3.73.67 \(\int \frac {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+(8 x^2-72 x^3) \log (x)+(x^2-18 x^3+81 x^4) \log ^2(x)+(-16 x^3 \log (x)+(-4 x^3+36 x^4) \log ^2(x)) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx\)

Optimal. Leaf size=24 \[ \frac {12}{x (4+\log (x) (1+x-2 x (5+\log (5 x))))} \]

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Rubi [F]  time = 1.81, 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 {-60+108 x+(-12+240 x) \log (x)+(24 x+48 x \log (x)) \log (5 x)}{16 x^2+\left (8 x^2-72 x^3\right ) \log (x)+\left (x^2-18 x^3+81 x^4\right ) \log ^2(x)+\left (-16 x^3 \log (x)+\left (-4 x^3+36 x^4\right ) \log ^2(x)\right ) \log (5 x)+4 x^4 \log ^2(x) \log ^2(5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-60 + 108*x + (-12 + 240*x)*Log[x] + (24*x + 48*x*Log[x])*Log[5*x])/(16*x^2 + (8*x^2 - 72*x^3)*Log[x] + (
x^2 - 18*x^3 + 81*x^4)*Log[x]^2 + (-16*x^3*Log[x] + (-4*x^3 + 36*x^4)*Log[x]^2)*Log[5*x] + 4*x^4*Log[x]^2*Log[
5*x]^2),x]

[Out]

48*Defer[Int][1/(x^2*(-4 - Log[x] + 9*x*Log[x] + 2*x*Log[x]*Log[5*x])^2), x] + 48*Defer[Int][1/(x^2*Log[x]*(-4
- Log[x] + 9*x*Log[x] + 2*x*Log[x]*Log[5*x])^2), x] + 12*Defer[Int][Log[x]/(x^2*(-4 - Log[x] + 9*x*Log[x] + 2
*x*Log[x]*Log[5*x])^2), x] + 24*Defer[Int][Log[x]/(x*(-4 - Log[x] + 9*x*Log[x] + 2*x*Log[x]*Log[5*x])^2), x] +
24*Defer[Int][1/(x^2*(-4 - Log[x] + 9*x*Log[x] + 2*x*Log[x]*Log[5*x])), x] + 12*Defer[Int][1/(x^2*Log[x]*(-4
- Log[x] + 9*x*Log[x] + 2*x*Log[x]*Log[5*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12 (-5+9 x+2 x \log (5 x)+\log (x) (-1+20 x+4 x \log (5 x)))}{x^2 (4-\log (x) (-1+9 x+2 x \log (5 x)))^2} \, dx\\ &=12 \int \frac {-5+9 x+2 x \log (5 x)+\log (x) (-1+20 x+4 x \log (5 x))}{x^2 (4-\log (x) (-1+9 x+2 x \log (5 x)))^2} \, dx\\ &=12 \int \left (\frac {4+4 \log (x)+\log ^2(x)+2 x \log ^2(x)}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2}+\frac {1+2 \log (x)}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))}\right ) \, dx\\ &=12 \int \frac {4+4 \log (x)+\log ^2(x)+2 x \log ^2(x)}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2} \, dx+12 \int \frac {1+2 \log (x)}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))} \, dx\\ &=12 \int \left (\frac {4}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2}+\frac {4}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2}+\frac {\log (x)}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2}+\frac {2 \log (x)}{x (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2}\right ) \, dx+12 \int \left (\frac {2}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))}+\frac {1}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))}\right ) \, dx\\ &=12 \int \frac {\log (x)}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2} \, dx+12 \int \frac {1}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))} \, dx+24 \int \frac {\log (x)}{x (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2} \, dx+24 \int \frac {1}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))} \, dx+48 \int \frac {1}{x^2 (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2} \, dx+48 \int \frac {1}{x^2 \log (x) (-4-\log (x)+9 x \log (x)+2 x \log (x) \log (5 x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.89, size = 24, normalized size = 1.00 \begin {gather*} -\frac {12}{x (-4+\log (x) (-1+9 x+2 x \log (5 x)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-60 + 108*x + (-12 + 240*x)*Log[x] + (24*x + 48*x*Log[x])*Log[5*x])/(16*x^2 + (8*x^2 - 72*x^3)*Log[
x] + (x^2 - 18*x^3 + 81*x^4)*Log[x]^2 + (-16*x^3*Log[x] + (-4*x^3 + 36*x^4)*Log[x]^2)*Log[5*x] + 4*x^4*Log[x]^
2*Log[5*x]^2),x]

[Out]

-12/(x*(-4 + Log[x]*(-1 + 9*x + 2*x*Log[5*x])))

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fricas [A]  time = 0.67, size = 36, normalized size = 1.50 \begin {gather*} -\frac {12}{2 \, x^{2} \log \relax (x)^{2} + {\left (2 \, x^{2} \log \relax (5) + 9 \, x^{2} - x\right )} \log \relax (x) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4*log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*
log(x)^2-16*x^3*log(x))*log(5*x)+(81*x^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="fri
cas")

