3.31.13 \(\int \frac {-1-5 e^{2-x}-\log (x \log (\frac {17}{7}))}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+(-10 e^{2-x} x+2 x \log (4)) \log (x \log (\frac {17}{7}))+x^2 \log ^2(x \log (\frac {17}{7}))} \, dx\) [3013]

Optimal. Leaf size=23 \[ \frac {1}{-5 e^{2-x}+\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )} \]

[Out]

1/(ln(x*ln(17/7))*x-5*exp(2-x)+2*ln(2))

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Rubi [F]
time = 6.36, 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 {-1-5 e^{2-x}-\log \left (x \log \left (\frac {17}{7}\right )\right )}{25 e^{4-2 x}-10 e^{2-x} \log (4)+\log ^2(4)+\left (-10 e^{2-x} x+2 x \log (4)\right ) \log \left (x \log \left (\frac {17}{7}\right )\right )+x^2 \log ^2\left (x \log \left (\frac {17}{7}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-1 - 5*E^(2 - x) - Log[x*Log[17/7]])/(25*E^(4 - 2*x) - 10*E^(2 - x)*Log[4] + Log[4]^2 + (-10*E^(2 - x)*x
+ 2*x*Log[4])*Log[x*Log[17/7]] + x^2*Log[x*Log[17/7]]^2),x]

[Out]

-5*(1 + Log[4])*Defer[Int][E^(2 + x)/((Log[4] + x*Log[x*Log[17/7]])*(-5*E^2 + E^x*Log[4] + E^x*x*Log[x*Log[17/
7]])^2), x] - 5*Defer[Int][(E^(2 + x)*Log[x*Log[17/7]])/((Log[4] + x*Log[x*Log[17/7]])*(-5*E^2 + E^x*Log[4] +
E^x*x*Log[x*Log[17/7]])^2), x] - 5*Defer[Int][(E^(2 + x)*x*Log[x*Log[17/7]])/((Log[4] + x*Log[x*Log[17/7]])*(-
5*E^2 + E^x*Log[4] + E^x*x*Log[x*Log[17/7]])^2), x] - Defer[Int][E^x/((Log[4] + x*Log[x*Log[17/7]])*(-5*E^2 +
E^x*Log[4] + E^x*x*Log[x*Log[17/7]])), x] - Defer[Int][(E^x*Log[x*Log[17/7]])/((Log[4] + x*Log[x*Log[17/7]])*(
-5*E^2 + E^x*Log[4] + E^x*x*Log[x*Log[17/7]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (-5 e^2-e^x-e^x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\\ &=\int \left (-\frac {5 e^{2+x} \left (1+\log (4)+\log \left (x \log \left (\frac {17}{7}\right )\right )+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}-\frac {e^x \left (1+\log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}\right ) \, dx\\ &=-\left (5 \int \frac {e^{2+x} \left (1+\log (4)+\log \left (x \log \left (\frac {17}{7}\right )\right )+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (5 e^2-e^x \log (4)-e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\right )-\int \frac {e^x \left (1+\log \left (x \log \left (\frac {17}{7}\right )\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx\\ &=-\left (5 \int \left (\frac {e^{2+x} (1+\log (4))}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}+\frac {e^{2+x} \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}+\frac {e^{2+x} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2}\right ) \, dx\right )-\int \left (\frac {e^x}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}+\frac {e^x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )}\right ) \, dx\\ &=-\left (5 \int \frac {e^{2+x} \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx\right )-5 \int \frac {e^{2+x} x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx-(5 (1+\log (4))) \int \frac {e^{2+x}}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )^2} \, dx-\int \frac {e^x}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx-\int \frac {e^x \log \left (x \log \left (\frac {17}{7}\right )\right )}{\left (\log (4)+x \log \left (x \log \left (\frac {17}{7}\right )\right )\right ) \left (-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.24, size = 30, normalized size = 1.30 \begin {gather*} \frac {e^x}{-5 e^2+e^x \log (4)+e^x x \log \left (x \log \left (\frac {17}{7}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 5*E^(2 - x) - Log[x*Log[17/7]])/(25*E^(4 - 2*x) - 10*E^(2 - x)*Log[4] + Log[4]^2 + (-10*E^(2 -
 x)*x + 2*x*Log[4])*Log[x*Log[17/7]] + x^2*Log[x*Log[17/7]]^2),x]

[Out]

E^x/(-5*E^2 + E^x*Log[4] + E^x*x*Log[x*Log[17/7]])

