3.30.15 \(\int \frac {-10 \log (5 x)+(-5 x-10 x^2 \log (5 x)) \log (x^2)-5 \log (x^2) \log (e^{x^2} \log (x^2))}{(4 x+4 x^2+x^3) \log (x^2)+(4 x+2 x^2) \log (5 x) \log (x^2) \log (e^{x^2} \log (x^2))+x \log ^2(5 x) \log (x^2) \log ^2(e^{x^2} \log (x^2))} \, dx\)

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

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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 {-10 \log (5 x)+\left (-5 x-10 x^2 \log (5 x)\right ) \log \left (x^2\right )-5 \log \left (x^2\right ) \log \left (e^{x^2} \log \left (x^2\right )\right )}{\left (4 x+4 x^2+x^3\right ) \log \left (x^2\right )+\left (4 x+2 x^2\right ) \log (5 x) \log \left (x^2\right ) \log \left (e^{x^2} \log \left (x^2\right )\right )+x \log ^2(5 x) \log \left (x^2\right ) \log ^2\left (e^{x^2} \log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 23, normalized size = 1.00 \begin {gather*} \frac {5}{2+x+\log (5 x) \log \left (e^{x^2} \log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

5/(2 + x + Log[5*x]*Log[E^x^2*Log[x^2]])

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fricas [A]  time = 0.72, size = 32, normalized size = 1.39 \begin {gather*} \frac {5}{\log \left (-2 \, e^{\left (x^{2}\right )} \log \relax (5) + 2 \, e^{\left (x^{2}\right )} \log \left (5 \, x\right )\right ) \log \left (5 \, x\right ) + x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x^2)*log(exp(x^2)*log(x^2))+(-10*x^2*log(5*x)-5*x)*log(x^2)-10*log(5*x))/(x*log(5*x)^2*log(x
^2)*log(exp(x^2)*log(x^2))^2+(2*x^2+4*x)*log(5*x)*log(x^2)*log(exp(x^2)*log(x^2))+(x^3+4*x^2+4*x)*log(x^2)),x,
algorithm="fricas")

[Out]

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

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giac [B]  time = 1.38, size = 697, normalized size = 30.30 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x^2)*log(exp(x^2)*log(x^2))+(-10*x^2*log(5*x)-5*x)*log(x^2)-10*log(5*x))/(x*log(5*x)^2*log(x
^2)*log(exp(x^2)*log(x^2))^2+(2*x^2+4*x)*log(5*x)*log(x^2)*log(exp(x^2)*log(x^2))+(x^3+4*x^2+4*x)*log(x^2)),x,
algorithm="giac")

[Out]

5*(2*x^2*log(5)^2*log(x^2)*log(x) + 4*x^2*log(5)*log(x^2)*log(x)^2 + 2*x^2*log(x^2)*log(x)^3 + x*log(5)*log(x^
2)*log(x) + x*log(x^2)*log(x)^2 + log(5)^2*log(x^2) - x*log(x^2)*log(x) + 2*log(5)*log(x^2)*log(x) + log(x^2)*
log(x)^2 - 2*log(x^2)*log(x))/(2*x^4*log(5)^3*log(x^2)*log(x) + 6*x^4*log(5)^2*log(x^2)*log(x)^2 + 6*x^4*log(5
)*log(x^2)*log(x)^3 + 2*x^4*log(x^2)*log(x)^4 + 2*x^2*log(5)^3*log(x^2)*log(x)*log(log(x^2)) + 6*x^2*log(5)^2*
log(x^2)*log(x)^2*log(log(x^2)) + 6*x^2*log(5)*log(x^2)*log(x)^3*log(log(x^2)) + 2*x^2*log(x^2)*log(x)^4*log(l
og(x^2)) + 3*x^3*log(5)^2*log(x^2)*log(x) + 6*x^3*log(5)*log(x^2)*log(x)^2 + 3*x^3*log(x^2)*log(x)^3 + 2*x^2*l
og(5)^3*log(x) - x^3*log(5)*log(x^2)*log(x) + 4*x^2*log(5)^2*log(x^2)*log(x) + 6*x^2*log(5)^2*log(x)^2 - x^3*l
og(x^2)*log(x)^2 + 8*x^2*log(5)*log(x^2)*log(x)^2 + 6*x^2*log(5)*log(x)^3 + 4*x^2*log(x^2)*log(x)^3 + 2*x^2*lo
g(x)^4 + x*log(5)^2*log(x^2)*log(x)*log(log(x^2)) + 2*x*log(5)*log(x^2)*log(x)^2*log(log(x^2)) + x*log(x^2)*lo
g(x)^3*log(log(x^2)) - x^2*log(5)*log(x^2)*log(x) - x^2*log(x^2)*log(x)^2 + 2*log(5)^3*log(x)*log(log(x^2)) -
x*log(5)*log(x^2)*log(x)*log(log(x^2)) + 6*log(5)^2*log(x)^2*log(log(x^2)) - x*log(x^2)*log(x)^2*log(log(x^2))
+ 6*log(5)*log(x)^3*log(log(x^2)) + 2*log(x)^4*log(log(x^2)) + 2*x*log(5)^2*log(x) - x^2*log(x^2)*log(x) + 2*
x*log(5)*log(x^2)*log(x) + 4*x*log(5)*log(x)^2 + 2*x*log(x^2)*log(x)^2 + 2*x*log(x)^3 - 2*log(5)*log(x^2)*log(
x)*log(log(x^2)) - 2*log(x^2)*log(x)^2*log(log(x^2)) + 4*log(5)^2*log(x) - 4*x*log(x^2)*log(x) + 8*log(5)*log(
x)^2 + 4*log(x)^3 - 4*log(x^2)*log(x))

