3.28.32 \(\int \frac {5-3 x+12 x \log (3)}{15 x-3 x^2+(-60 x+12 x^2) \log (3)+5 x \log (2 x)} \, dx\)

Optimal. Leaf size=18 \[ \log \left (-5+x+\frac {5 \log (2 x)}{-3+12 \log (3)}\right ) \]

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Rubi [F]  time = 0.29, 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 {5-3 x+12 x \log (3)}{15 x-3 x^2+\left (-60 x+12 x^2\right ) \log (3)+5 x \log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5 - 3*x + 12*x*Log[3])/(15*x - 3*x^2 + (-60*x + 12*x^2)*Log[3] + 5*x*Log[2*x]),x]

[Out]

-3*(1 - Log[81])*Defer[Int][(3*(-5 + x)*(-1 + Log[81]) + 5*Log[2*x])^(-1), x] + 5*Defer[Int][1/(x*(3*(-5 + x)*
(-1 + Log[81]) + 5*Log[2*x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5+x (-3+12 \log (3))}{15 x-3 x^2+\left (-60 x+12 x^2\right ) \log (3)+5 x \log (2 x)} \, dx\\ &=\int \left (\frac {5}{x (15 (1-\log (81))-3 x (1-\log (81))+5 \log (2 x))}+\frac {3 (-1+\log (81))}{15 (1-\log (81))-3 x (1-\log (81))+5 \log (2 x)}\right ) \, dx\\ &=5 \int \frac {1}{x (15 (1-\log (81))-3 x (1-\log (81))+5 \log (2 x))} \, dx-(3 (1-\log (81))) \int \frac {1}{15 (1-\log (81))-3 x (1-\log (81))+5 \log (2 x)} \, dx\\ &=5 \int \frac {1}{x (3 (-5+x) (-1+\log (81))+5 \log (2 x))} \, dx-(3 (1-\log (81))) \int \frac {1}{3 (-5+x) (-1+\log (81))+5 \log (2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 21, normalized size = 1.17 \begin {gather*} \log (15-3 x-15 \log (81)+3 x \log (81)+5 \log (2 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - 3*x + 12*x*Log[3])/(15*x - 3*x^2 + (-60*x + 12*x^2)*Log[3] + 5*x*Log[2*x]),x]

[Out]

Log[15 - 3*x - 15*Log[81] + 3*x*Log[81] + 5*Log[2*x]]

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fricas [A]  time = 0.46, size = 19, normalized size = 1.06 \begin {gather*} \log \left (12 \, {\left (x - 5\right )} \log \relax (3) - 3 \, x + 5 \, \log \left (2 \, x\right ) + 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x*log(3)-3*x+5)/(5*x*log(2*x)+(12*x^2-60*x)*log(3)-3*x^2+15*x),x, algorithm="fricas")

[Out]

log(12*(x - 5)*log(3) - 3*x + 5*log(2*x) + 15)

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giac [A]  time = 0.18, size = 21, normalized size = 1.17 \begin {gather*} \log \left (-12 \, x \log \relax (3) + 3 \, x + 60 \, \log \relax (3) - 5 \, \log \left (2 \, x\right ) - 15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x*log(3)-3*x+5)/(5*x*log(2*x)+(12*x^2-60*x)*log(3)-3*x^2+15*x),x, algorithm="giac")

[Out]

log(-12*x*log(3) + 3*x + 60*log(3) - 5*log(2*x) - 15)

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maple [A]  time = 0.04, size = 20, normalized size = 1.11

method	result	size
risch	\(\ln \left (\frac {12 x \ln \relax (3)}{5}-12 \ln \relax (3)-\frac {3 x}{5}+\ln \left (2 x \right )+3\right )\)	\(20\)
norman	\(\ln \left (12 x \ln \relax (3)-60 \ln \relax (3)+5 \ln \left (2 x \right )-3 x +15\right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x*ln(3)-3*x+5)/(5*x*ln(2*x)+(12*x^2-60*x)*ln(3)-3*x^2+15*x),x,method=_RETURNVERBOSE)

[Out]

ln(12/5*x*ln(3)-12*ln(3)-3/5*x+ln(2*x)+3)

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maxima [A]  time = 0.52, size = 20, normalized size = 1.11 \begin {gather*} \log \left (\frac {3}{5} \, x {\left (4 \, \log \relax (3) - 1\right )} - 12 \, \log \relax (3) + \log \relax (2) + \log \relax (x) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x*log(3)-3*x+5)/(5*x*log(2*x)+(12*x^2-60*x)*log(3)-3*x^2+15*x),x, algorithm="maxima")

[Out]

log(3/5*x*(4*log(3) - 1) - 12*log(3) + log(2) + log(x) + 3)

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mupad [B]  time = 2.25, size = 21, normalized size = 1.17 \begin {gather*} \ln \left (5\,\ln \left (2\,x\right )-3\,x-60\,\ln \relax (3)+12\,x\,\ln \relax (3)+15\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((12*x*log(3) - 3*x + 5)/(15*x + 5*x*log(2*x) - log(3)*(60*x - 12*x^2) - 3*x^2),x)

[Out]

log(5*log(2*x) - 3*x - 60*log(3) + 12*x*log(3) + 15)

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sympy [A]  time = 0.16, size = 26, normalized size = 1.44 \begin {gather*} \log {\left (- \frac {3 x}{5} + \frac {12 x \log {\relax (3 )}}{5} + \log {\left (2 x \right )} - 12 \log {\relax (3 )} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((12*x*ln(3)-3*x+5)/(5*x*ln(2*x)+(12*x**2-60*x)*ln(3)-3*x**2+15*x),x)

[Out]

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

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