\(\int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} (24 x^3+24 x^4+6 x^5)} \, dx\) [7601]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=3+\frac {\log (x \log (2))}{6 e^{20} x^2 (2+x)} \]

[Out]

3+1/6*ln(x*ln(2))/x^2/(2+x)/exp(5)^4

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(58\) vs. \(2(22)=44\).

Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.64, number of steps used = 14, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {12, 1608, 27, 6874, 46, 2404, 2341, 2351, 31} \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {\log (x \log (2))}{12 e^{20} x^2}+\frac {\log (x)}{48 e^{20}}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (x+2)} \]

[In]

Int[(2 + x + (-4 - 3*x)*Log[x*Log[2]])/(E^20*(24*x^3 + 24*x^4 + 6*x^5)),x]

[Out]

Log[x]/(48*E^20) + Log[x*Log[2]]/(12*E^20*x^2) - Log[x*Log[2]]/(24*E^20*x) - (x*Log[x*Log[2]])/(48*E^20*(2 + 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 1608

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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{24 x^3+24 x^4+6 x^5} \, dx}{e^{20}} \\ & = \frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{x^3 \left (24+24 x+6 x^2\right )} \, dx}{e^{20}} \\ & = \frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{6 x^3 (2+x)^2} \, dx}{e^{20}} \\ & = \frac {\int \frac {2+x+(-4-3 x) \log (x \log (2))}{x^3 (2+x)^2} \, dx}{6 e^{20}} \\ & = \frac {\int \left (\frac {1}{x^3 (2+x)}-\frac {(4+3 x) \log (x \log (2))}{x^3 (2+x)^2}\right ) \, dx}{6 e^{20}} \\ & = \frac {\int \frac {1}{x^3 (2+x)} \, dx}{6 e^{20}}-\frac {\int \frac {(4+3 x) \log (x \log (2))}{x^3 (2+x)^2} \, dx}{6 e^{20}} \\ & = \frac {\int \left (\frac {1}{2 x^3}-\frac {1}{4 x^2}+\frac {1}{8 x}-\frac {1}{8 (2+x)}\right ) \, dx}{6 e^{20}}-\frac {\int \left (\frac {\log (x \log (2))}{x^3}-\frac {\log (x \log (2))}{4 x^2}+\frac {\log (x \log (2))}{4 (2+x)^2}\right ) \, dx}{6 e^{20}} \\ & = -\frac {1}{24 e^{20} x^2}+\frac {1}{24 e^{20} x}+\frac {\log (x)}{48 e^{20}}-\frac {\log (2+x)}{48 e^{20}}+\frac {\int \frac {\log (x \log (2))}{x^2} \, dx}{24 e^{20}}-\frac {\int \frac {\log (x \log (2))}{(2+x)^2} \, dx}{24 e^{20}}-\frac {\int \frac {\log (x \log (2))}{x^3} \, dx}{6 e^{20}} \\ & = \frac {\log (x)}{48 e^{20}}-\frac {\log (2+x)}{48 e^{20}}+\frac {\log (x \log (2))}{12 e^{20} x^2}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (2+x)}+\frac {\int \frac {1}{2+x} \, dx}{48 e^{20}} \\ & = \frac {\log (x)}{48 e^{20}}+\frac {\log (x \log (2))}{12 e^{20} x^2}-\frac {\log (x \log (2))}{24 e^{20} x}-\frac {x \log (x \log (2))}{48 e^{20} (2+x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {\log (x \log (2))}{6 e^{20} x^2 (2+x)} \]

[In]

Integrate[(2 + x + (-4 - 3*x)*Log[x*Log[2]])/(E^20*(24*x^3 + 24*x^4 + 6*x^5)),x]

[Out]

Log[x*Log[2]]/(6*E^20*x^2*(2 + x))

