3.81.76 \(\int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (-25 x-100 x \log (x)-200 x \log ^2(x))}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+(100-40 \log (2)+4 \log ^2(2)) \log (x)} \, dx\)

Optimal. Leaf size=34 \[ \frac {x^2 \log (x)}{-1-e^{\frac {\frac {3}{4}+\log (x)}{3 \log (x)}}+\frac {\log (2)}{5}} \]

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Rubi [F]  time = 2.62, 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 {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-100*x + 20*x*Log[2])*Log[x] + (-200*x + 40*x*Log[2])*Log[x]^2 + E^((3 + 4*Log[x])/(12*Log[x]))*(-25*x -
100*x*Log[x] - 200*x*Log[x]^2))/(100*E^((3 + 4*Log[x])/(6*Log[x]))*Log[x] + E^((3 + 4*Log[x])/(12*Log[x]))*(2
00 - 40*Log[2])*Log[x] + (100 - 40*Log[2] + 4*Log[2]^2)*Log[x]),x]

[Out]

5*Defer[Int][x/(-5*E^(1/3 + 1/(4*Log[x])) - 5*(1 - Log[2]/5)), x] + (5*Defer[Int][x/((-5*E^(1/3 + 1/(4*Log[x])
) - 5*(1 - Log[2]/5))*Log[x]), x])/4 + (5*(5 - Log[2])*Defer[Int][x/((5*E^(1/3 + 1/(4*Log[x])) + 5*(1 - Log[2]
/5))^2*Log[x]), x])/4 + 10*Defer[Int][(x*Log[x])/(-5*E^(1/3 + 1/(4*Log[x])) - 5*(1 - Log[2]/5)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{4 \left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)} \, dx\\ &=\frac {1}{4} \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)} \, dx\\ &=\frac {1}{4} \int \left (\frac {5 x (5-\log (2))}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)}+\frac {5 x \left (-1-4 \log (x)-8 \log ^2(x)\right )}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right ) \log (x)}\right ) \, dx\\ &=\frac {5}{4} \int \frac {x \left (-1-4 \log (x)-8 \log ^2(x)\right )}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right ) \log (x)} \, dx+\frac {1}{4} (5 (5-\log (2))) \int \frac {x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)} \, dx\\ &=\frac {5}{4} \int \left (\frac {4 x}{-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )}+\frac {x}{\left (-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )\right ) \log (x)}+\frac {8 x \log (x)}{-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )}\right ) \, dx+\frac {1}{4} (5 (5-\log (2))) \int \frac {x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)} \, dx\\ &=\frac {5}{4} \int \frac {x}{\left (-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )\right ) \log (x)} \, dx+5 \int \frac {x}{-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )} \, dx+10 \int \frac {x \log (x)}{-5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-5 \left (1-\frac {\log (2)}{5}\right )} \, dx+\frac {1}{4} (5 (5-\log (2))) \int \frac {x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 31, normalized size = 0.91 \begin {gather*} -\frac {5 x^2 \log (x)}{5+5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-100*x + 20*x*Log[2])*Log[x] + (-200*x + 40*x*Log[2])*Log[x]^2 + E^((3 + 4*Log[x])/(12*Log[x]))*(-
25*x - 100*x*Log[x] - 200*x*Log[x]^2))/(100*E^((3 + 4*Log[x])/(6*Log[x]))*Log[x] + E^((3 + 4*Log[x])/(12*Log[x
]))*(200 - 40*Log[2])*Log[x] + (100 - 40*Log[2] + 4*Log[2]^2)*Log[x]),x]

[Out]

(-5*x^2*Log[x])/(5 + 5*E^(1/3 + 1/(4*Log[x])) - Log[2])

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fricas [A]  time = 0.84, size = 30, normalized size = 0.88 \begin {gather*} -\frac {5 \, x^{2} \log \relax (x)}{5 \, e^{\left (\frac {4 \, \log \relax (x) + 3}{12 \, \log \relax (x)}\right )} - \log \relax (2) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x))+(40*x*log(2)-200*x)*log(x)^2+(20*
x*log(2)-100*x)*log(x))/(100*log(x)*exp(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)
+3)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="fricas")

[Out]

-5*x^2*log(x)/(5*e^(1/12*(4*log(x) + 3)/log(x)) - log(2) + 5)

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giac [A]  time = 0.44, size = 26, normalized size = 0.76 \begin {gather*} -\frac {5 \, x^{2} \log \relax (x)}{5 \, e^{\left (\frac {1}{4 \, \log \relax (x)} + \frac {1}{3}\right )} - \log \relax (2) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x))+(40*x*log(2)-200*x)*log(x)^2+(20*
x*log(2)-100*x)*log(x))/(100*log(x)*exp(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)
+3)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="giac")

[Out]

-5*x^2*log(x)/(5*e^(1/4/log(x) + 1/3) - log(2) + 5)

