3.26.23 \(\int \frac {-12-48 x-48 x^2+e^4 (-2 x-6 x^2)}{(x^4+4 x^5+4 x^6) \log (\log (5))} \, dx\)

Optimal. Leaf size=29 \[ \frac {\frac {4}{x^2}+\frac {e^4}{x (1+2 x)}}{x \log (\log (5))} \]

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Rubi [A]  time = 0.08, antiderivative size = 53, normalized size of antiderivative = 1.83, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {12, 1594, 27, 1820} \begin {gather*} \frac {4}{x^3 \log (\log (5))}+\frac {e^4}{x^2 \log (\log (5))}+\frac {4 e^4}{(2 x+1) \log (\log (5))}-\frac {2 e^4}{x \log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-12 - 48*x - 48*x^2 + E^4*(-2*x - 6*x^2))/((x^4 + 4*x^5 + 4*x^6)*Log[Log[5]]),x]

[Out]

4/(x^3*Log[Log[5]]) + E^4/(x^2*Log[Log[5]]) - (2*E^4)/(x*Log[Log[5]]) + (4*E^4)/((1 + 2*x)*Log[Log[5]])

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 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 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-12-48 x-48 x^2+e^4 \left (-2 x-6 x^2\right )}{x^4+4 x^5+4 x^6} \, dx}{\log (\log (5))}\\ &=\frac {\int \frac {-12-48 x-48 x^2+e^4 \left (-2 x-6 x^2\right )}{x^4 \left (1+4 x+4 x^2\right )} \, dx}{\log (\log (5))}\\ &=\frac {\int \frac {-12-48 x-48 x^2+e^4 \left (-2 x-6 x^2\right )}{x^4 (1+2 x)^2} \, dx}{\log (\log (5))}\\ &=\frac {\int \left (-\frac {12}{x^4}-\frac {2 e^4}{x^3}+\frac {2 e^4}{x^2}-\frac {8 e^4}{(1+2 x)^2}\right ) \, dx}{\log (\log (5))}\\ &=\frac {4}{x^3 \log (\log (5))}+\frac {e^4}{x^2 \log (\log (5))}-\frac {2 e^4}{x \log (\log (5))}+\frac {4 e^4}{(1+2 x) \log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.86 \begin {gather*} \frac {4+\left (8+e^4\right ) x}{x^3 (1+2 x) \log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-12 - 48*x - 48*x^2 + E^4*(-2*x - 6*x^2))/((x^4 + 4*x^5 + 4*x^6)*Log[Log[5]]),x]

[Out]

(4 + (8 + E^4)*x)/(x^3*(1 + 2*x)*Log[Log[5]])

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fricas [A]  time = 1.02, size = 26, normalized size = 0.90 \begin {gather*} \frac {x e^{4} + 8 \, x + 4}{{\left (2 \, x^{4} + x^{3}\right )} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-2*x)*exp(4)-48*x^2-48*x-12)/(4*x^6+4*x^5+x^4)/log(log(5)),x, algorithm="fricas")

[Out]

(x*e^4 + 8*x + 4)/((2*x^4 + x^3)*log(log(5)))

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giac [A]  time = 0.17, size = 37, normalized size = 1.28 \begin {gather*} \frac {\frac {4 \, e^{4}}{2 \, x + 1} - \frac {2 \, x^{2} e^{4} - x e^{4} - 4}{x^{3}}}{\log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-2*x)*exp(4)-48*x^2-48*x-12)/(4*x^6+4*x^5+x^4)/log(log(5)),x, algorithm="giac")

[Out]

(4*e^4/(2*x + 1) - (2*x^2*e^4 - x*e^4 - 4)/x^3)/log(log(5))

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maple [A]  time = 0.06, size = 26, normalized size = 0.90

method	result	size
gosper	\(\frac {x \,{\mathrm e}^{4}+8 x +4}{x^{3} \ln \left (\ln \relax (5)\right ) \left (2 x +1\right )}\)	\(26\)
risch	\(\frac {4+2 \left (\frac {{\mathrm e}^{4}}{2}+4\right ) x}{x^{3} \ln \left (\ln \relax (5)\right ) \left (2 x +1\right )}\)	\(28\)
norman	\(\frac {\frac {\left ({\mathrm e}^{4}+8\right ) x}{\ln \left (\ln \relax (5)\right )}+\frac {4}{\ln \left (\ln \relax (5)\right )}}{x^{3} \left (2 x +1\right )}\)	\(31\)
default	\(\frac {-\frac {2 \,{\mathrm e}^{4}}{x}+\frac {4}{x^{3}}+\frac {{\mathrm e}^{4}}{x^{2}}+\frac {4 \,{\mathrm e}^{4}}{2 x +1}}{\ln \left (\ln \relax (5)\right )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-6*x^2-2*x)*exp(4)-48*x^2-48*x-12)/(4*x^6+4*x^5+x^4)/ln(ln(5)),x,method=_RETURNVERBOSE)

[Out]

1/x^3*(x*exp(4)+8*x+4)/ln(ln(5))/(2*x+1)

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maxima [A]  time = 0.37, size = 25, normalized size = 0.86 \begin {gather*} \frac {x {\left (e^{4} + 8\right )} + 4}{{\left (2 \, x^{4} + x^{3}\right )} \log \left (\log \relax (5)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x^2-2*x)*exp(4)-48*x^2-48*x-12)/(4*x^6+4*x^5+x^4)/log(log(5)),x, algorithm="maxima")

[Out]

(x*(e^4 + 8) + 4)/((2*x^4 + x^3)*log(log(5)))

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mupad [B]  time = 0.11, size = 27, normalized size = 0.93 \begin {gather*} \frac {x\,\left ({\mathrm {e}}^4+8\right )+4}{2\,\ln \left (\ln \relax (5)\right )\,x^4+\ln \left (\ln \relax (5)\right )\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(48*x + exp(4)*(2*x + 6*x^2) + 48*x^2 + 12)/(log(log(5))*(x^4 + 4*x^5 + 4*x^6)),x)

[Out]

(x*(exp(4) + 8) + 4)/(x^3*log(log(5)) + 2*x^4*log(log(5)))

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sympy [A]  time = 0.50, size = 29, normalized size = 1.00 \begin {gather*} - \frac {x \left (- e^{4} - 8\right ) - 4}{2 x^{4} \log {\left (\log {\relax (5 )} \right )} + x^{3} \log {\left (\log {\relax (5 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-6*x**2-2*x)*exp(4)-48*x**2-48*x-12)/(4*x**6+4*x**5+x**4)/ln(ln(5)),x)

[Out]

-(x*(-exp(4) - 8) - 4)/(2*x**4*log(log(5)) + x**3*log(log(5)))

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