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {-5 \ln \left (x^{2}\right ) \ln \left ({\mathrm e}^{x^{2}} \ln \left (x^{2}\right )\right )+\left (-10 x^{2} \ln \left (5 x \right )-5 x \right ) \ln \left (x^{2}\right )-10 \ln \left (5 x \right )}{x \ln \left (5 x \right )^{2} \ln \left (x^{2}\right ) \ln \left ({\mathrm e}^{x^{2}} \ln \left (x^{2}\right )\right )^{2}+\left (2 x^{2}+4 x \right ) \ln \left (5 x \right ) \ln \left (x^{2}\right ) \ln \left ({\mathrm e}^{x^{2}} \ln \left (x^{2}\right )\right )+\left (x^{3}+4 x^{2}+4 x \right ) \ln \left (x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5*ln(x^2)*ln(exp(x^2)*ln(x^2))+(-10*x^2*ln(5*x)-5*x)*ln(x^2)-10*ln(5*x))/(x*ln(5*x)^2*ln(x^2)*ln(exp(x^2
)*ln(x^2))^2+(2*x^2+4*x)*ln(5*x)*ln(x^2)*ln(exp(x^2)*ln(x^2))+(x^3+4*x^2+4*x)*ln(x^2)),x)

[Out]

int((-5*ln(x^2)*ln(exp(x^2)*ln(x^2))+(-10*x^2*ln(5*x)-5*x)*ln(x^2)-10*ln(5*x))/(x*ln(5*x)^2*ln(x^2)*ln(exp(x^2
)*ln(x^2))^2+(2*x^2+4*x)*ln(5*x)*ln(x^2)*ln(exp(x^2)*ln(x^2))+(x^3+4*x^2+4*x)*ln(x^2)),x)

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maxima [A]  time = 0.75, size = 36, normalized size = 1.57 \begin {gather*} \frac {5}{x^{2} \log \relax (5) + \log \relax (5) \log \relax (2) + {\left (x^{2} + \log \relax (2)\right )} \log \relax (x) + {\left (\log \relax (5) + \log \relax (x)\right )} \log \left (\log \relax (x)\right ) + x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*log(x^2)*log(exp(x^2)*log(x^2))+(-10*x^2*log(5*x)-5*x)*log(x^2)-10*log(5*x))/(x*log(5*x)^2*log(x
^2)*log(exp(x^2)*log(x^2))^2+(2*x^2+4*x)*log(5*x)*log(x^2)*log(exp(x^2)*log(x^2))+(x^3+4*x^2+4*x)*log(x^2)),x,
algorithm="maxima")

[Out]

5/(x^2*log(5) + log(5)*log(2) + (x^2 + log(2))*log(x) + (log(5) + log(x))*log(log(x)) + x + 2)

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mupad [B]  time = 2.42, size = 22, normalized size = 0.96 \begin {gather*} \frac {5}{x+\ln \left (5\,x\right )\,\ln \left (\ln \left (x^2\right )\,{\mathrm {e}}^{x^2}\right )+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*log(5*x) + log(x^2)*(5*x + 10*x^2*log(5*x)) + 5*log(x^2)*log(log(x^2)*exp(x^2)))/(log(x^2)*(4*x + 4*x
^2 + x^3) + x*log(5*x)^2*log(x^2)*log(log(x^2)*exp(x^2))^2 + log(5*x)*log(x^2)*log(log(x^2)*exp(x^2))*(4*x + 2
*x^2)),x)

[Out]

5/(x + log(5*x)*log(log(x^2)*exp(x^2)) + 2)

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sympy [A]  time = 0.59, size = 26, normalized size = 1.13 \begin {gather*} \frac {5}{x + \log {\left (5 x \right )} \log {\left (\left (2 \log {\left (5 x \right )} - \log {\left (25 \right )}\right ) e^{x^{2}} \right )} + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5*ln(x**2)*ln(exp(x**2)*ln(x**2))+(-10*x**2*ln(5*x)-5*x)*ln(x**2)-10*ln(5*x))/(x*ln(5*x)**2*ln(x**
2)*ln(exp(x**2)*ln(x**2))**2+(2*x**2+4*x)*ln(5*x)*ln(x**2)*ln(exp(x**2)*ln(x**2))+(x**3+4*x**2+4*x)*ln(x**2)),
x)

[Out]

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

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