3.95.48 \(\int \frac {(-3 e^6+e^3 (-5-2 x)) \log (4)+(-3 e^6+e^3 (-5-2 x)) \log (\log (4 \log (2)))}{25 x^2+9 e^6 x^2+10 x^3+x^4+e^3 (30 x^2+6 x^3)} \, dx\)

Optimal. Leaf size=24 \[ \frac {\log (4)+\log (\log (4 \log (2)))}{x \left (3+\frac {5+x}{e^3}\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 4, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 12, 1680, 261} \begin {gather*} \frac {e^3 \log (4 \log (\log (16)))}{x \left (x+3 e^3+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Log[Log[4*Log[2]]])/(25*x^2 + 9*E^6*x^2 + 10
*x^3 + x^4 + E^3*(30*x^2 + 6*x^3)),x]

[Out]

(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + 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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) \log (4)+\left (-3 e^6+e^3 (-5-2 x)\right ) \log (\log (4 \log (2)))}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx\\ &=\int \frac {\left (-3 e^6+e^3 (-5-2 x)\right ) (\log (4)+\log (\log (4 \log (2))))}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx\\ &=\log (4 \log (\log (16))) \int \frac {-3 e^6+e^3 (-5-2 x)}{\left (25+9 e^6\right ) x^2+10 x^3+x^4+e^3 \left (30 x^2+6 x^3\right )} \, dx\\ &=\log (4 \log (\log (16))) \operatorname {Subst}\left (\int -\frac {32 e^3 x}{\left (25+30 e^3+9 e^6-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (10+6 e^3\right )+x\right )\\ &=-\left (\left (32 e^3 \log (4 \log (\log (16)))\right ) \operatorname {Subst}\left (\int \frac {x}{\left (25+30 e^3+9 e^6-4 x^2\right )^2} \, dx,x,\frac {1}{4} \left (10+6 e^3\right )+x\right )\right )\\ &=\frac {e^3 \log (4 \log (\log (16)))}{x \left (5+3 e^3+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.96 \begin {gather*} \frac {e^3 \log (4 \log (\log (16)))}{x \left (5+3 e^3+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-3*E^6 + E^3*(-5 - 2*x))*Log[4] + (-3*E^6 + E^3*(-5 - 2*x))*Log[Log[4*Log[2]]])/(25*x^2 + 9*E^6*x^
2 + 10*x^3 + x^4 + E^3*(30*x^2 + 6*x^3)),x]

[Out]

(E^3*Log[4*Log[Log[16]]])/(x*(5 + 3*E^3 + x))

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fricas [A]  time = 0.76, size = 31, normalized size = 1.29 \begin {gather*} \frac {2 \, e^{3} \log \relax (2) + e^{3} \log \left (\log \left (4 \, \log \relax (2)\right )\right )}{x^{2} + 3 \, x e^{3} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: -(30*exp(6)*ln(2)+15*exp(6)*ln(ln(4*ln(2
)))-30*exp(3)^2*ln(2)-15*exp(3)^2*ln(ln(4*ln(2))))/(81*exp(6)^2+540*exp(6)*exp(3)+450*exp(6)+900*exp(3)^2+1500
*exp(3)+625)*ln(sageV

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maple [A]  time = 0.11, size = 27, normalized size = 1.12

method	result	size
gosper	\(\frac {{\mathrm e}^{3} \left (\ln \left (\ln \left (4 \ln \relax (2)\right )\right )+2 \ln \relax (2)\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(27\)
norman	\(\frac {2 \,{\mathrm e}^{3} \ln \relax (2)+{\mathrm e}^{3} \ln \left (2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(33\)
risch	\(\frac {2 \,{\mathrm e}^{3} \ln \relax (2)+{\mathrm e}^{3} \ln \left (2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )}{x \left (5+3 \,{\mathrm e}^{3}+x \right )}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*exp(3)^2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*ln(2))/(9*x^2*exp(3)^2+(6*x
^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 31, normalized size = 1.29 \begin {gather*} \frac {2 \, e^{3} \log \relax (2) + e^{3} \log \left (\log \left (4 \, \log \relax (2)\right )\right )}{x^{2} + x {\left (3 \, e^{3} + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(3)^2+(-2*x-5)*exp(3))*log(log(4*log(2)))+2*(-3*exp(3)^2+(-2*x-5)*exp(3))*log(2))/(9*x^2*exp
(3)^2+(6*x^3+30*x^2)*exp(3)+x^4+10*x^3+25*x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.30, size = 23, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {e}}^3\,\ln \left (4\,\ln \left (\ln \left (16\right )\right )\right )}{x^2+\left (3\,{\mathrm {e}}^3+5\right )\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(4*log(2)))*(3*exp(6) + exp(3)*(2*x + 5)) + 2*log(2)*(3*exp(6) + exp(3)*(2*x + 5)))/(exp(3)*(30*x
^2 + 6*x^3) + 9*x^2*exp(6) + 25*x^2 + 10*x^3 + x^4),x)

[Out]

(exp(3)*log(4*log(log(16))))/(x^2 + x*(3*exp(3) + 5))

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sympy [A]  time = 1.16, size = 37, normalized size = 1.54 \begin {gather*} - \frac {- 2 e^{3} \log {\relax (2 )} - e^{3} \log {\left (\log {\left (\log {\relax (2 )} \right )} + 2 \log {\relax (2 )} \right )}}{x^{2} + x \left (5 + 3 e^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*exp(3)**2+(-2*x-5)*exp(3))*ln(ln(4*ln(2)))+2*(-3*exp(3)**2+(-2*x-5)*exp(3))*ln(2))/(9*x**2*exp(
3)**2+(6*x**3+30*x**2)*exp(3)+x**4+10*x**3+25*x**2),x)

[Out]

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

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