∫ 4+x−2x log(−5x−5e5x+(2x+2e5x) log(2) 2 log(2) ) ___________________________________________________

Optimal. Leaf size=25 \[ \frac {\log \left (\frac {1}{2} \left (1+e^5\right ) x \left (2-\frac {5}{\log (2)}\right )\right )}{(4+x)^2} \]

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Rubi [A] time = 0.22, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6688, 6742, 44, 2319} \begin {gather*} \frac {\log \left (-\frac {\left (1+e^5\right ) x (5-\log (4))}{\log (4)}\right )}{(x+4)^2} \end {gather*}

Int[(4 + x - 2*x*Log[(-5*x - 5*E^5*x + (2*x + 2*E^5*x)*Log[2])/(2*Log[2])])/(64*x + 48*x^2 + 12*x^3 + x^4),x]

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] && L

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1

)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^

(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers

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

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Mathematica [A] time = 0.06, size = 22, normalized size = 0.88 \begin {gather*} \frac {\log \left (\frac {\left (1+e^5\right ) x (-5+\log (4))}{\log (4)}\right )}{(4+x)^2} \end {gather*}

Integrate[(4 + x - 2*x*Log[(-5*x - 5*E^5*x + (2*x + 2*E^5*x)*Log[2])/(2*Log[2])])/(64*x + 48*x^2 + 12*x^3 + x^

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fricas [A] time = 0.69, size = 37, normalized size = 1.48 \begin {gather*} \frac {\log \left (-\frac {5 \, x e^{5} - 2 \, {\left (x e^{5} + x\right )} \log \relax (2) + 5 \, x}{2 \, \log \relax (2)}\right )}{x^{2} + 8 \, x + 16} \end {gather*}

integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4+x)/(x^4+12*x^3+48*x^2+64*x),x, algo

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giac [A] time = 0.18, size = 42, normalized size = 1.68 \begin {gather*} -\frac {\log \relax (2) - \log \left (2 \, x e^{5} \log \relax (2) - 5 \, x e^{5} + 2 \, x \log \relax (2) - 5 \, x\right ) + \log \left (\log \relax (2)\right )}{x^{2} + 8 \, x + 16} \end {gather*}

integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4+x)/(x^4+12*x^3+48*x^2+64*x),x, algo

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method	result	size

norman	\(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \relax (2)-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \relax (2)}\right )}{\left (4+x \right )^{2}}\)	\(35\)
risch	\(\frac {\ln \left (\frac {\left (2 x \,{\mathrm e}^{5}+2 x \right ) \ln \relax (2)-5 x \,{\mathrm e}^{5}-5 x}{2 \ln \relax (2)}\right )}{x^{2}+8 x +16}\)	\(40\)
derivativedivides	\(\text {Expression too large to display}\)	\(150067\)
default	\(\text {Expression too large to display}\)	\(150067\)

int((-2*x*ln(1/2*((2*x*exp(5)+2*x)*ln(2)-5*x*exp(5)-5*x)/ln(2))+4+x)/(x^4+12*x^3+48*x^2+64*x),x,method=_RETURN

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maxima [B] time = 0.37, size = 176, normalized size = 7.04 \begin {gather*} -\frac {1}{16} \, {\left (\frac {\log \relax (2) \log \left (x + 4\right )}{2 \, {\left (e^{5} + 1\right )} \log \relax (2) - 5 \, e^{5} - 5} - \frac {\log \relax (2) \log \relax (x)}{2 \, {\left (e^{5} + 1\right )} \log \relax (2) - 5 \, e^{5} - 5} - \frac {4 \, \log \relax (2)}{{\left (2 \, {\left (e^{5} + 1\right )} \log \relax (2) - 5 \, e^{5} - 5\right )} x + 8 \, {\left (e^{5} + 1\right )} \log \relax (2) - 20 \, e^{5} - 20}\right )} {\left (\frac {5 \, e^{5}}{\log \relax (2)} + \frac {5}{\log \relax (2)} - 2 \, e^{5} - 2\right )} + \frac {x + 6}{4 \, {\left (x^{2} + 8 \, x + 16\right )}} + \frac {\log \left (x e^{5} + x - \frac {5 \, x e^{5}}{2 \, \log \relax (2)} - \frac {5 \, x}{2 \, \log \relax (2)}\right )}{x^{2} + 8 \, x + 16} - \frac {1}{2 \, {\left (x^{2} + 8 \, x + 16\right )}} - \frac {1}{16} \, \log \left (x + 4\right ) + \frac {1}{16} \, \log \relax (x) \end {gather*}

integrate((-2*x*log(1/2*((2*x*exp(5)+2*x)*log(2)-5*x*exp(5)-5*x)/log(2))+4+x)/(x^4+12*x^3+48*x^2+64*x),x, algo

-1/16*(log(2)*log(x + 4)/(2*(e^5 + 1)*log(2) - 5*e^5 - 5) - log(2)*log(x)/(2*(e^5 + 1)*log(2) - 5*e^5 - 5) - 4

*log(2)/((2*(e^5 + 1)*log(2) - 5*e^5 - 5)*x + 8*(e^5 + 1)*log(2) - 20*e^5 - 20))*(5*e^5/log(2) + 5/log(2) - 2*

e^5 - 2) + 1/4*(x + 6)/(x^2 + 8*x + 16) + log(x*e^5 + x - 5/2*x*e^5/log(2) - 5/2*x/log(2))/(x^2 + 8*x + 16) -

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mupad [B] time = 6.31, size = 49, normalized size = 1.96 \begin {gather*} \frac {x^2\,\left (\ln \left (\frac {\ln \relax (2)\,\left (2\,x+2\,x\,{\mathrm {e}}^5\right )}{2}-\frac {5\,x\,{\mathrm {e}}^5}{2}-\frac {5\,x}{2}\right )-\ln \left (\ln \relax (2)\right )\right )}{x^4+8\,x^3+16\,x^2} \end {gather*}

int((x - 2*x*log(-((5*x)/2 + (5*x*exp(5))/2 - (log(2)*(2*x + 2*x*exp(5)))/2)/log(2)) + 4)/(64*x + 48*x^2 + 12*

(x^2*(log((log(2)*(2*x + 2*x*exp(5)))/2 - (5*x*exp(5))/2 - (5*x)/2) - log(log(2))))/(16*x^2 + 8*x^3 + x^4)

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sympy [A] time = 0.16, size = 41, normalized size = 1.64 \begin {gather*} \frac {\log {\left (\frac {- \frac {5 x e^{5}}{2} - \frac {5 x}{2} + \frac {\left (2 x + 2 x e^{5}\right ) \log {\relax (2 )}}{2}}{\log {\relax (2 )}} \right )}}{x^{2} + 8 x + 16} \end {gather*}

integrate((-2*x*ln(1/2*((2*x*exp(5)+2*x)*ln(2)-5*x*exp(5)-5*x)/ln(2))+4+x)/(x**4+12*x**3+48*x**2+64*x),x)

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3.78.97 \(\int \frac {4+x-2 x \log (\frac {-5 x-5 e^5 x+(2 x+2 e^5 x) \log (2)}{2 \log (2)})}{64 x+48 x^2+12 x^3+x^4} \, dx\)