\(\int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx\) [5629]

Optimal result

Integrand size = 33, antiderivative size = 14 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\sqrt [5]{e}}}{(-3+\log (x))^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {12, 32} \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\sqrt [5]{e}}}{(3-\log (x))^2} \]

[In]

Int[(-2*E^E^(1/5))/(-27*x + 27*x*Log[x] - 9*x*Log[x]^2 + x*Log[x]^3),x]

[Out]

E^E^(1/5)/(3 - Log[x])^2

Rule 12

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\left (\left (2 e^{\sqrt [5]{e}}\right ) \int \frac {1}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx\right ) \\ & = -\left (\left (2 e^{\sqrt [5]{e}}\right ) \text {Subst}\left (\int \frac {1}{(-3+x)^3} \, dx,x,\log (x)\right )\right ) \\ & = \frac {e^{\sqrt [5]{e}}}{(3-\log (x))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\sqrt [5]{e}}}{(-3+\log (x))^2} \]

[In]

Integrate[(-2*E^E^(1/5))/(-27*x + 27*x*Log[x] - 9*x*Log[x]^2 + x*Log[x]^3),x]

[Out]

E^E^(1/5)/(-3 + Log[x])^2

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {1}{5}}}}{\left (\ln \left (x \right )-3\right )^{2}}\)	\(11\)
norman	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {1}{5}}}}{\left (\ln \left (x \right )-3\right )^{2}}\)	\(11\)
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {1}{5}}}}{\left (\ln \left (x \right )-3\right )^{2}}\)	\(11\)
parallelrisch	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {1}{5}}}}{\ln \left (x \right )^{2}-6 \ln \left (x \right )+9}\)	\(17\)

[In]

int(-2*exp(exp(1/5))/(x*ln(x)^3-9*x*ln(x)^2+27*x*ln(x)-27*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\left (e^{\frac {1}{5}}\right )}}{\log \left (x\right )^{2} - 6 \, \log \left (x\right ) + 9} \]

[In]

integrate(-2*exp(exp(1/5))/(x*log(x)^3-9*x*log(x)^2+27*x*log(x)-27*x),x, algorithm="fricas")

[Out]

e^(e^(1/5))/(log(x)^2 - 6*log(x) + 9)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{e^{\frac {1}{5}}}}{\log {\left (x \right )}^{2} - 6 \log {\left (x \right )} + 9} \]

[In]

integrate(-2*exp(exp(1/5))/(x*ln(x)**3-9*x*ln(x)**2+27*x*ln(x)-27*x),x)

[Out]

exp(exp(1/5))/(log(x)**2 - 6*log(x) + 9)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\left (e^{\frac {1}{5}}\right )}}{\log \left (x\right )^{2} - 6 \, \log \left (x\right ) + 9} \]

[In]

integrate(-2*exp(exp(1/5))/(x*log(x)^3-9*x*log(x)^2+27*x*log(x)-27*x),x, algorithm="maxima")

[Out]

e^(e^(1/5))/(log(x)^2 - 6*log(x) + 9)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {e^{\left (e^{\frac {1}{5}}\right )}}{{\left (\log \left (x\right ) - 3\right )}^{2}} \]

[In]

integrate(-2*exp(exp(1/5))/(x*log(x)^3-9*x*log(x)^2+27*x*log(x)-27*x),x, algorithm="giac")

[Out]

e^(e^(1/5))/(log(x) - 3)^2

Mupad [B] (verification not implemented)

Time = 11.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int -\frac {2 e^{\sqrt [5]{e}}}{-27 x+27 x \log (x)-9 x \log ^2(x)+x \log ^3(x)} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{1/5}}}{{\left (\ln \left (x\right )-3\right )}^2} \]

[In]

int((2*exp(exp(1/5)))/(27*x + 9*x*log(x)^2 - x*log(x)^3 - 27*x*log(x)),x)

[Out]

exp(exp(1/5))/(log(x) - 3)^2