\(\int \frac {-\frac {e^2}{x}+8100 x^4}{(e^2+2025 x^5) \log (\frac {12 e^2}{x}+24300 x^4)} \, dx\) [5562]

Optimal result

Integrand size = 43, antiderivative size = 17 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (12 \left (\frac {e^2}{x}+2025 x^4\right )\right )\right ) \]

[Out]

ln(ln(12*exp(-ln(x)+2)+24300*x^4))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {1607, 6816} \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (24300 x^4+\frac {12 e^2}{x}\right )\right ) \]

[In]

Int[(-(E^2/x) + 8100*x^4)/((E^2 + 2025*x^5)*Log[(12*E^2)/x + 24300*x^4]),x]

[Out]

Log[Log[(12*E^2)/x + 24300*x^4]]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-e^2+8100 x^5}{x \left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx \\ & = \log \left (\log \left (\frac {12 e^2}{x}+24300 x^4\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (\frac {12 \left (e^2+2025 x^5\right )}{x}\right )\right ) \]

[In]

Integrate[(-(E^2/x) + 8100*x^4)/((E^2 + 2025*x^5)*Log[(12*E^2)/x + 24300*x^4]),x]

[Out]

Log[Log[(12*(E^2 + 2025*x^5))/x]]

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
norman	\(\ln \left (\ln \left (\frac {12 \,{\mathrm e}^{2}}{x}+24300 x^{4}\right )\right )\)	\(16\)
risch	\(\ln \left (\ln \left (\frac {12 \,{\mathrm e}^{2}}{x}+24300 x^{4}\right )\right )\)	\(16\)
parallelrisch	\(\ln \left (\ln \left (12 \,{\mathrm e}^{-\ln \left (x \right )+2}+24300 x^{4}\right )\right )\)	\(18\)
default	\(\ln \left (\ln \left (3\right )+2 \ln \left (2\right )+\ln \left (\frac {{\mathrm e}^{2}+2025 x^{5}}{x}\right )\right )\)	\(22\)

[In]

int((-exp(-ln(x)+2)+8100*x^4)/(x*exp(-ln(x)+2)+2025*x^5)/ln(12*exp(-ln(x)+2)+24300*x^4),x,method=_RETURNVERBOS
E)

[Out]

ln(ln(12*exp(2)/x+24300*x^4))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (\frac {12 \, {\left (2025 \, x^{5} + e^{2}\right )}}{x}\right )\right ) \]

[In]

integrate((-exp(-log(x)+2)+8100*x^4)/(x*exp(-log(x)+2)+2025*x^5)/log(12*exp(-log(x)+2)+24300*x^4),x, algorithm
="fricas")

[Out]

log(log(12*(2025*x^5 + e^2)/x))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log {\left (\log {\left (24300 x^{4} + \frac {12 e^{2}}{x} \right )} \right )} \]

[In]

integrate((-exp(-ln(x)+2)+8100*x**4)/(x*exp(-ln(x)+2)+2025*x**5)/ln(12*exp(-ln(x)+2)+24300*x**4),x)

[Out]

log(log(24300*x**4 + 12*exp(2)/x))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (3\right ) + 2 \, \log \left (2\right ) + \log \left (2025 \, x^{5} + e^{2}\right ) - \log \left (x\right )\right ) \]

[In]

integrate((-exp(-log(x)+2)+8100*x^4)/(x*exp(-log(x)+2)+2025*x^5)/log(12*exp(-log(x)+2)+24300*x^4),x, algorithm
="maxima")

[Out]

log(log(3) + 2*log(2) + log(2025*x^5 + e^2) - log(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\log \left (\log \left (12\right ) + \log \left (2025 \, x^{5} + e^{2}\right ) - \log \left (x\right )\right ) \]

[In]

integrate((-exp(-log(x)+2)+8100*x^4)/(x*exp(-log(x)+2)+2025*x^5)/log(12*exp(-log(x)+2)+24300*x^4),x, algorithm
="giac")

[Out]

log(log(12) + log(2025*x^5 + e^2) - log(x))

Mupad [B] (verification not implemented)

Time = 11.76 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {-\frac {e^2}{x}+8100 x^4}{\left (e^2+2025 x^5\right ) \log \left (\frac {12 e^2}{x}+24300 x^4\right )} \, dx=\ln \left (\ln \left (\frac {12\,{\mathrm {e}}^2}{x}+24300\,x^4\right )\right ) \]

[In]

int(-(exp(2 - log(x)) - 8100*x^4)/(log(12*exp(2 - log(x)) + 24300*x^4)*(2025*x^5 + x*exp(2 - log(x)))),x)

[Out]

log(log((12*exp(2))/x + 24300*x^4))