3.72.21 \(\int \frac {1+5 x+2 x^2-x^3-x \log (x)+(6 x+x^3-x \log (x)) \log (3 x)}{12 x+24 x^2 \log (3 x)+12 x^3 \log ^2(3 x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {-5+x^2+\log (x)}{3 (4+4 x \log (3 x))} \]

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Rubi [F]  time = 1.30, 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 {1+5 x+2 x^2-x^3-x \log (x)+\left (6 x+x^3-x \log (x)\right ) \log (3 x)}{12 x+24 x^2 \log (3 x)+12 x^3 \log ^2(3 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + 5*x + 2*x^2 - x^3 - x*Log[x] + (6*x + x^3 - x*Log[x])*Log[3*x])/(12*x + 24*x^2*Log[3*x] + 12*x^3*Log[
3*x]^2),x]

[Out]

(5*Defer[Int][(1 + x*Log[3*x])^(-2), x])/12 - (5*Defer[Int][1/(x*(1 + x*Log[3*x])^2), x])/12 + Defer[Int][x/(1
+ x*Log[3*x])^2, x]/12 - Defer[Int][x^2/(1 + x*Log[3*x])^2, x]/12 - Defer[Int][Log[x]/(1 + x*Log[3*x])^2, x]/
12 + Defer[Int][Log[x]/(x*(1 + x*Log[3*x])^2), x]/12 + Defer[Int][1/(x*(1 + x*Log[3*x])), x]/2 + Defer[Int][x/
(1 + x*Log[3*x]), x]/12 - Defer[Int][Log[x]/(x*(1 + x*Log[3*x])), x]/12

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+5 x+2 x^2-x^3-x \log (x)+\left (6 x+x^3-x \log (x)\right ) \log (3 x)}{12 x (1+x \log (3 x))^2} \, dx\\ &=\frac {1}{12} \int \frac {1+5 x+2 x^2-x^3-x \log (x)+\left (6 x+x^3-x \log (x)\right ) \log (3 x)}{x (1+x \log (3 x))^2} \, dx\\ &=\frac {1}{12} \int \left (-\frac {(-1+x) \left (-5+x^2+\log (x)\right )}{x (1+x \log (3 x))^2}+\frac {6+x^2-\log (x)}{x (1+x \log (3 x))}\right ) \, dx\\ &=-\left (\frac {1}{12} \int \frac {(-1+x) \left (-5+x^2+\log (x)\right )}{x (1+x \log (3 x))^2} \, dx\right )+\frac {1}{12} \int \frac {6+x^2-\log (x)}{x (1+x \log (3 x))} \, dx\\ &=-\left (\frac {1}{12} \int \left (\frac {5-x^2-\log (x)}{x (1+x \log (3 x))^2}+\frac {-5+x^2+\log (x)}{(1+x \log (3 x))^2}\right ) \, dx\right )+\frac {1}{12} \int \left (\frac {6}{x (1+x \log (3 x))}+\frac {x}{1+x \log (3 x)}-\frac {\log (x)}{x (1+x \log (3 x))}\right ) \, dx\\ &=-\left (\frac {1}{12} \int \frac {5-x^2-\log (x)}{x (1+x \log (3 x))^2} \, dx\right )-\frac {1}{12} \int \frac {-5+x^2+\log (x)}{(1+x \log (3 x))^2} \, dx+\frac {1}{12} \int \frac {x}{1+x \log (3 x)} \, dx-\frac {1}{12} \int \frac {\log (x)}{x (1+x \log (3 x))} \, dx+\frac {1}{2} \int \frac {1}{x (1+x \log (3 x))} \, dx\\ &=\frac {1}{12} \int \frac {x}{1+x \log (3 x)} \, dx-\frac {1}{12} \int \frac {\log (x)}{x (1+x \log (3 x))} \, dx-\frac {1}{12} \int \left (-\frac {5}{(1+x \log (3 x))^2}+\frac {x^2}{(1+x \log (3 x))^2}+\frac {\log (x)}{(1+x \log (3 x))^2}\right ) \, dx-\frac {1}{12} \int \left (\frac {5}{x (1+x \log (3 x))^2}-\frac {x}{(1+x \log (3 x))^2}-\frac {\log (x)}{x (1+x \log (3 x))^2}\right ) \, dx+\frac {1}{2} \int \frac {1}{x (1+x \log (3 x))} \, dx\\ &=\frac {1}{12} \int \frac {x}{(1+x \log (3 x))^2} \, dx-\frac {1}{12} \int \frac {x^2}{(1+x \log (3 x))^2} \, dx-\frac {1}{12} \int \frac {\log (x)}{(1+x \log (3 x))^2} \, dx+\frac {1}{12} \int \frac {\log (x)}{x (1+x \log (3 x))^2} \, dx+\frac {1}{12} \int \frac {x}{1+x \log (3 x)} \, dx-\frac {1}{12} \int \frac {\log (x)}{x (1+x \log (3 x))} \, dx+\frac {5}{12} \int \frac {1}{(1+x \log (3 x))^2} \, dx-\frac {5}{12} \int \frac {1}{x (1+x \log (3 x))^2} \, dx+\frac {1}{2} \int \frac {1}{x (1+x \log (3 x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.26, size = 21, normalized size = 0.95 \begin {gather*} \frac {-5+x^2+\log (x)}{12 (1+x \log (3 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 5*x + 2*x^2 - x^3 - x*Log[x] + (6*x + x^3 - x*Log[x])*Log[3*x])/(12*x + 24*x^2*Log[3*x] + 12*x^
3*Log[3*x]^2),x]

