\(\int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log (\frac {2}{x})+15 x^4 \log (3) \log ^2(\frac {2}{x})} \, dx\) [3490]

Optimal result

Integrand size = 45, antiderivative size = 22 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4}{15 \log (3) \left (\frac {9}{x^2}+\log \left (\frac {2}{x}\right )\right )} \]

[Out]

4/15/(9/x^2+ln(2/x))/ln(3)

Rubi [F]

\[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx \]

[In]

Int[(72*x + 4*x^3)/(1215*Log[3] + 270*x^2*Log[3]*Log[2/x] + 15*x^4*Log[3]*Log[2/x]^2),x]

[Out]

(24*Defer[Int][x/(9 + x^2*Log[2/x])^2, x])/(5*Log[3]) + (4*Defer[Int][x^3/(9 + x^2*Log[2/x])^2, x])/(15*Log[3]
)

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (72+4 x^2\right )}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx \\ & = \int \frac {4 x \left (18+x^2\right )}{15 \log (3) \left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx \\ & = \frac {4 \int \frac {x \left (18+x^2\right )}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{15 \log (3)} \\ & = \frac {4 \int \left (\frac {18 x}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2}+\frac {x^3}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2}\right ) \, dx}{15 \log (3)} \\ & = \frac {4 \int \frac {x^3}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{15 \log (3)}+\frac {24 \int \frac {x}{\left (9+x^2 \log \left (\frac {2}{x}\right )\right )^2} \, dx}{5 \log (3)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4 x^2}{15 \log (3) \left (9+x^2 \log \left (\frac {2}{x}\right )\right )} \]

[In]

Integrate[(72*x + 4*x^3)/(1215*Log[3] + 270*x^2*Log[3]*Log[2/x] + 15*x^4*Log[3]*Log[2/x]^2),x]

[Out]

(4*x^2)/(15*Log[3]*(9 + x^2*Log[2/x]))

Maple [A] (verified)

Time = 4.29 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {16}{15 \ln \left (3\right ) \left (\frac {36}{x^{2}}+4 \ln \left (\frac {2}{x}\right )\right )}\)	\(23\)
default	\(\frac {16}{15 \ln \left (3\right ) \left (\frac {36}{x^{2}}+4 \ln \left (\frac {2}{x}\right )\right )}\)	\(23\)
norman	\(\frac {4 x^{2}}{15 \ln \left (3\right ) \left (x^{2} \ln \left (\frac {2}{x}\right )+9\right )}\)	\(24\)
risch	\(\frac {4 x^{2}}{15 \ln \left (3\right ) \left (x^{2} \ln \left (\frac {2}{x}\right )+9\right )}\)	\(24\)
parallelrisch	\(\frac {4 x^{2}}{15 \ln \left (3\right ) \left (x^{2} \ln \left (\frac {2}{x}\right )+9\right )}\)	\(24\)

[In]

int((4*x^3+72*x)/(15*x^4*ln(3)*ln(2/x)^2+270*x^2*ln(3)*ln(2/x)+1215*ln(3)),x,method=_RETURNVERBOSE)

[Out]

16/15/ln(3)/(36/x^2+4*ln(2/x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4 \, x^{2}}{15 \, {\left (x^{2} \log \left (3\right ) \log \left (\frac {2}{x}\right ) + 9 \, \log \left (3\right )\right )}} \]

[In]

integrate((4*x^3+72*x)/(15*x^4*log(3)*log(2/x)^2+270*x^2*log(3)*log(2/x)+1215*log(3)),x, algorithm="fricas")

[Out]

4/15*x^2/(x^2*log(3)*log(2/x) + 9*log(3))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4 x^{2}}{15 x^{2} \log {\left (3 \right )} \log {\left (\frac {2}{x} \right )} + 135 \log {\left (3 \right )}} \]

[In]

integrate((4*x**3+72*x)/(15*x**4*ln(3)*ln(2/x)**2+270*x**2*ln(3)*ln(2/x)+1215*ln(3)),x)

[Out]

4*x**2/(15*x**2*log(3)*log(2/x) + 135*log(3))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4 \, x^{2}}{15 \, {\left (x^{2} \log \left (3\right ) \log \left (2\right ) - x^{2} \log \left (3\right ) \log \left (x\right ) + 9 \, \log \left (3\right )\right )}} \]

[In]

integrate((4*x^3+72*x)/(15*x^4*log(3)*log(2/x)^2+270*x^2*log(3)*log(2/x)+1215*log(3)),x, algorithm="maxima")

[Out]

4/15*x^2/(x^2*log(3)*log(2) - x^2*log(3)*log(x) + 9*log(3))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4}{15 \, {\left (\log \left (3\right ) \log \left (\frac {2}{x}\right ) + \frac {9 \, \log \left (3\right )}{x^{2}}\right )}} \]

[In]

integrate((4*x^3+72*x)/(15*x^4*log(3)*log(2/x)^2+270*x^2*log(3)*log(2/x)+1215*log(3)),x, algorithm="giac")

[Out]

4/15/(log(3)*log(2/x) + 9*log(3)/x^2)

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {72 x+4 x^3}{1215 \log (3)+270 x^2 \log (3) \log \left (\frac {2}{x}\right )+15 x^4 \log (3) \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4\,x^2}{15\,\ln \left (3\right )\,\left (x^2\,\ln \left (\frac {2}{x}\right )+9\right )} \]

[In]

int((72*x + 4*x^3)/(1215*log(3) + 270*x^2*log(3)*log(2/x) + 15*x^4*log(3)*log(2/x)^2),x)

[Out]

(4*x^2)/(15*log(3)*(x^2*log(2/x) + 9))