3.56.37 \(\int \frac {e^{-\frac {(2 x+x^2) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} (e^{\frac {(2 x+x^2) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}} (4+4 x+x^2) \log (x)+(-20 x-20 x^2-5 x^3) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x)))}{(4 x+4 x^2+x^3) \log (x)} \, dx\)

Optimal. Leaf size=29 \[ 5 e^{-x \log (4)-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}}+\log (x) \]

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Rubi [B]  time = 7.94, antiderivative size = 59, normalized size of antiderivative = 2.03, number of steps used = 6, number of rules used = 5, integrand size = 136, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {1594, 27, 6688, 6742, 2288} \begin {gather*} \frac {5\ 4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{x+2}} \left (x^3 \log (x)+4 x^2 \log (x)+4 x \log (x)\right )}{x (x+2)^2 \log (x)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(((2*x + x^2)*Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4 + 4*x + x^2)*Log[x] + (-20*x - 20*x^2 - 5*x^
3)*Log[4]*Log[x] + (-20 - 10*x)*Log[3]^2*Log[Log[x]] + 5*x*Log[3]^2*Log[x]*Log[Log[x]]^2)/(E^(((2*x + x^2)*Log
[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4*x + 4*x^2 + x^3)*Log[x]),x]

[Out]

Log[x] + (5*(4*x*Log[x] + 4*x^2*Log[x] + x^3*Log[x]))/(4^x*E^((Log[3]^2*Log[Log[x]]^2)/(2 + x))*x*(2 + x)^2*Lo
g[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 1594

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}\right ) \left (\exp \left (\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}\right ) \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{x \left (4+4 x+x^2\right ) \log (x)} \, dx\\ &=\int \frac {\exp \left (-\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}\right ) \left (\exp \left (\frac {\left (2 x+x^2\right ) \log (4)+\log ^2(3) \log ^2(\log (x))}{2+x}\right ) \left (4+4 x+x^2\right ) \log (x)+\left (-20 x-20 x^2-5 x^3\right ) \log (4) \log (x)+(-20-10 x) \log ^2(3) \log (\log (x))+5 x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{x (2+x)^2 \log (x)} \, dx\\ &=\int \frac {4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}} \left (-10 (2+x) \log ^2(3) \log (\log (x))+\log (x) \left ((2+x)^2 \left (4^x e^{\frac {\log ^2(3) \log ^2(\log (x))}{2+x}}-5 x \log (4)\right )+5 x \log ^2(3) \log ^2(\log (x))\right )\right )}{x (2+x)^2 \log (x)} \, dx\\ &=\int \left (\frac {1}{x}-\frac {5\ 4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4 x \log (4) \log (x)+4 x^2 \log (4) \log (x)+x^3 \log (4) \log (x)+4 \log ^2(3) \log (\log (x))+2 x \log ^2(3) \log (\log (x))-x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{x (2+x)^2 \log (x)}\right ) \, dx\\ &=\log (x)-5 \int \frac {4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4 x \log (4) \log (x)+4 x^2 \log (4) \log (x)+x^3 \log (4) \log (x)+4 \log ^2(3) \log (\log (x))+2 x \log ^2(3) \log (\log (x))-x \log ^2(3) \log (x) \log ^2(\log (x))\right )}{x (2+x)^2 \log (x)} \, dx\\ &=\log (x)+\frac {5\ 4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}} \left (4 x \log (x)+4 x^2 \log (x)+x^3 \log (x)\right )}{x (2+x)^2 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 28, normalized size = 0.97 \begin {gather*} 5\ 4^{-x} e^{-\frac {\log ^2(3) \log ^2(\log (x))}{2+x}}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(((2*x + x^2)*Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4 + 4*x + x^2)*Log[x] + (-20*x - 20*x^2
- 5*x^3)*Log[4]*Log[x] + (-20 - 10*x)*Log[3]^2*Log[Log[x]] + 5*x*Log[3]^2*Log[x]*Log[Log[x]]^2)/(E^(((2*x + x^
2)*Log[4] + Log[3]^2*Log[Log[x]]^2)/(2 + x))*(4*x + 4*x^2 + x^3)*Log[x]),x]

[Out]

5/(4^x*E^((Log[3]^2*Log[Log[x]]^2)/(2 + x))) + Log[x]

