\(\int \frac {18 x-10 x^3+e^5 (-12-2 x^3)}{(e^{10} x^3-2 e^5 x^4+x^5) \log (\frac {\log ^2(5)}{4})} \, dx\) [8502]

Optimal result

Integrand size = 53, antiderivative size = 30 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {2 \left (5-\frac {3}{x^2}+x\right )}{\left (-e^5+x\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \]

[Out]

2*(x+5-3/x^2)/ln(1/4*ln(5)^2)/(-exp(5)+x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(75\) vs. \(2(30)=60\).

Time = 0.06 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.50, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {12, 1608, 27, 1634} \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {6}{e^5 x^2 \log \left (\frac {\log ^2(5)}{4}\right )}+\frac {6}{e^{10} x \log \left (\frac {\log ^2(5)}{4}\right )}-\frac {2 \left (3-5 e^{10}-e^{15}\right )}{e^{10} \left (e^5-x\right ) (\log (4)-2 \log (\log (5)))} \]

[In]

Int[(18*x - 10*x^3 + E^5*(-12 - 2*x^3))/((E^10*x^3 - 2*E^5*x^4 + x^5)*Log[Log[5]^2/4]),x]

[Out]

(-2*(3 - 5*E^10 - E^15))/(E^10*(E^5 - x)*(Log[4] - 2*Log[Log[5]])) + 6/(E^5*x^2*Log[Log[5]^2/4]) + 6/(E^10*x*L
og[Log[5]^2/4])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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}& = \frac {\int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{e^{10} x^3-2 e^5 x^4+x^5} \, dx}{\log \left (\frac {\log ^2(5)}{4}\right )} \\ & = \frac {\int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{x^3 \left (e^{10}-2 e^5 x+x^2\right )} \, dx}{\log \left (\frac {\log ^2(5)}{4}\right )} \\ & = \frac {\int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{x^3 \left (-e^5+x\right )^2} \, dx}{\log \left (\frac {\log ^2(5)}{4}\right )} \\ & = \frac {\int \left (-\frac {2 \left (-3+5 e^{10}+e^{15}\right )}{e^{10} \left (e^5-x\right )^2}-\frac {12}{e^5 x^3}-\frac {6}{e^{10} x^2}\right ) \, dx}{\log \left (\frac {\log ^2(5)}{4}\right )} \\ & = -\frac {2 \left (3-5 e^{10}-e^{15}\right )}{e^{10} \left (e^5-x\right ) (\log (4)-2 \log (\log (5)))}+\frac {6}{e^5 x^2 \log \left (\frac {\log ^2(5)}{4}\right )}+\frac {6}{e^{10} x \log \left (\frac {\log ^2(5)}{4}\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=-\frac {2 \left (-3+\left (5+e^5\right ) x^2\right )}{\left (e^5-x\right ) x^2 \log \left (\frac {\log ^2(5)}{4}\right )} \]

[In]

Integrate[(18*x - 10*x^3 + E^5*(-12 - 2*x^3))/((E^10*x^3 - 2*E^5*x^4 + x^5)*Log[Log[5]^2/4]),x]

[Out]

(-2*(-3 + (5 + E^5)*x^2))/((E^5 - x)*x^2*Log[Log[5]^2/4])

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20

method	result	size
gosper	\(-\frac {2 \left (x^{2} {\mathrm e}^{5}+5 x^{2}-3\right )}{x^{2} \ln \left (\frac {\ln \left (5\right )^{2}}{4}\right ) \left ({\mathrm e}^{5}-x \right )}\)	\(36\)
risch	\(\frac {\left (-2 \,{\mathrm e}^{5}-10\right ) x^{2}+6}{\left (-2 \ln \left (2\right )+2 \ln \left (\ln \left (5\right )\right )\right ) x^{2} \left ({\mathrm e}^{5}-x \right )}\)	\(37\)
parallelrisch	\(-\frac {-6+2 x^{2} {\mathrm e}^{5}+10 x^{2}}{\ln \left (\frac {\ln \left (5\right )^{2}}{4}\right ) x^{2} \left ({\mathrm e}^{5}-x \right )}\)	\(37\)
norman	\(\frac {\frac {\left ({\mathrm e}^{5}+5\right ) x^{2}}{\ln \left (2\right )-\ln \left (\ln \left (5\right )\right )}-\frac {3}{\ln \left (2\right )-\ln \left (\ln \left (5\right )\right )}}{x^{2} \left ({\mathrm e}^{5}-x \right )}\)	\(44\)

