\(\int \frac {(-x+e^{4+4 x+x^2} (-4 x-2 x^2)) \log (2 x)+(-e^{4+4 x+x^2}-x) \log (e^{4+4 x+x^2}+x)}{-5 e^{4+4 x+x^2} x-5 x^2+(e^{4+4 x+x^2} x+x^2) \log (2 x) \log (e^{4+4 x+x^2}+x)} \, dx\) [9014]

Optimal result

Integrand size = 116, antiderivative size = 24 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=\log (5)-\log \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \]

[Out]

ln(5)-ln(5-ln(x+exp((2+x)^2))*ln(2*x))

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6820, 6840, 31} \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right ) \]

[In]

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

[Out]

-Log[5 - Log[2*x]*Log[E^(2 + x)^2 + x]]

Rule 31

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

Rule 6820

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

Rule 6840

Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])
]}, Dist[c, Subst[Int[(a + b*x^p)^m, x], x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (1+2 e^{(2+x)^2} (2+x)\right ) \log (2 x)+\left (e^{(2+x)^2}+x\right ) \log \left (e^{(2+x)^2}+x\right )}{x \left (e^{(2+x)^2}+x\right ) \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right )} \, dx \\ & = \text {Subst}\left (\int \frac {1}{5-x} \, dx,x,\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \\ & = -\log \left (5-\log (2 x) \log \left (e^{(2+x)^2}+x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (5-\log (2 x) \log \left (e^{4+4 x+x^2}+x\right )\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(-\ln \left (\ln \left (2 x \right ) \ln \left ({\mathrm e}^{x^{2}+4 x +4}+x \right )-5\right )\)	\(23\)
risch	\(-\ln \left (\ln \left (2 x \right )\right )-\ln \left (\ln \left (x +{\mathrm e}^{\left (2+x \right )^{2}}\right )-\frac {5}{\ln \left (2 x \right )}\right )\)	\(30\)

[In]

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

[Out]

-ln(ln(2*x)*ln(exp(x^2+4*x+4)+x)-5)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2 \, x\right )}\right ) - \log \left (\log \left (2 \, x\right )\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=- \log {\left (\log {\left (x + e^{x^{2} + 4 x + 4} \right )} - \frac {5}{\log {\left (2 x \right )}} \right )} - \log {\left (\log {\left (2 x \right )} \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\frac {{\left (\log \left (2\right ) + \log \left (x\right )\right )} \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \left (2\right ) + \log \left (x\right )}\right ) - \log \left (\log \left (2\right ) + \log \left (x\right )\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\log \left (\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {\left (-x+e^{4+4 x+x^2} \left (-4 x-2 x^2\right )\right ) \log (2 x)+\left (-e^{4+4 x+x^2}-x\right ) \log \left (e^{4+4 x+x^2}+x\right )}{-5 e^{4+4 x+x^2} x-5 x^2+\left (e^{4+4 x+x^2} x+x^2\right ) \log (2 x) \log \left (e^{4+4 x+x^2}+x\right )} \, dx=-\ln \left (\frac {\ln \left (2\,x\right )\,\ln \left (x+{\mathrm {e}}^{{\left (x+2\right )}^2}\right )-5}{\ln \left (2\,x\right )}\right )-\ln \left (\ln \left (2\,x\right )\right ) \]

[In]

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

[Out]

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