3.100.54 \(\int \frac {-10 x-10 x \log (\frac {100}{2401 x^2}) \log (5 \log (\frac {100}{2401 x^2}))+30 x \log (\frac {100}{2401 x^2}) \log ^2(5 \log (\frac {100}{2401 x^2}))}{\log (\frac {100}{2401 x^2}) \log ^2(5 \log (\frac {100}{2401 x^2}))} \, dx\)

Optimal. Leaf size=23 \[ -5+x^2 \left (15-\frac {5}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}\right ) \]

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Rubi [F]  time = 0.25, 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 {-10 x-10 x \log \left (\frac {100}{2401 x^2}\right ) \log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )+30 x \log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}{\log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10*x - 10*x*Log[100/(2401*x^2)]*Log[5*Log[100/(2401*x^2)]] + 30*x*Log[100/(2401*x^2)]*Log[5*Log[100/(240
1*x^2)]]^2)/(Log[100/(2401*x^2)]*Log[5*Log[100/(2401*x^2)]]^2),x]

[Out]

15*x^2 - 5*Defer[Subst][Defer[Int][1/(Log[100/(2401*x)]*Log[5*Log[100/(2401*x)]]^2), x], x, x^2] - 5*Defer[Sub
st][Defer[Int][Log[5*Log[100/(2401*x)]]^(-1), x], x, x^2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 10 x \left (3-\frac {1}{\log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}-\frac {1}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}\right ) \, dx\\ &=10 \int x \left (3-\frac {1}{\log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}-\frac {1}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}\right ) \, dx\\ &=10 \int \left (3 x-\frac {x}{\log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}-\frac {x}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )}\right ) \, dx\\ &=15 x^2-10 \int \frac {x}{\log \left (\frac {100}{2401 x^2}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x^2}\right )\right )} \, dx-10 \int \frac {x}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )} \, dx\\ &=15 x^2-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (\frac {100}{2401 x}\right ) \log ^2\left (5 \log \left (\frac {100}{2401 x}\right )\right )} \, dx,x,x^2\right )-5 \operatorname {Subst}\left (\int \frac {1}{\log \left (5 \log \left (\frac {100}{2401 x}\right )\right )} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 24, normalized size = 1.04 \begin {gather*} 15 x^2-\frac {5 x^2}{\log \left (5 \log \left (\frac {100}{2401 x^2}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x - 10*x*Log[100/(2401*x^2)]*Log[5*Log[100/(2401*x^2)]] + 30*x*Log[100/(2401*x^2)]*Log[5*Log[10
0/(2401*x^2)]]^2)/(Log[100/(2401*x^2)]*Log[5*Log[100/(2401*x^2)]]^2),x]

[Out]

15*x^2 - (5*x^2)/Log[5*Log[100/(2401*x^2)]]

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fricas [A]  time = 0.55, size = 33, normalized size = 1.43 \begin {gather*} \frac {5 \, {\left (3 \, x^{2} \log \left (5 \, \log \left (\frac {100}{2401 \, x^{2}}\right )\right ) - x^{2}\right )}}{\log \left (5 \, \log \left (\frac {100}{2401 \, x^{2}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))^2-10*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))-1
0*x)/log(100/2401/x^2)/log(5*log(100/2401/x^2))^2,x, algorithm="fricas")

[Out]

5*(3*x^2*log(5*log(100/2401/x^2)) - x^2)/log(5*log(100/2401/x^2))

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giac [B]  time = 0.54, size = 79, normalized size = 3.43 \begin {gather*} 15 \, x^{2} - \frac {5 \, x^{2} \log \left (\frac {100}{2401 \, x^{2}}\right )}{2 \, \log \relax (5)^{2} + 2 \, \log \relax (5) \log \relax (2) - \log \relax (5) \log \left (2401 \, x^{2}\right ) + 2 \, \log \relax (5) \log \left (\log \left (\frac {100}{2401 \, x^{2}}\right )\right ) + 2 \, \log \relax (2) \log \left (\log \left (\frac {100}{2401 \, x^{2}}\right )\right ) - \log \left (2401 \, x^{2}\right ) \log \left (\log \left (\frac {100}{2401 \, x^{2}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))^2-10*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))-1
0*x)/log(100/2401/x^2)/log(5*log(100/2401/x^2))^2,x, algorithm="giac")

[Out]

15*x^2 - 5*x^2*log(100/2401/x^2)/(2*log(5)^2 + 2*log(5)*log(2) - log(5)*log(2401*x^2) + 2*log(5)*log(log(100/2
401/x^2)) + 2*log(2)*log(log(100/2401/x^2)) - log(2401*x^2)*log(log(100/2401/x^2)))

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maple [A]  time = 0.03, size = 33, normalized size = 1.43

method	result	size
norman	\(\frac {-5 x^{2}+15 x^{2} \ln \left (5 \ln \left (\frac {100}{2401 x^{2}}\right )\right )}{\ln \left (5 \ln \left (\frac {100}{2401 x^{2}}\right )\right )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((30*x*ln(100/2401/x^2)*ln(5*ln(100/2401/x^2))^2-10*x*ln(100/2401/x^2)*ln(5*ln(100/2401/x^2))-10*x)/ln(100/
2401/x^2)/ln(5*ln(100/2401/x^2))^2,x,method=_RETURNVERBOSE)

[Out]

(-5*x^2+15*x^2*ln(5*ln(100/2401/x^2)))/ln(5*ln(100/2401/x^2))

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maxima [C]  time = 0.48, size = 43, normalized size = 1.87 \begin {gather*} 15 \, x^{2} - \frac {10 \, x^{2}}{2 i \, \pi + 2 \, \log \relax (5) + 2 \, \log \relax (2) + 2 \, \log \left (2 \, \log \relax (7) - \log \relax (5) - \log \relax (2) + \log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))^2-10*x*log(100/2401/x^2)*log(5*log(100/2401/x^2))-1
0*x)/log(100/2401/x^2)/log(5*log(100/2401/x^2))^2,x, algorithm="maxima")

[Out]

15*x^2 - 10*x^2/(2*I*pi + 2*log(5) + 2*log(2) + 2*log(2*log(7) - log(5) - log(2) + log(x)))

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mupad [B]  time = 7.09, size = 22, normalized size = 0.96 \begin {gather*} 15\,x^2-\frac {5\,x^2}{\ln \left (5\,\ln \left (\frac {100}{2401\,x^2}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x - 30*x*log(5*log(100/(2401*x^2)))^2*log(100/(2401*x^2)) + 10*x*log(5*log(100/(2401*x^2)))*log(100/(
2401*x^2)))/(log(5*log(100/(2401*x^2)))^2*log(100/(2401*x^2))),x)

[Out]

15*x^2 - (5*x^2)/log(5*log(100/(2401*x^2)))

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sympy [A]  time = 0.28, size = 20, normalized size = 0.87 \begin {gather*} 15 x^{2} - \frac {5 x^{2}}{\log {\left (5 \log {\left (\frac {100}{2401 x^{2}} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((30*x*ln(100/2401/x**2)*ln(5*ln(100/2401/x**2))**2-10*x*ln(100/2401/x**2)*ln(5*ln(100/2401/x**2))-10
*x)/ln(100/2401/x**2)/ln(5*ln(100/2401/x**2))**2,x)

[Out]

15*x**2 - 5*x**2/log(5*log(100/(2401*x**2)))

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