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

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

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Rubi [A]  time = 1.36, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 116, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6688, 6708, 31} \begin {gather*} -\log \left (5-\log (2 x) \log \left (x+e^{(x+2)^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6688

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

Rule 6708

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 {gather*} \begin {aligned} \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\\ &=\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.58, size = 24, normalized size = 1.00 \begin {gather*} -\log \left (5-\log (2 x) \log \left (e^{4+4 x+x^2}+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.53, size = 37, normalized size = 1.54 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.31, size = 22, normalized size = 0.92 \begin {gather*} -\log \left (\log \left (2 \, x\right ) \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 30, normalized size = 1.25

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

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.50, size = 40, normalized size = 1.67 \begin {gather*} -\log \left (\frac {{\left (\log \relax (2) + \log \relax (x)\right )} \log \left (x + e^{\left (x^{2} + 4 \, x + 4\right )}\right ) - 5}{\log \relax (2) + \log \relax (x)}\right ) - \log \left (\log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 7.88, size = 34, normalized size = 1.42 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.98, size = 29, normalized size = 1.21 \begin {gather*} - \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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