[In]

int(((-2*x^3-12)*exp(5)-10*x^3+18*x)/(x^3*exp(5)^2-2*x^4*exp(5)+x^5)/ln(1/4*ln(5)^2),x,method=_RETURNVERBOSE)

[Out]

-2/x^2*(x^2*exp(5)+5*x^2-3)/ln(1/4*ln(5)^2)/(exp(5)-x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {2 \, {\left (x^{2} e^{5} + 5 \, x^{2} - 3\right )}}{{\left (x^{3} - x^{2} e^{5}\right )} \log \left (\frac {1}{4} \, \log \left (5\right )^{2}\right )} \]

[In]

integrate(((-2*x^3-12)*exp(5)-10*x^3+18*x)/(x^3*exp(5)^2-2*x^4*exp(5)+x^5)/log(1/4*log(5)^2),x, algorithm="fri
cas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {x^{2} \left (- e^{5} - 5\right ) + 3}{x^{3} \left (- \log {\left (\log {\left (5 \right )} \right )} + \log {\left (2 \right )}\right ) + x^{2} \left (- e^{5} \log {\left (2 \right )} + e^{5} \log {\left (\log {\left (5 \right )} \right )}\right )} \]

[In]

integrate(((-2*x**3-12)*exp(5)-10*x**3+18*x)/(x**3*exp(5)**2-2*x**4*exp(5)+x**5)/ln(1/4*ln(5)**2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {2 \, {\left (x^{2} {\left (e^{5} + 5\right )} - 3\right )}}{{\left (x^{3} - x^{2} e^{5}\right )} \log \left (\frac {1}{4} \, \log \left (5\right )^{2}\right )} \]

[In]

integrate(((-2*x^3-12)*exp(5)-10*x^3+18*x)/(x^3*exp(5)^2-2*x^4*exp(5)+x^5)/log(1/4*log(5)^2),x, algorithm="max
ima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {2 \, {\left (\frac {{\left (e^{15} + 5 \, e^{10} - 3\right )} e^{\left (-10\right )}}{x - e^{5}} + \frac {3 \, {\left (x + e^{5}\right )} e^{\left (-10\right )}}{x^{2}}\right )}}{\log \left (\frac {1}{4} \, \log \left (5\right )^{2}\right )} \]

[In]

integrate(((-2*x^3-12)*exp(5)-10*x^3+18*x)/(x^3*exp(5)^2-2*x^4*exp(5)+x^5)/log(1/4*log(5)^2),x, algorithm="gia
c")

[Out]

2*((e^15 + 5*e^10 - 3)*e^(-10)/(x - e^5) + 3*(x + e^5)*e^(-10)/x^2)/log(1/4*log(5)^2)

Mupad [B] (verification not implemented)

Time = 12.93 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.37 \[ \int \frac {18 x-10 x^3+e^5 \left (-12-2 x^3\right )}{\left (e^{10} x^3-2 e^5 x^4+x^5\right ) \log \left (\frac {\log ^2(5)}{4}\right )} \, dx=\frac {x^2\,\left (2\,{\mathrm {e}}^5+10\right )-6}{x^3\,\ln \left (\frac {{\ln \left (5\right )}^2}{4}\right )-x^2\,{\mathrm {e}}^5\,\ln \left (\frac {{\ln \left (5\right )}^2}{4}\right )} \]

[In]

int(-(exp(5)*(2*x^3 + 12) - 18*x + 10*x^3)/(log(log(5)^2/4)*(x^3*exp(10) - 2*x^4*exp(5) + x^5)),x)

[Out]

(x^2*(2*exp(5) + 10) - 6)/(x^3*log(log(5)^2/4) - x^2*exp(5)*log(log(5)^2/4))