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

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

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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 = {} $$\begin{array}{c}\int \frac{-60+108x+(-12+240x)\log (x)+(24x+48x\log (x))\log (5x)}{16x^2+(8x^2-72x^3)\log (x)+(x^2-18x^3+81x^4){\log }^2(x)+(-16x^3\log (x)+(-4x^3+36x^4){\log }^2(x))\log (5x)+4x^4{\log }^2(x){\log }^2(5x)}dx\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{12(-5+9x+2x\log (5x)+\log (x)(-1+20x+4x\log (5x)))}{x^2(4-\log (x)(-1+9x+2x\log (5x)){)}^2}dx\\ & =12\int \frac{-5+9x+2x\log (5x)+\log (x)(-1+20x+4x\log (5x))}{x^2(4-\log (x)(-1+9x+2x\log (5x)){)}^2}dx\\ & =12\int (\frac{4+4\log (x)+{\log }^2(x)+2x{\log }^2(x)}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}+\frac{1+2\log (x)}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))})dx\\ & =12\int \frac{4+4\log (x)+{\log }^2(x)+2x{\log }^2(x)}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}dx+12\int \frac{1+2\log (x)}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))}dx\\ & =12\int (\frac{4}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}+\frac{4}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}+\frac{\log (x)}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}+\frac{2\log (x)}{x(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2})dx+12\int (\frac{2}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))}+\frac{1}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))})dx\\ & =12\int \frac{\log (x)}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}dx+12\int \frac{1}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))}dx+24\int \frac{\log (x)}{x(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}dx+24\int \frac{1}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x))}dx+48\int \frac{1}{x^2(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}dx+48\int \frac{1}{x^2\log (x)(-4-\log (x)+9x\log (x)+2x\log (x)\log (5x){)}^2}dx\end{array}\end{array}$$

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

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 $$\begin{array}{c}-\frac{12}{2x^2\log (x{)}^2+(2x^2\log (5)+9x^2-x)\log (x)-4x}\end{array}$$

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 $$\begin{array}{c}-\frac{12}{2x^2\log (5)\log (x)+2x^2\log (x{)}^2+9x^2\log (x)-x\log (x)-4x}\end{array}$$

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	$-\frac{12i}{x(2i\ln (5)x\ln (x)+2ix\ln (x{)}^2+9ix\ln (x)-i\ln (x)-4i)}$	$39$

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 $$\begin{array}{c}-\frac{12}{2x^2\log (x{)}^2+(x^2(2\log (5)+9)-x)\log (x)-4x}\end{array}$$

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 $$\begin{array}{c}\int \frac{108x+\ln (x)(240x-12)+\ln \left(5x\right)(24x+48x\ln (x))-60}{\ln (x)(8x^2-72x^3)+{\ln (x)}^2(81x^4-18x^3+x^2)-\ln \left(5x\right)(16x^3\ln (x)+{\ln (x)}^2(4x^3-36x^4))+16x^2+4x^4{\ln \left(5x\right)}^2{\ln (x)}^2}dx\end{array}$$

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 $$\begin{array}{c}-\frac{12}{2x^2\log {(x)}^2-4x+(2x^2\log (5)+9x^2-x)\log (x)}\end{array}$$

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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