3.44.59 \(\int \frac {2-6 x+e^{18} (-100 x-20 e x-e^2 x)}{-6 x^2+e^{18} (-100 x^2-20 e x^2-e^2 x^2)+2 x \log (x)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (3 x+\frac {1}{2} e^{18} (10+e)^2 x-\log (x)\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 3, number of rules used = 3, integrand size = 61, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6741, 2561, 6684} \begin {gather*} \log \left (\left (6+e^{18} (10+e)^2\right ) x-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 6*x + E^18*(-100*x - 20*E*x - E^2*x))/(-6*x^2 + E^18*(-100*x^2 - 20*E*x^2 - E^2*x^2) + 2*x*Log[x]),x]

[Out]

Log[(6 + E^18*(10 + E)^2)*x - 2*Log[x]]

Rule 2561

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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2+\left (6+100 e^{18}+20 e^{19}+e^{20}\right ) x}{6 \left (1+\frac {1}{6} e^{18} (10+e)^2\right ) x^2-2 x \log (x)} \, dx\\ &=\int \frac {-2+\left (6+100 e^{18}+20 e^{19}+e^{20}\right ) x}{x \left (6 \left (1+\frac {1}{6} e^{18} (10+e)^2\right ) x-2 \log (x)\right )} \, dx\\ &=\log \left (\left (6+e^{18} (10+e)^2\right ) x-2 \log (x)\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 26, normalized size = 1.18 \begin {gather*} \log \left (6 x+100 e^{18} x+20 e^{19} x+e^{20} x-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 6*x + E^18*(-100*x - 20*E*x - E^2*x))/(-6*x^2 + E^18*(-100*x^2 - 20*E*x^2 - E^2*x^2) + 2*x*Log[
x]),x]

[Out]

Log[6*x + 100*E^18*x + 20*E^19*x + E^20*x - 2*Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 24, normalized size = 1.09 \begin {gather*} \log \left (-x e^{20} - 20 \, x e^{19} - 100 \, x e^{18} - 6 \, x + 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(1)^2-20*x*exp(1)-100*x)*exp(9)^2-6*x+2)/(2*x*log(x)+(-x^2*exp(1)^2-20*x^2*exp(1)-100*x^2)*e
xp(9)^2-6*x^2),x, algorithm="fricas")

[Out]

log(-x*e^20 - 20*x*e^19 - 100*x*e^18 - 6*x + 2*log(x))

________________________________________________________________________________________

giac [A]  time = 0.16, size = 24, normalized size = 1.09 \begin {gather*} \log \left (-x e^{20} - 20 \, x e^{19} - 100 \, x e^{18} - 6 \, x + 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(1)^2-20*x*exp(1)-100*x)*exp(9)^2-6*x+2)/(2*x*log(x)+(-x^2*exp(1)^2-20*x^2*exp(1)-100*x^2)*e
xp(9)^2-6*x^2),x, algorithm="giac")

[Out]

log(-x*e^20 - 20*x*e^19 - 100*x*e^18 - 6*x + 2*log(x))

________________________________________________________________________________________

maple [A]  time = 0.04, size = 23, normalized size = 1.05

method	result	size
risch	\(\ln \left (-\frac {x \,{\mathrm e}^{20}}{2}-10 \,{\mathrm e}^{19} x -50 \,{\mathrm e}^{18} x -3 x +\ln \relax (x )\right )\)	\(23\)
norman	\(\ln \left ({\mathrm e}^{2} {\mathrm e}^{18} x +20 \,{\mathrm e} \,{\mathrm e}^{18} x +100 \,{\mathrm e}^{18} x +6 x -2 \ln \relax (x )\right )\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(1)^2-20*x*exp(1)-100*x)*exp(9)^2-6*x+2)/(2*x*ln(x)+(-x^2*exp(1)^2-20*x^2*exp(1)-100*x^2)*exp(9)^2
-6*x^2),x,method=_RETURNVERBOSE)

[Out]

ln(-1/2*x*exp(20)-10*exp(19)*x-50*exp(18)*x-3*x+ln(x))

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 19, normalized size = 0.86 \begin {gather*} \log \left (-\frac {1}{2} \, x {\left (e^{20} + 20 \, e^{19} + 100 \, e^{18} + 6\right )} + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(1)^2-20*x*exp(1)-100*x)*exp(9)^2-6*x+2)/(2*x*log(x)+(-x^2*exp(1)^2-20*x^2*exp(1)-100*x^2)*e
xp(9)^2-6*x^2),x, algorithm="maxima")

[Out]

log(-1/2*x*(e^20 + 20*e^19 + 100*e^18 + 6) + log(x))

________________________________________________________________________________________

mupad [B]  time = 3.39, size = 23, normalized size = 1.05 \begin {gather*} \ln \left (6\,x-2\,\ln \relax (x)+100\,x\,{\mathrm {e}}^{18}+20\,x\,{\mathrm {e}}^{19}+x\,{\mathrm {e}}^{20}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + exp(18)*(100*x + 20*x*exp(1) + x*exp(2)) - 2)/(exp(18)*(20*x^2*exp(1) + x^2*exp(2) + 100*x^2) - 2*x
*log(x) + 6*x^2),x)

[Out]

log(6*x - 2*log(x) + 100*x*exp(18) + 20*x*exp(19) + x*exp(20))

________________________________________________________________________________________

sympy [A]  time = 0.19, size = 27, normalized size = 1.23 \begin {gather*} \log {\left (- 50 x e^{18} - 10 x e^{19} - \frac {x e^{20}}{2} - 3 x + \log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(1)**2-20*x*exp(1)-100*x)*exp(9)**2-6*x+2)/(2*x*ln(x)+(-x**2*exp(1)**2-20*x**2*exp(1)-100*x*
*2)*exp(9)**2-6*x**2),x)

[Out]

log(-50*x*exp(18) - 10*x*exp(19) - x*exp(20)/2 - 3*x + log(x))

________________________________________________________________________________________