3.46.59 \(\int \frac {2 x^2+\log ^{e^{2 x} x}(3) (-4+8 x-4 x^2+e^{2 x} (4 x-12 x^3+8 x^4) \log (\log (3)))}{x^2-2 x^3+x^4} \, dx\)

Optimal. Leaf size=28 \[ \frac {2 x}{x-x^2}+\frac {4 \log ^{e^{2 x} x}(3)}{x} \]

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Rubi [A]  time = 0.69, antiderivative size = 51, normalized size of antiderivative = 1.82, number of steps used = 5, number of rules used = 4, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1594, 27, 6688, 2288} \begin {gather*} \frac {2}{1-x}+\frac {4 e^{2 x} (2 x+1) \log ^{e^{2 x} x}(3)}{x \left (2 e^{2 x} x+e^{2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + Log[3]^(E^(2*x)*x)*(-4 + 8*x - 4*x^2 + E^(2*x)*(4*x - 12*x^3 + 8*x^4)*Log[Log[3]]))/(x^2 - 2*x^3
+ x^4),x]

[Out]

2/(1 - x) + (4*E^(2*x)*(1 + 2*x)*Log[3]^(E^(2*x)*x))/(x*(E^(2*x) + 2*E^(2*x)*x))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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 {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{x^2 \left (1-2 x+x^2\right )} \, dx\\ &=\int \frac {2 x^2+\log ^{e^{2 x} x}(3) \left (-4+8 x-4 x^2+e^{2 x} \left (4 x-12 x^3+8 x^4\right ) \log (\log (3))\right )}{(-1+x)^2 x^2} \, dx\\ &=\int \left (\frac {2}{(-1+x)^2}+\frac {4 \log ^{e^{2 x} x}(3) \left (-1+e^{2 x} x (1+2 x) \log (\log (3))\right )}{x^2}\right ) \, dx\\ &=\frac {2}{1-x}+4 \int \frac {\log ^{e^{2 x} x}(3) \left (-1+e^{2 x} x (1+2 x) \log (\log (3))\right )}{x^2} \, dx\\ &=\frac {2}{1-x}+\frac {4 e^{2 x} (1+2 x) \log ^{e^{2 x} x}(3)}{x \left (e^{2 x}+2 e^{2 x} x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 49, normalized size = 1.75 \begin {gather*} -\frac {2}{-1+x}+\frac {4 e^{2 x} (1+2 x) \log ^{e^{2 x} x}(3)}{x \left (e^{2 x}+2 e^{2 x} x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + Log[3]^(E^(2*x)*x)*(-4 + 8*x - 4*x^2 + E^(2*x)*(4*x - 12*x^3 + 8*x^4)*Log[Log[3]]))/(x^2 -
2*x^3 + x^4),x]

[Out]

-2/(-1 + x) + (4*E^(2*x)*(1 + 2*x)*Log[3]^(E^(2*x)*x))/(x*(E^(2*x) + 2*E^(2*x)*x))

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fricas [A]  time = 0.63, size = 29, normalized size = 1.04 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x - 1\right )} \log \relax (3)^{x e^{\left (2 \, x\right )}} - x\right )}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-12*x^3+4*x)*exp(x)^2*log(log(3))-4*x^2+8*x-4)*exp(x*exp(x)^2*log(log(3)))+2*x^2)/(x^4-2*x^3
+x^2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (x^{2} + 2 \, {\left ({\left (2 \, x^{4} - 3 \, x^{3} + x\right )} e^{\left (2 \, x\right )} \log \left (\log \relax (3)\right ) - x^{2} + 2 \, x - 1\right )} \log \relax (3)^{x e^{\left (2 \, x\right )}}\right )}}{x^{4} - 2 \, x^{3} + x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-12*x^3+4*x)*exp(x)^2*log(log(3))-4*x^2+8*x-4)*exp(x*exp(x)^2*log(log(3)))+2*x^2)/(x^4-2*x^3
+x^2),x, algorithm="giac")

[Out]

integrate(2*(x^2 + 2*((2*x^4 - 3*x^3 + x)*e^(2*x)*log(log(3)) - x^2 + 2*x - 1)*log(3)^(x*e^(2*x)))/(x^4 - 2*x^
3 + x^2), x)

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maple [A]  time = 0.13, size = 23, normalized size = 0.82

method	result	size
risch	\(-\frac {2}{x -1}+\frac {4 \ln \relax (3)^{x \,{\mathrm e}^{2 x}}}{x}\)	\(23\)
norman	\(\frac {-2 x +4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \relax (3)\right )} x -4 \,{\mathrm e}^{x \,{\mathrm e}^{2 x} \ln \left (\ln \relax (3)\right )}}{x \left (x -1\right )}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((8*x^4-12*x^3+4*x)*exp(x)^2*ln(ln(3))-4*x^2+8*x-4)*exp(x*exp(x)^2*ln(ln(3)))+2*x^2)/(x^4-2*x^3+x^2),x,me
thod=_RETURNVERBOSE)

[Out]

-2/(x-1)+4/x*ln(3)^(x*exp(2*x))

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maxima [A]  time = 0.50, size = 29, normalized size = 1.04 \begin {gather*} \frac {2 \, {\left (2 \, {\left (x - 1\right )} \log \relax (3)^{x e^{\left (2 \, x\right )}} - x\right )}}{x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x^4-12*x^3+4*x)*exp(x)^2*log(log(3))-4*x^2+8*x-4)*exp(x*exp(x)^2*log(log(3)))+2*x^2)/(x^4-2*x^3
+x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.38, size = 22, normalized size = 0.79 \begin {gather*} \frac {4\,{\ln \relax (3)}^{x\,{\mathrm {e}}^{2\,x}}}{x}-\frac {2}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x*exp(2*x)*log(log(3)))*(8*x - 4*x^2 + exp(2*x)*log(log(3))*(4*x - 12*x^3 + 8*x^4) - 4) + 2*x^2)/(x^2
- 2*x^3 + x^4),x)

[Out]

(4*log(3)^(x*exp(2*x)))/x - 2/(x - 1)

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sympy [A]  time = 0.21, size = 20, normalized size = 0.71 \begin {gather*} - \frac {2}{x - 1} + \frac {4 e^{x e^{2 x} \log {\left (\log {\relax (3 )} \right )}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((8*x**4-12*x**3+4*x)*exp(x)**2*ln(ln(3))-4*x**2+8*x-4)*exp(x*exp(x)**2*ln(ln(3)))+2*x**2)/(x**4-2*
x**3+x**2),x)

[Out]

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

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