3.61.93 \(\int \frac {-1+e^2-3 x+(-1+e^2) \log (x)}{(-x+e^2 x-3 x^2) \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (\frac {x \log (x)}{1-e^2+3 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 19, normalized size of antiderivative = 1.12, number of steps used = 9, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {6, 1593, 6742, 36, 31, 29, 2302} \begin {gather*} \log (x)-\log \left (3 x-e^2+1\right )+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - Log[1 - E^2 + 3*x] + Log[Log[x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1593

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

Rule 2302

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 6742

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 19, normalized size = 1.12 \begin {gather*} \log (x)-\log \left (1-e^2+3 x\right )+\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[x] - Log[1 - E^2 + 3*x] + Log[Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 18, normalized size = 1.06 \begin {gather*} -\log \left (3 \, x - e^{2} + 1\right ) + \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.15, size = 18, normalized size = 1.06 \begin {gather*} -\log \left (3 \, x - e^{2} + 1\right ) + \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.38, size = 17, normalized size = 1.00

method	result	size
norman	\(\ln \relax (x )-\ln \left ({\mathrm e}^{2}-1-3 x \right )+\ln \left (\ln \relax (x )\right )\)	\(17\)
risch	\(-\ln \left (1+3 x -{\mathrm e}^{2}\right )+\ln \relax (x )+\ln \left (\ln \relax (x )\right )\)	\(19\)
default	\(\ln \left (\ln \relax (x )\right )-\frac {{\mathrm e}^{2} \ln \left (1+3 x -{\mathrm e}^{2}\right )}{{\mathrm e}^{2}-1}+\frac {{\mathrm e}^{2} \ln \relax (x )}{{\mathrm e}^{2}-1}+\frac {\ln \left (1+3 x -{\mathrm e}^{2}\right )}{{\mathrm e}^{2}-1}-\frac {\ln \relax (x )}{{\mathrm e}^{2}-1}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 18, normalized size = 1.06 \begin {gather*} -\log \left (3 \, x - e^{2} + 1\right ) + \log \relax (x) + \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 4.56, size = 18, normalized size = 1.06 \begin {gather*} \ln \left (\ln \relax (x)\right )-\ln \left (3\,x-{\mathrm {e}}^2+1\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.33, size = 141, normalized size = 8.29 \begin {gather*} \left (1 - e^{2}\right ) \left (\frac {\log {\left (x - \frac {e^{4}}{6 \left (-1 + e\right ) \left (1 + e\right )} - \frac {e^{2}}{6} - \frac {1}{6 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6} + \frac {e^{2}}{3 \left (-1 + e\right ) \left (1 + e\right )} \right )}}{\left (-1 + e\right ) \left (1 + e\right )} - \frac {\log {\left (x - \frac {e^{2}}{6} - \frac {e^{2}}{3 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6 \left (-1 + e\right ) \left (1 + e\right )} + \frac {1}{6} + \frac {e^{4}}{6 \left (-1 + e\right ) \left (1 + e\right )} \right )}}{\left (-1 + e\right ) \left (1 + e\right )}\right ) + \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - exp(2))*(log(x - exp(4)/(6*(-1 + E)*(1 + E)) - exp(2)/6 - 1/(6*(-1 + E)*(1 + E)) + 1/6 + exp(2)/(3*(-1 +
E)*(1 + E)))/((-1 + E)*(1 + E)) - log(x - exp(2)/6 - exp(2)/(3*(-1 + E)*(1 + E)) + 1/(6*(-1 + E)*(1 + E)) + 1/
6 + exp(4)/(6*(-1 + E)*(1 + E)))/((-1 + E)*(1 + E))) + log(log(x))

________________________________________________________________________________________