3.51.17 \(\int \frac {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} (24-e^5+10 x+x^2+(-6-2 x) \log (8))}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx\)

Optimal. Leaf size=28 \[ -4+e^{\frac {-3+e^5+(3+x)^2}{5 (5+x-\log (8))}}+x \]

________________________________________________________________________________________

Rubi [F]  time = 2.30, 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 {125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )}{125+50 x+5 x^2+(-50-10 x) \log (8)+5 \log ^2(8)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2 + E^((-6 - E^5 - 6*x - x^2)/(-25 - 5*x + 5*Log[8]))
*(24 - E^5 + 10*x + x^2 + (-6 - 2*x)*Log[8]))/(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2),x]

[Out]

x + (10*(5 - Log[8]))/(5 + x - Log[8]) - (5 - Log[8])^2/(5 + x - Log[8]) + (2*Log[8]^2)/(5 + x - Log[8]) - (25
+ Log[8]^2)/(5 + x - Log[8]) + 10*Log[5 + x - Log[8]] - 2*(5 - Log[8])*Log[5 + x - Log[8]] - 2*Log[8]*Log[5 +
x - Log[8]] + Defer[Int][E^((6 + E^5 + 6*x + x^2)/(5*(5 + x - Log[8]))), x]/5 - ((1 + E^5 - 4*Log[8] + Log[8]
^2)*Defer[Int][E^((6 + E^5 + 6*x + x^2)/(5*(5 + x - Log[8])))/(5 + x - Log[8])^2, x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{5 x^2+10 x (5-\log (8))+5 (5-\log (8))^2} \, dx\\ &=\int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{5 (5+x-\log (8))^2} \, dx\\ &=\frac {1}{5} \int \frac {50 x+5 x^2+(-50-10 x) \log (8)+e^{\frac {-6-e^5-6 x-x^2}{-25-5 x+5 \log (8)}} \left (24-e^5+10 x+x^2+(-6-2 x) \log (8)\right )+125 \left (1+\frac {\log ^2(8)}{25}\right )}{(5+x-\log (8))^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {50 x}{(5+x-\log (8))^2}+\frac {5 x^2}{(5+x-\log (8))^2}+\frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (24-e^5+x^2+2 x (5-\log (8))-6 \log (8)\right )}{(5+x-\log (8))^2}-\frac {10 (5+x) \log (8)}{(5+x-\log (8))^2}+\frac {5 \left (25+\log ^2(8)\right )}{(5+x-\log (8))^2}\right ) \, dx\\ &=-\frac {25+\log ^2(8)}{5+x-\log (8)}+\frac {1}{5} \int \frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (24-e^5+x^2+2 x (5-\log (8))-6 \log (8)\right )}{(5+x-\log (8))^2} \, dx+10 \int \frac {x}{(5+x-\log (8))^2} \, dx-(2 \log (8)) \int \frac {5+x}{(5+x-\log (8))^2} \, dx+\int \frac {x^2}{(5+x-\log (8))^2} \, dx\\ &=-\frac {25+\log ^2(8)}{5+x-\log (8)}+\frac {1}{5} \int \left (e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}}+\frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \left (-1-e^5+4 \log (8)-\log ^2(8)\right )}{(5+x-\log (8))^2}\right ) \, dx+10 \int \left (\frac {1}{5+x-\log (8)}+\frac {-5+\log (8)}{(5+x-\log (8))^2}\right ) \, dx-(2 \log (8)) \int \left (\frac {1}{5+x-\log (8)}+\frac {\log (8)}{(5+x-\log (8))^2}\right ) \, dx+\int \left (1+\frac {2 (-5+\log (8))}{5+x-\log (8)}+\frac {(-5+\log (8))^2}{(5+x-\log (8))^2}\right ) \, dx\\ &=x+\frac {10 (5-\log (8))}{5+x-\log (8)}-\frac {(5-\log (8))^2}{5+x-\log (8)}+\frac {2 \log ^2(8)}{5+x-\log (8)}-\frac {25+\log ^2(8)}{5+x-\log (8)}+10 \log (5+x-\log (8))-2 (5-\log (8)) \log (5+x-\log (8))-2 \log (8) \log (5+x-\log (8))+\frac {1}{5} \int e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}} \, dx+\frac {1}{5} \left (-1-e^5+4 \log (8)-\log ^2(8)\right ) \int \frac {e^{\frac {6+e^5+6 x+x^2}{5 (5+x-\log (8))}}}{(5+x-\log (8))^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.38, size = 63, normalized size = 2.25 \begin {gather*} \frac {1}{5} \left (5\ 2^{-\frac {3 \log (8)}{5 (5+x-\log (8))}} e^{\frac {6+e^5+6 x+x^2+\log ^2(8)}{5 (5+x-\log (8))}}+5 (5+x-\log (8))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2 + E^((-6 - E^5 - 6*x - x^2)/(-25 - 5*x + 5*Lo
g[8]))*(24 - E^5 + 10*x + x^2 + (-6 - 2*x)*Log[8]))/(125 + 50*x + 5*x^2 + (-50 - 10*x)*Log[8] + 5*Log[8]^2),x]