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
risch	\(\frac {\ln \left (x \ln \left (2\right )\right ) {\mathrm e}^{-20}}{6 x^{2} \left (2+x \right )}\)	\(18\)
norman	\(\frac {\ln \left (x \ln \left (2\right )\right ) {\mathrm e}^{-20}}{6 x^{2} \left (2+x \right )}\)	\(20\)
parallelrisch	\(\frac {\ln \left (x \ln \left (2\right )\right ) {\mathrm e}^{-20}}{6 x^{2} \left (2+x \right )}\)	\(20\)
derivativedivides	\(\frac {{\mathrm e}^{-20} \left (\frac {\ln \left (x \ln \left (2\right )\right ) \ln \left (2\right )}{8}-\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{4 x}-\frac {\ln \left (2\right )^{2} \ln \left (x \ln \left (2\right )\right ) x}{8 \left (2 \ln \left (2\right )+x \ln \left (2\right )\right )}+\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{2 x^{2}}\right )}{6 \ln \left (2\right )}\)	\(68\)
default	\(\frac {{\mathrm e}^{-20} \left (\frac {\ln \left (x \ln \left (2\right )\right ) \ln \left (2\right )}{8}-\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{4 x}-\frac {\ln \left (2\right )^{2} \ln \left (x \ln \left (2\right )\right ) x}{8 \left (2 \ln \left (2\right )+x \ln \left (2\right )\right )}+\frac {\ln \left (2\right ) \ln \left (x \ln \left (2\right )\right )}{2 x^{2}}\right )}{6 \ln \left (2\right )}\)	\(68\)
parts	\(\frac {{\mathrm e}^{-20} \left (-\frac {1}{4 x^{2}}+\frac {1}{4 x}+\frac {\ln \left (x \right )}{8}-\frac {\ln \left (2+x \right )}{8}\right )}{6}+\frac {{\mathrm e}^{-20} \ln \left (2 \ln \left (2\right )+x \ln \left (2\right )\right )}{48}-\frac {{\mathrm e}^{-20} \ln \left (2\right ) \ln \left (x \ln \left (2\right )\right ) x}{48 \left (2 \ln \left (2\right )+x \ln \left (2\right )\right )}+\frac {{\mathrm e}^{-20} \ln \left (x \ln \left (2\right )\right )}{12 x^{2}}+\frac {{\mathrm e}^{-20}}{24 x^{2}}-\frac {{\mathrm e}^{-20} \ln \left (x \ln \left (2\right )\right )}{24 x}-\frac {{\mathrm e}^{-20}}{24 x}\)	\(116\)

[In]

int(((-3*x-4)*ln(x*ln(2))+2+x)/(6*x^5+24*x^4+24*x^3)/exp(5)^4,x,method=_RETURNVERBOSE)

[Out]

1/6*ln(x*ln(2))/x^2/(2+x)*exp(-20)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {e^{\left (-20\right )} \log \left (x \log \left (2\right )\right )}{6 \, {\left (x^{3} + 2 \, x^{2}\right )}} \]

[In]

integrate(((-3*x-4)*log(x*log(2))+2+x)/(6*x^5+24*x^4+24*x^3)/exp(5)^4,x, algorithm="fricas")

[Out]

1/6*e^(-20)*log(x*log(2))/(x^3 + 2*x^2)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {\log {\left (x \log {\left (2 \right )} \right )}}{6 x^{3} e^{20} + 12 x^{2} e^{20}} \]

[In]

integrate(((-3*x-4)*ln(x*ln(2))+2+x)/(6*x**5+24*x**4+24*x**3)/exp(5)**4,x)

[Out]

log(x*log(2))/(6*x**3*exp(20) + 12*x**2*exp(20))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (19) = 38\).

Time = 0.33 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.86 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {1}{48} \, {\left (\frac {2 \, {\left (3 \, x^{2} + 3 \, x - 2\right )}}{x^{3} + 2 \, x^{2}} - \frac {2 \, x^{2} + {\left (x^{3} + 2 \, x^{2} - 8\right )} \log \left (x\right ) + 2 \, x - 8 \, \log \left (\log \left (2\right )\right ) - 4}{x^{3} + 2 \, x^{2}} - \frac {4 \, {\left (x + 1\right )}}{x^{2} + 2 \, x} + \log \left (x\right )\right )} e^{\left (-20\right )} \]

[In]

integrate(((-3*x-4)*log(x*log(2))+2+x)/(6*x^5+24*x^4+24*x^3)/exp(5)^4,x, algorithm="maxima")

[Out]

1/48*(2*(3*x^2 + 3*x - 2)/(x^3 + 2*x^2) - (2*x^2 + (x^3 + 2*x^2 - 8)*log(x) + 2*x - 8*log(log(2)) - 4)/(x^3 +
2*x^2) - 4*(x + 1)/(x^2 + 2*x) + log(x))*e^(-20)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {1}{24} \, {\left (\frac {1}{x + 2} - \frac {x - 2}{x^{2}}\right )} e^{\left (-20\right )} \log \left (x \log \left (2\right )\right ) \]

[In]

integrate(((-3*x-4)*log(x*log(2))+2+x)/(6*x^5+24*x^4+24*x^3)/exp(5)^4,x, algorithm="giac")

[Out]

1/24*(1/(x + 2) - (x - 2)/x^2)*e^(-20)*log(x*log(2))

Mupad [B] (verification not implemented)

Time = 13.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {2+x+(-4-3 x) \log (x \log (2))}{e^{20} \left (24 x^3+24 x^4+6 x^5\right )} \, dx=\frac {\ln \left (\ln \left (2\right )\right )+\ln \left (x\right )}{6\,{\mathrm {e}}^{20}\,x^3+12\,{\mathrm {e}}^{20}\,x^2} \]

[In]

int((exp(-20)*(x - log(x*log(2))*(3*x + 4) + 2))/(24*x^3 + 24*x^4 + 6*x^5),x)

[Out]

(log(log(2)) + log(x))/(12*x^2*exp(20) + 6*x^3*exp(20))