3.55.27 \(\int \frac {(6-6 x-6 x^2+e^x (-2+4 x-x^2)) \log (\log (4))}{16 x^2-16 x^3+4 x^4} \, dx\)

Optimal. Leaf size=27 \[ \frac {\left (3-e^x+6 x\right ) \log (\log (4))}{2 x (-4+2 x)} \]

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Rubi [B]  time = 0.49, antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 16, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1594, 27, 6742, 2177, 2178, 893} \begin {gather*} \frac {e^x \log (\log (4))}{8 x}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {15 \log (\log (4))}{8 (2-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((6 - 6*x - 6*x^2 + E^x*(-2 + 4*x - x^2))*Log[Log[4]])/(16*x^2 - 16*x^3 + 4*x^4),x]

[Out]

(-15*Log[Log[4]])/(8*(2 - x)) + (E^x*Log[Log[4]])/(8*(2 - x)) - (3*Log[Log[4]])/(8*x) + (E^x*Log[Log[4]])/(8*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 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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{16 x^2-16 x^3+4 x^4} \, dx\\ &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{x^2 \left (16-16 x+4 x^2\right )} \, dx\\ &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{4 (-2+x)^2 x^2} \, dx\\ &=\frac {1}{4} \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{(-2+x)^2 x^2} \, dx\\ &=\frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2}-\frac {6 \left (-1+x+x^2\right )}{(-2+x)^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \log (\log (4)) \int \frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2} \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \frac {-1+x+x^2}{(-2+x)^2 x^2} \, dx\\ &=-\left (\frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x}{2 (-2+x)^2}+\frac {e^x}{2 (-2+x)}+\frac {e^x}{2 x^2}-\frac {e^x}{2 x}\right ) \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \left (\frac {5}{4 (-2+x)^2}-\frac {1}{4 x^2}\right ) \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{(-2+x)^2} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x^2} \, dx+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x}-\frac {1}{8} e^2 \text {Ei}(-2+x) \log (\log (4))+\frac {1}{8} \text {Ei}(x) \log (\log (4))+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 23, normalized size = 0.85 \begin {gather*} -\frac {\left (-3+e^x-6 x\right ) \log (\log (4))}{4 (-2+x) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((6 - 6*x - 6*x^2 + E^x*(-2 + 4*x - x^2))*Log[Log[4]])/(16*x^2 - 16*x^3 + 4*x^4),x]

[Out]

-1/4*((-3 + E^x - 6*x)*Log[Log[4]])/((-2 + x)*x)

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fricas [A]  time = 0.59, size = 25, normalized size = 0.93 \begin {gather*} \frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+4*x-2)*exp(x)-6*x^2-6*x+6)*log(2*log(2))/(4*x^4-16*x^3+16*x^2),x, algorithm="fricas")

[Out]

1/4*(6*x - e^x + 3)*log(2*log(2))/(x^2 - 2*x)

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giac [A]  time = 0.26, size = 25, normalized size = 0.93 \begin {gather*} \frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+4*x-2)*exp(x)-6*x^2-6*x+6)*log(2*log(2))/(4*x^4-16*x^3+16*x^2),x, algorithm="giac")

[Out]

1/4*(6*x - e^x + 3)*log(2*log(2))/(x^2 - 2*x)

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maple [A]  time = 0.07, size = 37, normalized size = 1.37

method	result	size
default	\(\frac {\ln \left (2 \ln \relax (2)\right ) \left (\frac {15}{2 \left (x -2\right )}-\frac {3}{2 x}+\frac {{\mathrm e}^{x}}{2 x}-\frac {{\mathrm e}^{x}}{2 \left (x -2\right )}\right )}{4}\)	\(37\)
risch	\(\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (\frac {3 x}{2}+\frac {3}{4}\right )}{\left (x -2\right ) x}-\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{x}}{4 \left (x -2\right ) x}\)	\(40\)
norman	\(\frac {\left (-\frac {\ln \relax (2)}{4}-\frac {\ln \left (\ln \relax (2)\right )}{4}\right ) {\mathrm e}^{x}+\left (\frac {3 \ln \relax (2)}{2}+\frac {3 \ln \left (\ln \relax (2)\right )}{2}\right ) x +\frac {3 \ln \left (\ln \relax (2)\right )}{4}+\frac {3 \ln \relax (2)}{4}}{\left (x -2\right ) x}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+4*x-2)*exp(x)-6*x^2-6*x+6)*ln(2*ln(2))/(4*x^4-16*x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(2*ln(2))*(15/2/(x-2)-3/2/x+1/2*exp(x)/x-1/2*exp(x)/(x-2))

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maxima [A]  time = 0.48, size = 41, normalized size = 1.52 \begin {gather*} -\frac {1}{4} \, {\left (\frac {3 \, {\left (x - 1\right )}}{x^{2} - 2 \, x} + \frac {e^{x}}{x^{2} - 2 \, x} - \frac {9}{x - 2}\right )} \log \left (2 \, \log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+4*x-2)*exp(x)-6*x^2-6*x+6)*log(2*log(2))/(4*x^4-16*x^3+16*x^2),x, algorithm="maxima")

[Out]

-1/4*(3*(x - 1)/(x^2 - 2*x) + e^x/(x^2 - 2*x) - 9/(x - 2))*log(2*log(2))

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mupad [B]  time = 3.56, size = 34, normalized size = 1.26 \begin {gather*} \frac {3\,\ln \left (\ln \relax (4)\right )+x^2\,\ln \left ({\ln \relax (4)}^3\right )-\ln \left (2\,\ln \relax (2)\right )\,{\mathrm {e}}^x}{4\,x\,\left (x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*log(2))*(6*x + exp(x)*(x^2 - 4*x + 2) + 6*x^2 - 6))/(16*x^2 - 16*x^3 + 4*x^4),x)

[Out]

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

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sympy [B]  time = 0.46, size = 56, normalized size = 2.07 \begin {gather*} - \frac {x \left (- 6 \log {\relax (2 )} - 6 \log {\left (\log {\relax (2 )} \right )}\right ) - 3 \log {\relax (2 )} - 3 \log {\left (\log {\relax (2 )} \right )}}{4 x^{2} - 8 x} + \frac {\left (- \log {\relax (2 )} - \log {\left (\log {\relax (2 )} \right )}\right ) e^{x}}{4 x^{2} - 8 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+4*x-2)*exp(x)-6*x**2-6*x+6)*ln(2*ln(2))/(4*x**4-16*x**3+16*x**2),x)

[Out]

-(x*(-6*log(2) - 6*log(log(2))) - 3*log(2) - 3*log(log(2)))/(4*x**2 - 8*x) + (-log(2) - log(log(2)))*exp(x)/(4
*x**2 - 8*x)

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