[Out]

((5*E^((6 + E^5 + 6*x + x^2 + Log[8]^2)/(5*(5 + x - Log[8]))))/2^((3*Log[8])/(5*(5 + x - Log[8]))) + 5*(5 + x
- Log[8]))/5

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 24, normalized size = 0.86 \begin {gather*} x + e^{\left (\frac {x^{2} + 6 \, x + e^{5} + 6}{5 \, {\left (x - 3 \, \log \relax (2) + 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/(15*log(2)-5*x-25))+45*log(2)^2+3*(-
10*x-50)*log(2)+5*x^2+50*x+125)/(45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm="fricas")

[Out]

x + e^(1/5*(x^2 + 6*x + e^5 + 6)/(x - 3*log(2) + 5))

________________________________________________________________________________________

giac [B]  time = 0.47, size = 67, normalized size = 2.39 \begin {gather*} x + e^{\left (\frac {3 \, x^{2} \log \relax (2) - 5 \, x^{2} + x e^{5} + 18 \, x \log \relax (2) - 24 \, x}{5 \, {\left (3 \, x \log \relax (2) - 9 \, \log \relax (2)^{2} - 5 \, x + 30 \, \log \relax (2) - 25\right )}} - \frac {e^{5} + 6}{5 \, {\left (3 \, \log \relax (2) - 5\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/(15*log(2)-5*x-25))+45*log(2)^2+3*(-
10*x-50)*log(2)+5*x^2+50*x+125)/(45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm="giac")

[Out]

x + e^(1/5*(3*x^2*log(2) - 5*x^2 + x*e^5 + 18*x*log(2) - 24*x)/(3*x*log(2) - 9*log(2)^2 - 5*x + 30*log(2) - 25
) - 1/5*(e^5 + 6)/(3*log(2) - 5))

________________________________________________________________________________________

maple [A]  time = 0.35, size = 27, normalized size = 0.96

method	result	size
risch	\(x +{\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{5}+6 x +6}{5 \left (3 \ln \relax (2)-x -5\right )}}\)	\(27\)
norman	\(\frac {\left (3 \ln \relax (2)-5\right ) {\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \relax (2)-5 x -25}}-x^{2}-x \,{\mathrm e}^{\frac {-{\mathrm e}^{5}-x^{2}-6 x -6}{15 \ln \relax (2)-5 x -25}}+9 \ln \relax (2)^{2}-30 \ln \relax (2)+25}{3 \ln \relax (2)-x -5}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*(-2*x-6)*ln(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/(15*ln(2)-5*x-25))+45*ln(2)^2+3*(-10*x-50)*
ln(2)+5*x^2+50*x+125)/(45*ln(2)^2+3*(-10*x-50)*ln(2)+5*x^2+50*x+125),x,method=_RETURNVERBOSE)

[Out]

x+exp(-1/5*(x^2+exp(5)+6*x+6)/(3*ln(2)-x-5))