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Maple [A]
time = 0.14, size = 28, normalized size = 1.22

method result size
risch \(\frac {1}{x \ln \left (x \left (\ln \left (17\right )-\ln \left (7\right )\right )\right )+2 \ln \left (2\right )-5 \,{\mathrm e}^{2-x}}\) \(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x*ln(17/7))-5*exp(2-x)-1)/(x^2*ln(x*ln(17/7))^2+(-10*x*exp(2-x)+4*x*ln(2))*ln(x*ln(17/7))+25*exp(2-x)
^2-20*exp(2-x)*ln(2)+4*ln(2)^2),x,method=_RETURNVERBOSE)

[Out]

1/(x*ln(x*(ln(17)-ln(7)))+2*ln(2)-5*exp(2-x))

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Maxima [A]
time = 0.57, size = 32, normalized size = 1.39 \begin {gather*} \frac {e^{x}}{{\left (x \log \left (x\right ) + x \log \left (\log \left (17\right ) - \log \left (7\right )\right ) + 2 \, \log \left (2\right )\right )} e^{x} - 5 \, e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x*exp(2-x)+4*x*log(2))*log(x*log(17/7)
)+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*log(2)^2),x, algorithm="maxima")

[Out]

e^x/((x*log(x) + x*log(log(17) - log(7)) + 2*log(2))*e^x - 5*e^2)

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Fricas [A]
time = 0.50, size = 22, normalized size = 0.96 \begin {gather*} \frac {1}{x \log \left (x \log \left (\frac {17}{7}\right )\right ) - 5 \, e^{\left (-x + 2\right )} + 2 \, \log \left (2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x*exp(2-x)+4*x*log(2))*log(x*log(17/7)
)+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*log(2)^2),x, algorithm="fricas")

[Out]

1/(x*log(x*log(17/7)) - 5*e^(-x + 2) + 2*log(2))

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Sympy [A]
time = 0.09, size = 24, normalized size = 1.04 \begin {gather*} - \frac {1}{- x \log {\left (x \log {\left (\frac {17}{7} \right )} \right )} + 5 e^{2 - x} - 2 \log {\left (2 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x*ln(17/7))-5*exp(2-x)-1)/(x**2*ln(x*ln(17/7))**2+(-10*x*exp(2-x)+4*x*ln(2))*ln(x*ln(17/7))+25*
exp(2-x)**2-20*exp(2-x)*ln(2)+4*ln(2)**2),x)

[Out]

-1/(-x*log(x*log(17/7)) + 5*exp(2 - x) - 2*log(2))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (22) = 44\).
time = 0.44, size = 45, normalized size = 1.96 \begin {gather*} \frac {1}{{\left (x - 2\right )} \log \left ({\left (x - 2\right )} \log \left (\frac {17}{7}\right ) + 2 \, \log \left (\frac {17}{7}\right )\right ) - 5 \, e^{\left (-x + 2\right )} + 2 \, \log \left (2\right ) + 2 \, \log \left ({\left (x - 2\right )} \log \left (\frac {17}{7}\right ) + 2 \, \log \left (\frac {17}{7}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x*log(17/7))-5*exp(2-x)-1)/(x^2*log(x*log(17/7))^2+(-10*x*exp(2-x)+4*x*log(2))*log(x*log(17/7)
)+25*exp(2-x)^2-20*exp(2-x)*log(2)+4*log(2)^2),x, algorithm="giac")

[Out]

1/((x - 2)*log((x - 2)*log(17/7) + 2*log(17/7)) - 5*e^(-x + 2) + 2*log(2) + 2*log((x - 2)*log(17/7) + 2*log(17
/7)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {5\,{\mathrm {e}}^{2-x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )+1}{25\,{\mathrm {e}}^{4-2\,x}+\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )\,\left (4\,x\,\ln \left (2\right )-10\,x\,{\mathrm {e}}^{2-x}\right )+x^2\,{\ln \left (x\,\ln \left (\frac {17}{7}\right )\right )}^2-20\,{\mathrm {e}}^{2-x}\,\ln \left (2\right )+4\,{\ln \left (2\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*exp(2 - x) + log(x*log(17/7)) + 1)/(25*exp(4 - 2*x) + log(x*log(17/7))*(4*x*log(2) - 10*x*exp(2 - x))
+ x^2*log(x*log(17/7))^2 - 20*exp(2 - x)*log(2) + 4*log(2)^2),x)

[Out]

int(-(5*exp(2 - x) + log(x*log(17/7)) + 1)/(25*exp(4 - 2*x) + log(x*log(17/7))*(4*x*log(2) - 10*x*exp(2 - x))
+ x^2*log(x*log(17/7))^2 - 20*exp(2 - x)*log(2) + 4*log(2)^2), x)

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