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maple [A]  time = 0.19, size = 29, normalized size = 0.85

method	result	size
norman	\(\frac {5 x^{2} \ln \relax (x )}{-5+\ln \relax (2)-5 \,{\mathrm e}^{\frac {4 \ln \relax (x )+3}{12 \ln \relax (x )}}}\)	\(29\)
risch	\(\frac {5 x^{2} \ln \relax (x )}{-5+\ln \relax (2)-5 \,{\mathrm e}^{\frac {4 \ln \relax (x )+3}{12 \ln \relax (x )}}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-200*x*ln(x)^2-100*x*ln(x)-25*x)*exp(1/12*(4*ln(x)+3)/ln(x))+(40*x*ln(2)-200*x)*ln(x)^2+(20*x*ln(2)-100*
x)*ln(x))/(100*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))^2+(-40*ln(2)+200)*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))+(4*ln(2)^
2-40*ln(2)+100)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

5*x^2*ln(x)/(-5+ln(2)-5*exp(1/12*(4*ln(x)+3)/ln(x)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {5}{4} \, \int \frac {8 \, {\left (x \log \relax (2) - 5 \, x\right )} \log \relax (x)^{2} - 5 \, {\left (8 \, x \log \relax (x)^{2} + 4 \, x \log \relax (x) + x\right )} e^{\left (\frac {4 \, \log \relax (x) + 3}{12 \, \log \relax (x)}\right )} + 4 \, {\left (x \log \relax (2) - 5 \, x\right )} \log \relax (x)}{10 \, {\left (\log \relax (2) - 5\right )} e^{\left (\frac {4 \, \log \relax (x) + 3}{12 \, \log \relax (x)}\right )} \log \relax (x) - {\left (\log \relax (2)^{2} - 10 \, \log \relax (2) + 25\right )} \log \relax (x) - 25 \, e^{\left (\frac {4 \, \log \relax (x) + 3}{6 \, \log \relax (x)}\right )} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x))+(40*x*log(2)-200*x)*log(x)^2+(20*
x*log(2)-100*x)*log(x))/(100*log(x)*exp(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)
+3)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="maxima")

[Out]

-5/4*integrate((8*(x*log(2) - 5*x)*log(x)^2 - 5*(8*x*log(x)^2 + 4*x*log(x) + x)*e^(1/12*(4*log(x) + 3)/log(x))
+ 4*(x*log(2) - 5*x)*log(x))/(10*(log(2) - 5)*e^(1/12*(4*log(x) + 3)/log(x))*log(x) - (log(2)^2 - 10*log(2) +
25)*log(x) - 25*e^(1/6*(4*log(x) + 3)/log(x))*log(x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {\frac {\ln \relax (x)}{3}+\frac {1}{4}}{\ln \relax (x)}}\,\left (200\,x\,{\ln \relax (x)}^2+100\,x\,\ln \relax (x)+25\,x\right )+{\ln \relax (x)}^2\,\left (200\,x-40\,x\,\ln \relax (2)\right )+\ln \relax (x)\,\left (100\,x-20\,x\,\ln \relax (2)\right )}{\ln \relax (x)\,\left (4\,{\ln \relax (2)}^2-40\,\ln \relax (2)+100\right )+100\,{\mathrm {e}}^{\frac {2\,\left (\frac {\ln \relax (x)}{3}+\frac {1}{4}\right )}{\ln \relax (x)}}\,\ln \relax (x)-{\mathrm {e}}^{\frac {\frac {\ln \relax (x)}{3}+\frac {1}{4}}{\ln \relax (x)}}\,\ln \relax (x)\,\left (40\,\ln \relax (2)-200\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((log(x)/3 + 1/4)/log(x))*(25*x + 200*x*log(x)^2 + 100*x*log(x)) + log(x)^2*(200*x - 40*x*log(2)) + l
og(x)*(100*x - 20*x*log(2)))/(log(x)*(4*log(2)^2 - 40*log(2) + 100) + 100*exp((2*(log(x)/3 + 1/4))/log(x))*log
(x) - exp((log(x)/3 + 1/4)/log(x))*log(x)*(40*log(2) - 200)),x)

[Out]

int(-(exp((log(x)/3 + 1/4)/log(x))*(25*x + 200*x*log(x)^2 + 100*x*log(x)) + log(x)^2*(200*x - 40*x*log(2)) + l
og(x)*(100*x - 20*x*log(2)))/(log(x)*(4*log(2)^2 - 40*log(2) + 100) + 100*exp((2*(log(x)/3 + 1/4))/log(x))*log
(x) - exp((log(x)/3 + 1/4)/log(x))*log(x)*(40*log(2) - 200)), x)

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sympy [A]  time = 0.33, size = 29, normalized size = 0.85 \begin {gather*} - \frac {5 x^{2} \log {\relax (x )}}{5 e^{\frac {\frac {\log {\relax (x )}}{3} + \frac {1}{4}}{\log {\relax (x )}}} - \log {\relax (2 )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-200*x*ln(x)**2-100*x*ln(x)-25*x)*exp(1/12*(4*ln(x)+3)/ln(x))+(40*x*ln(2)-200*x)*ln(x)**2+(20*x*ln
(2)-100*x)*ln(x))/(100*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))**2+(-40*ln(2)+200)*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))+
(4*ln(2)**2-40*ln(2)+100)*ln(x)),x)

[Out]

-5*x**2*log(x)/(5*exp((log(x)/3 + 1/4)/log(x)) - log(2) + 5)

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