________________________________________________________________________________________

maxima [B]  time = 0.61, size = 200, normalized size = 7.14 \begin {gather*} 6 \, {\left (\frac {3 \, \log \relax (2) - 5}{x - 3 \, \log \relax (2) + 5} - \log \left (x - 3 \, \log \relax (2) + 5\right )\right )} \log \relax (2) + 2 \, {\left (3 \, \log \relax (2) - 5\right )} \log \left (x - 3 \, \log \relax (2) + 5\right ) + 2^{\frac {3}{5}} e^{\left (\frac {1}{5} \, x + \frac {9 \, \log \relax (2)^{2}}{5 \, {\left (x - 3 \, \log \relax (2) + 5\right )}} + \frac {e^{5}}{5 \, {\left (x - 3 \, \log \relax (2) + 5\right )}} - \frac {12 \, \log \relax (2)}{5 \, {\left (x - 3 \, \log \relax (2) + 5\right )}} + \frac {1}{5 \, {\left (x - 3 \, \log \relax (2) + 5\right )}} + \frac {1}{5}\right )} + x - \frac {9 \, \log \relax (2)^{2}}{x - 3 \, \log \relax (2) + 5} - \frac {9 \, \log \relax (2)^{2} - 30 \, \log \relax (2) + 25}{x - 3 \, \log \relax (2) + 5} - \frac {10 \, {\left (3 \, \log \relax (2) - 5\right )}}{x - 3 \, \log \relax (2) + 5} + \frac {30 \, \log \relax (2)}{x - 3 \, \log \relax (2) + 5} - \frac {25}{x - 3 \, \log \relax (2) + 5} + 10 \, \log \left (x - 3 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(-2*x-6)*log(2)-exp(5)+x^2+10*x+24)*exp((-exp(5)-x^2-6*x-6)/(15*log(2)-5*x-25))+45*log(2)^2+3*(-
10*x-50)*log(2)+5*x^2+50*x+125)/(45*log(2)^2+3*(-10*x-50)*log(2)+5*x^2+50*x+125),x, algorithm="maxima")

[Out]

6*((3*log(2) - 5)/(x - 3*log(2) + 5) - log(x - 3*log(2) + 5))*log(2) + 2*(3*log(2) - 5)*log(x - 3*log(2) + 5)
+ 2^(3/5)*e^(1/5*x + 9/5*log(2)^2/(x - 3*log(2) + 5) + 1/5*e^5/(x - 3*log(2) + 5) - 12/5*log(2)/(x - 3*log(2)
+ 5) + 1/5/(x - 3*log(2) + 5) + 1/5) + x - 9*log(2)^2/(x - 3*log(2) + 5) - (9*log(2)^2 - 30*log(2) + 25)/(x -
3*log(2) + 5) - 10*(3*log(2) - 5)/(x - 3*log(2) + 5) + 30*log(2)/(x - 3*log(2) + 5) - 25/(x - 3*log(2) + 5) +
10*log(x - 3*log(2) + 5)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {50\,x-3\,\ln \relax (2)\,\left (10\,x+50\right )+45\,{\ln \relax (2)}^2+{\mathrm {e}}^{\frac {x^2+6\,x+{\mathrm {e}}^5+6}{5\,x-15\,\ln \relax (2)+25}}\,\left (10\,x-{\mathrm {e}}^5-3\,\ln \relax (2)\,\left (2\,x+6\right )+x^2+24\right )+5\,x^2+125}{50\,x-3\,\ln \relax (2)\,\left (10\,x+50\right )+45\,{\ln \relax (2)}^2+5\,x^2+125} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + exp((6*x + exp(5) + x^2 + 6)/(5*x - 15*log(2) + 25))*(10*x -
exp(5) - 3*log(2)*(2*x + 6) + x^2 + 24) + 5*x^2 + 125)/(50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + 5*x^2 + 12
5),x)

[Out]

int((50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + exp((6*x + exp(5) + x^2 + 6)/(5*x - 15*log(2) + 25))*(10*x -
exp(5) - 3*log(2)*(2*x + 6) + x^2 + 24) + 5*x^2 + 125)/(50*x - 3*log(2)*(10*x + 50) + 45*log(2)^2 + 5*x^2 + 12
5), x)

________________________________________________________________________________________

sympy [A]  time = 0.52, size = 26, normalized size = 0.93 \begin {gather*} x + e^{\frac {- x^{2} - 6 x - e^{5} - 6}{- 5 x - 25 + 15 \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*(-2*x-6)*ln(2)-exp(5)+x**2+10*x+24)*exp((-exp(5)-x**2-6*x-6)/(15*ln(2)-5*x-25))+45*ln(2)**2+3*(-
10*x-50)*ln(2)+5*x**2+50*x+125)/(45*ln(2)**2+3*(-10*x-50)*ln(2)+5*x**2+50*x+125),x)

[Out]

x + exp((-x**2 - 6*x - exp(5) - 6)/(-5*x - 25 + 15*log(2)))

________________________________________________________________________________________