[Out]

(-5 + x^2 + Log[x])/(12*(1 + x*Log[3*x]))

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fricas [A]  time = 0.90, size = 21, normalized size = 0.95 \begin {gather*} \frac {x^{2} + \log \relax (x) - 5}{12 \, {\left (x \log \relax (3) + x \log \relax (x) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)+x^3+6*x)*log(3*x)-x*log(x)-x^3+2*x^2+5*x+1)/(12*x^3*log(3*x)^2+24*x^2*log(3*x)+12*x),x,
algorithm="fricas")

[Out]

1/12*(x^2 + log(x) - 5)/(x*log(3) + x*log(x) + 1)

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giac [A]  time = 0.16, size = 37, normalized size = 1.68 \begin {gather*} \frac {x^{3} - x \log \relax (3) - 5 \, x - 1}{12 \, {\left (x^{2} \log \relax (3) + x^{2} \log \relax (x) + x\right )}} + \frac {1}{12 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)+x^3+6*x)*log(3*x)-x*log(x)-x^3+2*x^2+5*x+1)/(12*x^3*log(3*x)^2+24*x^2*log(3*x)+12*x),x,
algorithm="giac")

[Out]

1/12*(x^3 - x*log(3) - 5*x - 1)/(x^2*log(3) + x^2*log(x) + x) + 1/12/x

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maple [A]  time = 0.08, size = 41, normalized size = 1.86

method	result	size
risch	\(\frac {1}{12 x}+\frac {-2+2 x^{3}-2 x \ln \relax (3)-10 x}{12 x \left (2+2 x \ln \relax (3)+2 x \ln \relax (x )\right )}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(x)+x^3+6*x)*ln(3*x)-x*ln(x)-x^3+2*x^2+5*x+1)/(12*x^3*ln(3*x)^2+24*x^2*ln(3*x)+12*x),x,method=_RETU
RNVERBOSE)

[Out]

1/12/x+1/12*(-2+2*x^3-2*x*ln(3)-10*x)/x/(2+2*x*ln(3)+2*x*ln(x))

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maxima [A]  time = 0.52, size = 21, normalized size = 0.95 \begin {gather*} \frac {x^{2} + \log \relax (x) - 5}{12 \, {\left (x \log \relax (3) + x \log \relax (x) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)+x^3+6*x)*log(3*x)-x*log(x)-x^3+2*x^2+5*x+1)/(12*x^3*log(3*x)^2+24*x^2*log(3*x)+12*x),x,
algorithm="maxima")

[Out]

1/12*(x^2 + log(x) - 5)/(x*log(3) + x*log(x) + 1)

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mupad [B]  time = 4.47, size = 19, normalized size = 0.86 \begin {gather*} \frac {\ln \relax (x)+x^2-5}{12\,\left (x\,\ln \left (3\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + log(3*x)*(6*x - x*log(x) + x^3) - x*log(x) + 2*x^2 - x^3 + 1)/(12*x + 24*x^2*log(3*x) + 12*x^3*log(
3*x)^2),x)

[Out]

(log(x) + x^2 - 5)/(12*(x*log(3*x) + 1))

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sympy [B]  time = 0.35, size = 37, normalized size = 1.68 \begin {gather*} \frac {x^{3} - 5 x - x \log {\relax (3 )} - 1}{12 x^{2} \log {\relax (x )} + 12 x^{2} \log {\relax (3 )} + 12 x} + \frac {1}{12 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(x)+x**3+6*x)*ln(3*x)-x*ln(x)-x**3+2*x**2+5*x+1)/(12*x**3*ln(3*x)**2+24*x**2*ln(3*x)+12*x),x)

[Out]

(x**3 - 5*x - x*log(3) - 1)/(12*x**2*log(x) + 12*x**2*log(3) + 12*x) + 1/(12*x)

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