\(\int \frac {10 e+2 x^2-5 x^3+(-2 e+x^3) \log (\frac {e^2}{3 x^2})}{-5 x^3+x^3 \log (\frac {e^2}{3 x^2})} \, dx\) [974]

Optimal result

Integrand size = 57, antiderivative size = 26 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=3+\frac {e}{x^2}+x-\log \left (5-\log \left (\frac {e^2}{3 x^2}\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2641, 6874, 14, 2339, 29} \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=\frac {e}{x^2}-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right )+x \]

[In]

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

[Out]

E/x^2 + x - Log[3 - Log[1/(3*x^2)]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2641

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

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{x^3 \left (-5+\log \left (\frac {e^2}{3 x^2}\right )\right )} \, dx \\ & = \int \left (\frac {-2 e+x^3}{x^3}+\frac {2}{x \left (-3+\log \left (\frac {1}{3 x^2}\right )\right )}\right ) \, dx \\ & = 2 \int \frac {1}{x \left (-3+\log \left (\frac {1}{3 x^2}\right )\right )} \, dx+\int \frac {-2 e+x^3}{x^3} \, dx \\ & = \int \left (1-\frac {2 e}{x^3}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,-3+\log \left (\frac {1}{3 x^2}\right )\right ) \\ & = \frac {e}{x^2}+x-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=\frac {e}{x^2}+x-\log \left (3-\log \left (\frac {1}{3 x^2}\right )\right ) \]

[In]

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

[Out]

E/x^2 + x - Log[3 - Log[1/(3*x^2)]]

Maple [A] (verified)

Time = 1.76 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
norman	\(\frac {x^{3}+{\mathrm e}}{x^{2}}-\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{3 x^{2}}\right )-5\right )\)	\(25\)
risch	\(\frac {x^{3}+{\mathrm e}}{x^{2}}-\ln \left (\ln \left (\frac {{\mathrm e}^{2}}{3 x^{2}}\right )-5\right )\)	\(25\)
parallelrisch	\(\frac {-2 \ln \left (\ln \left (\frac {{\mathrm e}^{2}}{3 x^{2}}\right )-5\right ) x^{2}+2 x^{3}+2 \,{\mathrm e}}{2 x^{2}}\)	\(32\)

[In]

int(((-2*exp(1)+x^3)*ln(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*ln(1/3*exp(2)/x^2)-5*x^3),x,method=_RETURN
VERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=\frac {x^{3} - x^{2} \log \left (\log \left (\frac {e^{2}}{3 \, x^{2}}\right ) - 5\right ) + e}{x^{2}} \]

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=x - \log {\left (\log {\left (\frac {e^{2}}{3 x^{2}} \right )} - 5 \right )} + \frac {e}{x^{2}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=\frac {x^{3} + e}{x^{2}} - \log \left (\frac {1}{2} \, \log \left (3\right ) + \log \left (x\right ) + \frac {3}{2}\right ) \]

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=\frac {x^{3} - x^{2} \log \left (\log \left (3 \, x^{2}\right ) + 3\right ) + e}{x^{2}} \]

[In]

integrate(((-2*exp(1)+x^3)*log(1/3*exp(2)/x^2)+10*exp(1)-5*x^3+2*x^2)/(x^3*log(1/3*exp(2)/x^2)-5*x^3),x, algor
ithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {10 e+2 x^2-5 x^3+\left (-2 e+x^3\right ) \log \left (\frac {e^2}{3 x^2}\right )}{-5 x^3+x^3 \log \left (\frac {e^2}{3 x^2}\right )} \, dx=x-\ln \left (\ln \left (\frac {1}{3\,x^2}\right )-3\right )+\frac {\mathrm {e}}{x^2} \]

[In]

int((10*exp(1) - log(exp(2)/(3*x^2))*(2*exp(1) - x^3) + 2*x^2 - 5*x^3)/(x^3*log(exp(2)/(3*x^2)) - 5*x^3),x)

[Out]

x - log(log(1/(3*x^2)) - 3) + exp(1)/x^2