3.30.13 \(\int \frac {(50+(100+100 x) \log (2)+(100+100 x) \log (x)) \log (4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x))+(-50 \log (2)-50 \log (x)) \log ^2(4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x))}{x^3 \log (2)+x^3 \log (x)} \, dx\)

Optimal. Leaf size=23 \[ \frac {25 \log ^2\left (4 e^{2 x} x^2 (\log (2)+\log (x))\right )}{x^2} \]

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Rubi [F]  time = 2.20, 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 {(50+(100+100 x) \log (2)+(100+100 x) \log (x)) \log \left (4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x)\right )+(-50 \log (2)-50 \log (x)) \log ^2\left (4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x)\right )}{x^3 \log (2)+x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((50 + (100 + 100*x)*Log[2] + (100 + 100*x)*Log[x])*Log[4*E^(2*x)*x^2*Log[2] + 4*E^(2*x)*x^2*Log[x]] + (-5
0*Log[2] - 50*Log[x])*Log[4*E^(2*x)*x^2*Log[2] + 4*E^(2*x)*x^2*Log[x]]^2)/(x^3*Log[2] + x^3*Log[x]),x]

[Out]

50*(1 + Log[4])*Defer[Int][Log[4*E^(2*x)*x^2*Log[2*x]]/(x^3*Log[2*x]), x] + 50*Log[4]*Defer[Int][Log[4*E^(2*x)
*x^2*Log[2*x]]/(x^2*Log[2*x]), x] + 100*Defer[Int][(Log[x]*Log[4*E^(2*x)*x^2*Log[2*x]])/(x^3*Log[2*x]), x] + 1
00*Defer[Int][(Log[x]*Log[4*E^(2*x)*x^2*Log[2*x]])/(x^2*Log[2*x]), x] - 50*Defer[Int][Log[4*E^(2*x)*x^2*Log[2*
x]]^2/x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {(50+(100+100 x) \log (2)+(100+100 x) \log (x)) \log \left (4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x)\right )+(-50 \log (2)-50 \log (x)) \log ^2\left (4 e^{2 x} x^2 \log (2)+4 e^{2 x} x^2 \log (x)\right )}{x^3 (\log (2)+\log (x))} \, dx\\ &=\int \frac {50 \log \left (4 e^{2 x} x^2 \log (2 x)\right ) \left (1+\log (4)+x \log (4)+2 (1+x) \log (x)-\log (2 x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )\right )}{x^3 \log (2 x)} \, dx\\ &=50 \int \frac {\log \left (4 e^{2 x} x^2 \log (2 x)\right ) \left (1+\log (4)+x \log (4)+2 (1+x) \log (x)-\log (2 x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )\right )}{x^3 \log (2 x)} \, dx\\ &=50 \int \left (\frac {(1+\log (4)+x \log (4)+2 \log (x)+2 x \log (x)) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)}-\frac {\log ^2\left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3}\right ) \, dx\\ &=50 \int \frac {(1+\log (4)+x \log (4)+2 \log (x)+2 x \log (x)) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)} \, dx-50 \int \frac {\log ^2\left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3} \, dx\\ &=-\left (50 \int \frac {\log ^2\left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3} \, dx\right )+50 \int \left (\frac {\log (4) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^2 \log (2 x)}+\frac {(1+\log (4)) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)}+\frac {2 \log (x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)}+\frac {2 \log (x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^2 \log (2 x)}\right ) \, dx\\ &=-\left (50 \int \frac {\log ^2\left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3} \, dx\right )+100 \int \frac {\log (x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)} \, dx+100 \int \frac {\log (x) \log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^2 \log (2 x)} \, dx+(50 \log (4)) \int \frac {\log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^2 \log (2 x)} \, dx+(50 (1+\log (4))) \int \frac {\log \left (4 e^{2 x} x^2 \log (2 x)\right )}{x^3 \log (2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 28, normalized size = 1.22 \begin {gather*} 50 \left (-2+\frac {\log ^2\left (4 e^{2 x} x^2 \log (2 x)\right )}{2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((50 + (100 + 100*x)*Log[2] + (100 + 100*x)*Log[x])*Log[4*E^(2*x)*x^2*Log[2] + 4*E^(2*x)*x^2*Log[x]]
+ (-50*Log[2] - 50*Log[x])*Log[4*E^(2*x)*x^2*Log[2] + 4*E^(2*x)*x^2*Log[x]]^2)/(x^3*Log[2] + x^3*Log[x]),x]

[Out]

50*(-2 + Log[4*E^(2*x)*x^2*Log[2*x]]^2/(2*x^2))

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fricas [A]  time = 1.12, size = 31, normalized size = 1.35 \begin {gather*} \frac {25 \, \log \left (4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (2) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)\right )^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*log(x)-50*log(2))*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2)^2+((100*x+100)*log(x)+(100*
x+100)*log(2)+50)*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2))/(x^3*log(x)+x^3*log(2)),x, algorithm="fric
as")

[Out]

