^{x} \log \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/log(x*log(3) - e^x*log(3))

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maple [A]  time = 0.07, size = 17, normalized size = 0.89

method	result	size
norman	\(-\frac {2}{5 \ln \left (-\ln \relax (3) {\mathrm e}^{x}+x \ln \relax (3)\right )}\)	\(17\)
risch	\(-\frac {2}{5 \ln \left (-\ln \relax (3) {\mathrm e}^{x}+x \ln \relax (3)\right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 15, normalized size = 0.79 \begin {gather*} -\frac {2}{5 \, {\left (\log \left (x - e^{x}\right ) + \log \left (\log \relax (3)\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5/(log(x - e^x) + log(log(3)))

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mupad [B]  time = 2.59, size = 16, normalized size = 0.84 \begin {gather*} -\frac {2}{5\,\ln \left (x\,\ln \relax (3)-{\mathrm {e}}^x\,\ln \relax (3)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(5*log(x*log(3) - exp(x)*log(3)))

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sympy [A]  time = 0.14, size = 17, normalized size = 0.89 \begin {gather*} - \frac {2}{5 \log {\left (x \log {\relax (3 )} - e^{x} \log {\relax (3 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(5*log(x*log(3) - exp(x)*log(3)))

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