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fricas [B]  time = 0.53, size = 65, normalized size = 2.24 \begin {gather*} {\left (e^{\left (\frac {\log \relax (3)^{2} \log \left (\log \relax (x)\right )^{2} + 2 \, {\left (x^{2} + 2 \, x\right )} \log \relax (2)}{x + 2}\right )} \log \relax (x) + 5\right )} e^{\left (-\frac {\log \relax (3)^{2} \log \left (\log \relax (x)\right )^{2} + 2 \, {\left (x^{2} + 2 \, x\right )} \log \relax (2)}{x + 2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x))+5*x*log(3)^2*log(x)*log(l
og(x))^2+(-10*x-20)*log(3)^2*log(log(x))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log
(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 1.05, size = 45, normalized size = 1.55 \begin {gather*} 5 \, e^{\left (-\frac {\log \relax (3)^{2} \log \left (\log \relax (x)\right )^{2}}{x + 2} - \frac {2 \, x^{2} \log \relax (2)}{x + 2} - \frac {4 \, x \log \relax (2)}{x + 2}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x))+5*x*log(3)^2*log(x)*log(l
og(x))^2+(-10*x-20)*log(3)^2*log(log(x))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log
(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.18, size = 37, normalized size = 1.28

method	result	size
risch	\(\ln \relax (x )+5 \,{\mathrm e}^{-\frac {\ln \relax (3)^{2} \ln \left (\ln \relax (x )\right )^{2}+2 x^{2} \ln \relax (2)+4 x \ln \relax (2)}{2+x}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+4*x+4)*ln(x)*exp((ln(3)^2*ln(ln(x))^2+2*(x^2+2*x)*ln(2))/(2+x))+5*x*ln(3)^2*ln(x)*ln(ln(x))^2+(-10*x
-20)*ln(3)^2*ln(ln(x))+2*(-5*x^3-20*x^2-20*x)*ln(2)*ln(x))/(x^3+4*x^2+4*x)/ln(x)/exp((ln(3)^2*ln(ln(x))^2+2*(x
^2+2*x)*ln(2))/(2+x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.74, size = 28, normalized size = 0.97 \begin {gather*} 5 \, e^{\left (-\frac {\log \relax (3)^{2} \log \left (\log \relax (x)\right )^{2}}{x + 2} - 2 \, x \log \relax (2)\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+4*x+4)*log(x)*exp((log(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x))+5*x*log(3)^2*log(x)*log(l
og(x))^2+(-10*x-20)*log(3)^2*log(log(x))+2*(-5*x^3-20*x^2-20*x)*log(2)*log(x))/(x^3+4*x^2+4*x)/log(x)/exp((log
(3)^2*log(log(x))^2+2*(x^2+2*x)*log(2))/(2+x)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.94, size = 48, normalized size = 1.66 \begin {gather*} \ln \relax (x)+\frac {5\,{\mathrm {e}}^{-\frac {{\ln \left (\ln \relax (x)\right )}^2\,{\ln \relax (3)}^2}{x+2}}}{2^{\frac {4\,x}{x+2}}\,2^{\frac {2\,x^2}{x+2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*log(2)*(2*x + x^2) + log(log(x))^2*log(3)^2)/(x + 2))*(2*log(2)*log(x)*(20*x + 20*x^2 + 5*x^3) -
exp((2*log(2)*(2*x + x^2) + log(log(x))^2*log(3)^2)/(x + 2))*log(x)*(4*x + x^2 + 4) + log(log(x))*log(3)^2*(1
0*x + 20) - 5*x*log(log(x))^2*log(3)^2*log(x)))/(log(x)*(4*x + 4*x^2 + x^3)),x)

[Out]

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

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sympy [A]  time = 1.00, size = 32, normalized size = 1.10 \begin {gather*} \log {\relax (x )} + 5 e^{- \frac {\left (2 x^{2} + 4 x\right ) \log {\relax (2 )} + \log {\relax (3 )}^{2} \log {\left (\log {\relax (x )} \right )}^{2}}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+4*x+4)*ln(x)*exp((ln(3)**2*ln(ln(x))**2+2*(x**2+2*x)*ln(2))/(2+x))+5*x*ln(3)**2*ln(x)*ln(ln(x
))**2+(-10*x-20)*ln(3)**2*ln(ln(x))+2*(-5*x**3-20*x**2-20*x)*ln(2)*ln(x))/(x**3+4*x**2+4*x)/ln(x)/exp((ln(3)**
2*ln(ln(x))**2+2*(x**2+2*x)*ln(2))/(2+x)),x)

[Out]

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

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