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

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

________________________________________________________________________________________

Rubi [F]  time = 6.22, 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*}

________________________________________________________________________________________

Mathematica [A]  time = 1.35, 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]])

________________________________________________________________________________________

fricas [A]  time = 0.53, 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 \relax (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="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 2.15, size = 808, normalized size = 35.13 result too large to display

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]

(x^2*e^x*log(x)^2 + 2*x^2*e^x*log(x)*log(log(17) - log(7)) + x^2*e^x*log(log(17) - log(7))^2 + 4*x*e^x*log(2)*
log(x) + x*e^x*log(x)^2 + 4*x*e^x*log(2)*log(log(17) - log(7)) + 2*x*e^x*log(x)*log(log(17) - log(7)) + x*e^x*
log(log(17) - log(7))^2 + 4*e^x*log(2)^2 - 5*x*e^2*log(x) + x*e^x*log(x) + 2*e^x*log(2)*log(x) - 5*x*e^2*log(l
og(17) - log(7)) + x*e^x*log(log(17) - log(7)) + 2*e^x*log(2)*log(log(17) - log(7)) - 10*e^2*log(2) + 2*e^x*lo
g(2) - 5*e^2*log(x) - 5*e^2*log(log(17) - log(7)) - 5*e^2)/(x^3*e^x*log(x)^3 + 3*x^3*e^x*log(x)^2*log(log(17)
- log(7)) + 3*x^3*e^x*log(x)*log(log(17) - log(7))^2 + x^3*e^x*log(log(17) - log(7))^3 + 6*x^2*e^x*log(2)*log(
x)^2 + x^2*e^x*log(x)^3 + 12*x^2*e^x*log(2)*log(x)*log(log(17) - log(7)) + 3*x^2*e^x*log(x)^2*log(log(17) - lo
g(7)) + 6*x^2*e^x*log(2)*log(log(17) - log(7))^2 + 3*x^2*e^x*log(x)*log(log(17) - log(7))^2 + x^2*e^x*log(log(
17) - log(7))^3 + 12*x*e^x*log(2)^2*log(x) - 10*x^2*e^2*log(x)^2 + x^2*e^x*log(x)^2 + 4*x*e^x*log(2)*log(x)^2
+ 12*x*e^x*log(2)^2*log(log(17) - log(7)) - 20*x^2*e^2*log(x)*log(log(17) - log(7)) + 2*x^2*e^x*log(x)*log(log
(17) - log(7)) + 8*x*e^x*log(2)*log(x)*log(log(17) - log(7)) - 10*x^2*e^2*log(log(17) - log(7))^2 + x^2*e^x*lo
g(log(17) - log(7))^2 + 4*x*e^x*log(2)*log(log(17) - log(7))^2 + 8*e^x*log(2)^3 - 40*x*e^2*log(2)*log(x) + 4*x
*e^x*log(2)*log(x) + 4*e^x*log(2)^2*log(x) - 10*x*e^2*log(x)^2 - 40*x*e^2*log(2)*log(log(17) - log(7)) + 4*x*e
^x*log(2)*log(log(17) - log(7)) + 4*e^x*log(2)^2*log(log(17) - log(7)) - 20*x*e^2*log(x)*log(log(17) - log(7))
- 10*x*e^2*log(log(17) - log(7))^2 - 40*e^2*log(2)^2 + 4*e^x*log(2)^2 - 10*x*e^2*log(x) + 25*x*e^(-x + 4)*log
(x) - 20*e^2*log(2)*log(x) - 10*x*e^2*log(log(17) - log(7)) + 25*x*e^(-x + 4)*log(log(17) - log(7)) - 20*e^2*l
og(2)*log(log(17) - log(7)) - 20*e^2*log(2) + 50*e^(-x + 4)*log(2) + 25*e^(-x + 4)*log(x) + 25*e^(-x + 4)*log(
log(17) - log(7)) + 25*e^(-x + 4))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 28, normalized size = 1.22

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

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))

________________________________________________________________________________________

maxima [A]  time = 0.59, size = 32, normalized size = 1.39 \begin {gather*} \frac {e^{x}}{{\left (x \log \relax (x) + x \log \left (\log \left (17\right ) - \log \relax (7)\right ) + 2 \, \log \relax (2)\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)

________________________________________________________________________________________

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 \relax (2)-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 \relax (2)+4\,{\ln \relax (2)}^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)

________________________________________________________________________________________

sympy [A]  time = 0.31, 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 {\relax (2 )}} \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))

________________________________________________________________________________________