[Out]

-12/(2*x^2*log(x)^2 + (2*x^2*log(5) + 9*x^2 - x)*log(x) - 4*x)

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giac [A]  time = 0.26, size = 38, normalized size = 1.58 \begin {gather*} -\frac {12}{2 \, x^{2} \log \relax (5) \log \relax (x) + 2 \, x^{2} \log \relax (x)^{2} + 9 \, x^{2} \log \relax (x) - x \log \relax (x) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4*log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*
log(x)^2-16*x^3*log(x))*log(5*x)+(81*x^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="gia
c")

[Out]

-12/(2*x^2*log(5)*log(x) + 2*x^2*log(x)^2 + 9*x^2*log(x) - x*log(x) - 4*x)

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maple [C]  time = 0.21, size = 39, normalized size = 1.62

method	result	size
risch	\(-\frac {12 i}{x \left (2 i \ln \relax (5) x \ln \relax (x )+2 i x \ln \relax (x )^{2}+9 i x \ln \relax (x )-i \ln \relax (x )-4 i\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((48*x*ln(x)+24*x)*ln(5*x)+(240*x-12)*ln(x)+108*x-60)/(4*x^4*ln(x)^2*ln(5*x)^2+((36*x^4-4*x^3)*ln(x)^2-16*
x^3*ln(x))*ln(5*x)+(81*x^4-18*x^3+x^2)*ln(x)^2+(-72*x^3+8*x^2)*ln(x)+16*x^2),x,method=_RETURNVERBOSE)

[Out]

-12*I/x/(2*I*ln(5)*x*ln(x)+2*I*x*ln(x)^2+9*I*x*ln(x)-I*ln(x)-4*I)

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maxima [A]  time = 0.48, size = 34, normalized size = 1.42 \begin {gather*} -\frac {12}{2 \, x^{2} \log \relax (x)^{2} + {\left (x^{2} {\left (2 \, \log \relax (5) + 9\right )} - x\right )} \log \relax (x) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*log(x)+24*x)*log(5*x)+(240*x-12)*log(x)+108*x-60)/(4*x^4*log(x)^2*log(5*x)^2+((36*x^4-4*x^3)*
log(x)^2-16*x^3*log(x))*log(5*x)+(81*x^4-18*x^3+x^2)*log(x)^2+(-72*x^3+8*x^2)*log(x)+16*x^2),x, algorithm="max
ima")

[Out]

-12/(2*x^2*log(x)^2 + (x^2*(2*log(5) + 9) - x)*log(x) - 4*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {108\,x+\ln \relax (x)\,\left (240\,x-12\right )+\ln \left (5\,x\right )\,\left (24\,x+48\,x\,\ln \relax (x)\right )-60}{\ln \relax (x)\,\left (8\,x^2-72\,x^3\right )+{\ln \relax (x)}^2\,\left (81\,x^4-18\,x^3+x^2\right )-\ln \left (5\,x\right )\,\left (16\,x^3\,\ln \relax (x)+{\ln \relax (x)}^2\,\left (4\,x^3-36\,x^4\right )\right )+16\,x^2+4\,x^4\,{\ln \left (5\,x\right )}^2\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((108*x + log(x)*(240*x - 12) + log(5*x)*(24*x + 48*x*log(x)) - 60)/(log(x)*(8*x^2 - 72*x^3) + log(x)^2*(x^
2 - 18*x^3 + 81*x^4) - log(5*x)*(16*x^3*log(x) + log(x)^2*(4*x^3 - 36*x^4)) + 16*x^2 + 4*x^4*log(5*x)^2*log(x)
^2),x)

[Out]

int((108*x + log(x)*(240*x - 12) + log(5*x)*(24*x + 48*x*log(x)) - 60)/(log(x)*(8*x^2 - 72*x^3) + log(x)^2*(x^
2 - 18*x^3 + 81*x^4) - log(5*x)*(16*x^3*log(x) + log(x)^2*(4*x^3 - 36*x^4)) + 16*x^2 + 4*x^4*log(5*x)^2*log(x)
^2), x)

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sympy [A]  time = 0.44, size = 34, normalized size = 1.42 \begin {gather*} - \frac {12}{2 x^{2} \log {\relax (x )}^{2} - 4 x + \left (2 x^{2} \log {\relax (5 )} + 9 x^{2} - x\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x*ln(x)+24*x)*ln(5*x)+(240*x-12)*ln(x)+108*x-60)/(4*x**4*ln(x)**2*ln(5*x)**2+((36*x**4-4*x**3)*
ln(x)**2-16*x**3*ln(x))*ln(5*x)+(81*x**4-18*x**3+x**2)*ln(x)**2+(-72*x**3+8*x**2)*ln(x)+16*x**2),x)

[Out]

-12/(2*x**2*log(x)**2 - 4*x + (2*x**2*log(5) + 9*x**2 - x)*log(x))

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