25*log(4*x^2*e^(2*x)*log(2) + 4*x^2*e^(2*x)*log(x))^2/x^2

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giac [B]  time = 0.45, size = 72, normalized size = 3.13 \begin {gather*} 100 \, {\left (\frac {x + \log \relax (2)}{x^{2}} + \frac {\log \relax (x)}{x^{2}}\right )} \log \left (\log \relax (2) + \log \relax (x)\right ) + \frac {200 \, {\left (x + \log \relax (2)\right )} \log \relax (x)}{x^{2}} + \frac {100 \, \log \relax (x)^{2}}{x^{2}} + \frac {25 \, \log \left (\log \relax (2) + \log \relax (x)\right )^{2}}{x^{2}} + \frac {100 \, {\left (2 \, x \log \relax (2) + \log \relax (2)^{2}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*log(x)-50*log(2))*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2)^2+((100*x+100)*log(x)+(100*
x+100)*log(2)+50)*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2))/(x^3*log(x)+x^3*log(2)),x, algorithm="giac
")

[Out]

100*((x + log(2))/x^2 + log(x)/x^2)*log(log(2) + log(x)) + 200*(x + log(2))*log(x)/x^2 + 100*log(x)^2/x^2 + 25
*log(log(2) + log(x))^2/x^2 + 100*(2*x*log(2) + log(2)^2)/x^2

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maple [C]  time = 0.87, size = 5485, normalized size = 238.48

method	result	size
risch	\(\text {Expression too large to display}\)	\(5485\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-50*ln(x)-50*ln(2))*ln(4*x^2*exp(x)^2*ln(x)+4*x^2*ln(2)*exp(x)^2)^2+((100*x+100)*ln(x)+(100*x+100)*ln(2)
+50)*ln(4*x^2*exp(x)^2*ln(x)+4*x^2*ln(2)*exp(x)^2))/(x^3*ln(x)+x^3*ln(2)),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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maxima [B]  time = 0.61, size = 53, normalized size = 2.30 \begin {gather*} \frac {25 \, {\left (8 \, x \log \relax (2) + 4 \, \log \relax (2)^{2} + 8 \, {\left (x + \log \relax (2)\right )} \log \relax (x) + 4 \, \log \relax (x)^{2} + 4 \, {\left (x + \log \relax (2) + \log \relax (x)\right )} \log \left (\log \relax (2) + \log \relax (x)\right ) + \log \left (\log \relax (2) + \log \relax (x)\right )^{2}\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*log(x)-50*log(2))*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2)^2+((100*x+100)*log(x)+(100*
x+100)*log(2)+50)*log(4*x^2*exp(x)^2*log(x)+4*x^2*log(2)*exp(x)^2))/(x^3*log(x)+x^3*log(2)),x, algorithm="maxi
ma")

[Out]

25*(8*x*log(2) + 4*log(2)^2 + 8*(x + log(2))*log(x) + 4*log(x)^2 + 4*(x + log(2) + log(x))*log(log(2) + log(x)
) + log(log(2) + log(x))^2)/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\ln \left (4\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)+4\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\right )}^2\,\left (50\,\ln \relax (2)+50\,\ln \relax (x)\right )-\ln \left (4\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (x)+4\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (2)\right )\,\left (\ln \relax (2)\,\left (100\,x+100\right )+\ln \relax (x)\,\left (100\,x+100\right )+50\right )}{x^3\,\ln \relax (x)+x^3\,\ln \relax (2)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(4*x^2*exp(2*x)*log(x) + 4*x^2*exp(2*x)*log(2))^2*(50*log(2) + 50*log(x)) - log(4*x^2*exp(2*x)*log(x)
+ 4*x^2*exp(2*x)*log(2))*(log(2)*(100*x + 100) + log(x)*(100*x + 100) + 50))/(x^3*log(x) + x^3*log(2)),x)

[Out]

int(-(log(4*x^2*exp(2*x)*log(x) + 4*x^2*exp(2*x)*log(2))^2*(50*log(2) + 50*log(x)) - log(4*x^2*exp(2*x)*log(x)
+ 4*x^2*exp(2*x)*log(2))*(log(2)*(100*x + 100) + log(x)*(100*x + 100) + 50))/(x^3*log(x) + x^3*log(2)), x)

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sympy [A]  time = 0.59, size = 34, normalized size = 1.48 \begin {gather*} \frac {25 \log {\left (4 x^{2} e^{2 x} \log {\relax (x )} + 4 x^{2} e^{2 x} \log {\relax (2 )} \right )}^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-50*ln(x)-50*ln(2))*ln(4*x**2*exp(x)**2*ln(x)+4*x**2*ln(2)*exp(x)**2)**2+((100*x+100)*ln(x)+(100*x
+100)*ln(2)+50)*ln(4*x**2*exp(x)**2*ln(x)+4*x**2*ln(2)*exp(x)**2))/(x**3*ln(x)+x**3*ln(2)),x)

[Out]

25*log(4*x**2*exp(2*x)*log(x) + 4*x**2*exp(2*x)*log(2))**2/x**2

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