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

Optimal. Leaf size=18 \[ \frac {3}{1-\frac {\log (x)}{\sqrt [5]{e} x}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {3}{1-\frac {\sqrt [5]{e} x}{\log (x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3/(1 - (E^(1/5)*x)/Log[x])

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]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 21, normalized size = 1.17 \begin {gather*} -\frac {3 \sqrt [5]{e} x}{-\sqrt [5]{e} x+\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*E^(1/5)*x)/(-(E^(1/5)*x) + Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 16, normalized size = 0.89 \begin {gather*} \frac {3 \, x e^{\frac {1}{5}}}{x e^{\frac {1}{5}} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x*e^(1/5)/(x*e^(1/5) - log(x))

________________________________________________________________________________________

giac [A]  time = 0.15, size = 16, normalized size = 0.89 \begin {gather*} \frac {3 \, x e^{\frac {1}{5}}}{x e^{\frac {1}{5}} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x*e^(1/5)/(x*e^(1/5) - log(x))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 17, normalized size = 0.94

method	result	size
norman	\(\frac {3 \,{\mathrm e}^{\frac {1}{5}} x}{x \,{\mathrm e}^{\frac {1}{5}}-\ln \relax (x )}\)	\(17\)
risch	\(\frac {3 \,{\mathrm e}^{\frac {1}{5}} x}{x \,{\mathrm e}^{\frac {1}{5}}-\ln \relax (x )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*exp(1/5)*x/(x*exp(1/5)-ln(x))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 18, normalized size = 1.00 \begin {gather*} \frac {3 \, x e^{\frac {2}{5}}}{x e^{\frac {2}{5}} - e^{\frac {1}{5}} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.31, size = 15, normalized size = 0.83 \begin {gather*} -\frac {3\,x\,{\mathrm {e}}^{1/5}}{\ln \relax (x)-x\,{\mathrm {e}}^{1/5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*x*exp(1/5))/(log(x) - x*exp(1/5))

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 19, normalized size = 1.06 \begin {gather*} - \frac {3 x e^{\frac {1}{5}}}{- x e^{\frac {1}{5}} + \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x*exp(1/5)/(-x*exp(1/5) + log(x))

________________________________________________________________________________________