\(\int \frac {-30 e^3 x+e^6 x^4+(-15+2 e^3 x^3) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx\) [614]

Optimal result

Integrand size = 68, antiderivative size = 24 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x-\frac {15 \left (2 x-\frac {1}{e^3 x+\log (\log (2))}\right )}{x} \]

[Out]

x-15*(2*x-1/(x*exp(3)+ln(ln(2))))/x

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {1608, 27, 1634} \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )}+\frac {15}{x \log (\log (2))} \]

[In]

Int[(-30*E^3*x + E^6*x^4 + (-15 + 2*E^3*x^3)*Log[Log[2]] + x^2*Log[Log[2]]^2)/(E^6*x^4 + 2*E^3*x^3*Log[Log[2]]
+ x^2*Log[Log[2]]^2),x]

[Out]

x + 15/(x*Log[Log[2]]) - (15*E^3)/(Log[Log[2]]*(E^3*x + Log[Log[2]]))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1608

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

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{x^2 \left (e^6 x^2+2 e^3 x \log (\log (2))+\log ^2(\log (2))\right )} \, dx \\ & = \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{x^2 \left (e^3 x+\log (\log (2))\right )^2} \, dx \\ & = \int \left (1-\frac {15}{x^2 \log (\log (2))}+\frac {15 e^6}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )^2}\right ) \, dx \\ & = x+\frac {15}{x \log (\log (2))}-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x+\frac {15}{x \log (\log (2))}-\frac {15 e^3}{\log (\log (2)) \left (e^3 x+\log (\log (2))\right )} \]

[In]

Integrate[(-30*E^3*x + E^6*x^4 + (-15 + 2*E^3*x^3)*Log[Log[2]] + x^2*Log[Log[2]]^2)/(E^6*x^4 + 2*E^3*x^3*Log[L
og[2]] + x^2*Log[Log[2]]^2),x]

[Out]

x + 15/(x*Log[Log[2]]) - (15*E^3)/(Log[Log[2]]*(E^3*x + Log[Log[2]]))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
risch	\(x +\frac {15}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\)	\(18\)
norman	\(\frac {15+x^{3} {\mathrm e}^{3}+x^{2} \ln \left (\ln \left (2\right )\right )}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\)	\(30\)
gosper	\(\frac {\left (x^{3} {\mathrm e}^{6}-x \ln \left (\ln \left (2\right )\right )^{2}+15 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\)	\(40\)
parallelrisch	\(\frac {\left (x^{3} {\mathrm e}^{6}-x \ln \left (\ln \left (2\right )\right )^{2}+15 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-3}}{x \left (x \,{\mathrm e}^{3}+\ln \left (\ln \left (2\right )\right )\right )}\)	\(40\)

[In]

int((x^2*ln(ln(2))^2+(2*x^3*exp(3)-15)*ln(ln(2))+x^4*exp(3)^2-30*x*exp(3))/(x^2*ln(ln(2))^2+2*x^3*exp(3)*ln(ln
(2))+x^4*exp(3)^2),x,method=_RETURNVERBOSE)

[Out]

x+15/x/(x*exp(3)+ln(ln(2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=\frac {x^{3} e^{3} + x^{2} \log \left (\log \left (2\right )\right ) + 15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]

[In]

integrate((x^2*log(log(2))^2+(2*x^3*exp(3)-15)*log(log(2))+x^4*exp(3)^2-30*x*exp(3))/(x^2*log(log(2))^2+2*x^3*
exp(3)*log(log(2))+x^4*exp(3)^2),x, algorithm="fricas")

[Out]

(x^3*e^3 + x^2*log(log(2)) + 15)/(x^2*e^3 + x*log(log(2)))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log {\left (\log {\left (2 \right )} \right )}} \]

[In]

integrate((x**2*ln(ln(2))**2+(2*x**3*exp(3)-15)*ln(ln(2))+x**4*exp(3)**2-30*x*exp(3))/(x**2*ln(ln(2))**2+2*x**
3*exp(3)*ln(ln(2))+x**4*exp(3)**2),x)

[Out]

x + 15/(x**2*exp(3) + x*log(log(2)))

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]

[In]

integrate((x^2*log(log(2))^2+(2*x^3*exp(3)-15)*log(log(2))+x^4*exp(3)^2-30*x*exp(3))/(x^2*log(log(2))^2+2*x^3*
exp(3)*log(log(2))+x^4*exp(3)^2),x, algorithm="maxima")

[Out]

x + 15/(x^2*e^3 + x*log(log(2)))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x + \frac {15}{x^{2} e^{3} + x \log \left (\log \left (2\right )\right )} \]

[In]

integrate((x^2*log(log(2))^2+(2*x^3*exp(3)-15)*log(log(2))+x^4*exp(3)^2-30*x*exp(3))/(x^2*log(log(2))^2+2*x^3*
exp(3)*log(log(2))+x^4*exp(3)^2),x, algorithm="giac")

[Out]

x + 15/(x^2*e^3 + x*log(log(2)))

Mupad [B] (verification not implemented)

Time = 9.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-30 e^3 x+e^6 x^4+\left (-15+2 e^3 x^3\right ) \log (\log (2))+x^2 \log ^2(\log (2))}{e^6 x^4+2 e^3 x^3 \log (\log (2))+x^2 \log ^2(\log (2))} \, dx=x+\frac {15}{x\,\left (\ln \left (\ln \left (2\right )\right )+x\,{\mathrm {e}}^3\right )} \]

[In]

int((log(log(2))*(2*x^3*exp(3) - 15) - 30*x*exp(3) + x^2*log(log(2))^2 + x^4*exp(6))/(x^2*log(log(2))^2 + x^4*
exp(6) + 2*x^3*exp(3)*log(log(2))),x)

[Out]

x + 15/(x*(log(log(2)) + x*exp(3)))