3.56.54 \(\int \frac {-4-e^4+e^2 (4-4 x)+8 x}{16 x+4 e^4 x+32 x^2+16 x^3+e^2 (-16 x-16 x^2)+(-16 x-4 e^4 x-16 x^2+e^2 (16 x+8 x^2)) \log (x)+(4 x-4 e^2 x+e^4 x) \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {1}{-2-\frac {4 x}{2-e^2}+\log (x)} \]

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 35, normalized size of antiderivative = 1.94, number of steps used = 4, number of rules used = 4, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {6, 6688, 12, 6686} \begin {gather*} -\frac {2-e^2}{4 x-\left (2-e^2\right ) \log (x)+2 \left (2-e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - E^4 + E^2*(4 - 4*x) + 8*x)/(16*x + 4*E^4*x + 32*x^2 + 16*x^3 + E^2*(-16*x - 16*x^2) + (-16*x - 4*E^4
*x - 16*x^2 + E^2*(16*x + 8*x^2))*Log[x] + (4*x - 4*E^2*x + E^4*x)*Log[x]^2),x]

[Out]

-((2 - E^2)/(2*(2 - E^2) + 4*x - (2 - E^2)*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 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-e^4+e^2 (4-4 x)+8 x}{\left (16+4 e^4\right ) x+32 x^2+16 x^3+e^2 \left (-16 x-16 x^2\right )+\left (-16 x-4 e^4 x-16 x^2+e^2 \left (16 x+8 x^2\right )\right ) \log (x)+\left (4 x-4 e^2 x+e^4 x\right ) \log ^2(x)} \, dx\\ &=\int \frac {\left (2-e^2\right ) \left (-2+e^2+4 x\right )}{x \left (4 \left (1-\frac {e^2}{2}\right )+4 x+\left (-2+e^2\right ) \log (x)\right )^2} \, dx\\ &=\left (2-e^2\right ) \int \frac {-2+e^2+4 x}{x \left (4 \left (1-\frac {e^2}{2}\right )+4 x+\left (-2+e^2\right ) \log (x)\right )^2} \, dx\\ &=-\frac {2-e^2}{2 \left (2-e^2\right )+4 x-\left (2-e^2\right ) \log (x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 26, normalized size = 1.44 \begin {gather*} \frac {-2+e^2}{4-2 e^2+4 x+\left (-2+e^2\right ) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - E^4 + E^2*(4 - 4*x) + 8*x)/(16*x + 4*E^4*x + 32*x^2 + 16*x^3 + E^2*(-16*x - 16*x^2) + (-16*x -
4*E^4*x - 16*x^2 + E^2*(16*x + 8*x^2))*Log[x] + (4*x - 4*E^2*x + E^4*x)*Log[x]^2),x]

[Out]

(-2 + E^2)/(4 - 2*E^2 + 4*x + (-2 + E^2)*Log[x])

________________________________________________________________________________________

fricas [A]  time = 0.77, size = 23, normalized size = 1.28 \begin {gather*} \frac {e^{2} - 2}{{\left (e^{2} - 2\right )} \log \relax (x) + 4 \, x - 2 \, e^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)^2+(-4*x+4)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*log(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*
exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2+(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="fricas")

[Out]

(e^2 - 2)/((e^2 - 2)*log(x) + 4*x - 2*e^2 + 4)

________________________________________________________________________________________

giac [A]  time = 0.17, size = 25, normalized size = 1.39 \begin {gather*} \frac {e^{2} - 2}{e^{2} \log \relax (x) + 4 \, x - 2 \, e^{2} - 2 \, \log \relax (x) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)^2+(-4*x+4)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*log(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*
exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2+(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="giac")

[Out]

(e^2 - 2)/(e^2*log(x) + 4*x - 2*e^2 - 2*log(x) + 4)

________________________________________________________________________________________

maple [A]  time = 0.14, size = 26, normalized size = 1.44

method	result	size
norman	\(\frac {{\mathrm e}^{2}-2}{{\mathrm e}^{2} \ln \relax (x )-2 \,{\mathrm e}^{2}-2 \ln \relax (x )+4 x +4}\)	\(26\)
risch	\(\frac {{\mathrm e}^{2}}{{\mathrm e}^{2} \ln \relax (x )-2 \,{\mathrm e}^{2}-2 \ln \relax (x )+4 x +4}-\frac {2}{{\mathrm e}^{2} \ln \relax (x )-2 \,{\mathrm e}^{2}-2 \ln \relax (x )+4 x +4}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-exp(2)^2+(-4*x+4)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*ln(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*exp(2)-
16*x^2-16*x)*ln(x)+4*x*exp(2)^2+(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

(exp(2)-2)/(exp(2)*ln(x)-2*exp(2)-2*ln(x)+4*x+4)

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 23, normalized size = 1.28 \begin {gather*} \frac {e^{2} - 2}{{\left (e^{2} - 2\right )} \log \relax (x) + 4 \, x - 2 \, e^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)^2+(-4*x+4)*exp(2)+8*x-4)/((x*exp(2)^2-4*exp(2)*x+4*x)*log(x)^2+(-4*x*exp(2)^2+(8*x^2+16*x)*
exp(2)-16*x^2-16*x)*log(x)+4*x*exp(2)^2+(-16*x^2-16*x)*exp(2)+16*x^3+32*x^2+16*x),x, algorithm="maxima")

[Out]

(e^2 - 2)/((e^2 - 2)*log(x) + 4*x - 2*e^2 + 4)

________________________________________________________________________________________

mupad [B]  time = 4.31, size = 25, normalized size = 1.39 \begin {gather*} \frac {{\mathrm {e}}^2-2}{4\,x-2\,{\mathrm {e}}^2-2\,\ln \relax (x)+{\mathrm {e}}^2\,\ln \relax (x)+4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4) - 8*x + exp(2)*(4*x - 4) + 4)/(16*x - log(x)*(16*x - exp(2)*(16*x + 8*x^2) + 4*x*exp(4) + 16*x^2)
- exp(2)*(16*x + 16*x^2) + 4*x*exp(4) + log(x)^2*(4*x - 4*x*exp(2) + x*exp(4)) + 32*x^2 + 16*x^3),x)

[Out]

(exp(2) - 2)/(4*x - 2*exp(2) - 2*log(x) + exp(2)*log(x) + 4)

________________________________________________________________________________________

sympy [A]  time = 0.18, size = 22, normalized size = 1.22 \begin {gather*} \frac {-2 + e^{2}}{4 x + \left (-2 + e^{2}\right ) \log {\relax (x )} - 2 e^{2} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-exp(2)**2+(-4*x+4)*exp(2)+8*x-4)/((x*exp(2)**2-4*exp(2)*x+4*x)*ln(x)**2+(-4*x*exp(2)**2+(8*x**2+16
*x)*exp(2)-16*x**2-16*x)*ln(x)+4*x*exp(2)**2+(-16*x**2-16*x)*exp(2)+16*x**3+32*x**2+16*x),x)

[Out]

(-2 + exp(2))/(4*x + (-2 + exp(2))*log(x) - 2*exp(2) + 4)

________________________________________________________________________________________