\(\int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} (4 x^2+6 x^3+(2 x^2+3 x^3) \log (x)) \log ^2(2 x+x \log (x)))}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx\) [10167]

Optimal result

Integrand size = 120, antiderivative size = 24 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=-25+5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x \left (x+x^2\right ) \]

[Out]

x*(x^2+x)+5/exp(5-1/ln((ln(x)+2)*x))-25

Rubi [A] (verified)

Time = 1.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2641, 6874, 6820, 45, 6838} \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^3+x^2+5 e^{\frac {1}{\log (x (\log (x)+2))}-5} \]

[In]

Int[(-15 - 5*Log[x] + E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(4*x^2 + 6*x^3 + (2*x^2 + 3*x^3)*Lo
g[x])*Log[2*x + x*Log[x]]^2)/(E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(2*x + x*Log[x])*Log[2*x +
x*Log[x]]^2),x]

[Out]

5*E^(-5 + Log[x*(2 + Log[x])]^(-1)) + x^2 + x^3

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2641

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) \left (-15-5 \log (x)+\exp \left (\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx \\ & = \int \left (\exp \left (5-\frac {1}{\log (x (2+\log (x)))}-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) x (2+3 x)+\frac {5 \exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))}\right ) \, dx \\ & = 5 \int \frac {\exp \left (-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx+\int \exp \left (5-\frac {1}{\log (x (2+\log (x)))}-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}\right ) x (2+3 x) \, dx \\ & = 5 \int \frac {e^{-5+\frac {1}{\log (x (2+\log (x)))}} (-3-\log (x))}{x (2+\log (x)) \log ^2(2 x+x \log (x))} \, dx+\int x (2+3 x) \, dx \\ & = 5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+\int \left (2 x+3 x^2\right ) \, dx \\ & = 5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x^2+x^3 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\frac {5 e^{\frac {1}{\log (x (2+\log (x)))}}+e^5 x^2 (1+x)}{e^5} \]

[In]

Integrate[(-15 - 5*Log[x] + E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(4*x^2 + 6*x^3 + (2*x^2 + 3*x
^3)*Log[x])*Log[2*x + x*Log[x]]^2)/(E^((-1 + 5*Log[2*x + x*Log[x]])/Log[2*x + x*Log[x]])*(2*x + x*Log[x])*Log[
2*x + x*Log[x]]^2),x]

[Out]

(5*E^Log[x*(2 + Log[x])]^(-1) + E^5*x^2*(1 + x))/E^5

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(113\) vs. \(2(27)=54\).

Time = 7.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 4.75

method	result	size
parallelrisch	\(\frac {\left (4 \,{\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}} x^{3} \ln \left (\left (\ln \left (x \right )+2\right ) x \right )+4 \ln \left (\left (\ln \left (x \right )+2\right ) x \right ) {\mathrm e}^{\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}} x^{2}+20 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )\right ) {\mathrm e}^{-\frac {5 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )-1}{\ln \left (\left (\ln \left (x \right )+2\right ) x \right )}}}{4 \ln \left (\left (\ln \left (x \right )+2\right ) x \right )}\)	\(114\)
risch	\(x^{3}+x^{2}+5 \,{\mathrm e}^{-\frac {-5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+5 i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-5 i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+10 \ln \left (x \right )+10 \ln \left (\ln \left (x \right )+2\right )-2}{-i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{3}+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right )^{2} \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )-i \pi \,\operatorname {csgn}\left (i x \left (\ln \left (x \right )+2\right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\ln \left (x \right )+2\right )\right )+2 \ln \left (x \right )+2 \ln \left (\ln \left (x \right )+2\right )}}\)	\(208\)

[In]

int((((3*x^3+2*x^2)*ln(x)+6*x^3+4*x^2)*ln(x*ln(x)+2*x)^2*exp((5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x))-5*ln(x)-15
)/(x*ln(x)+2*x)/ln(x*ln(x)+2*x)^2/exp((5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x)),x,method=_RETURNVERBOSE)

[Out]

1/4*(4*exp((5*ln((ln(x)+2)*x)-1)/ln((ln(x)+2)*x))*x^3*ln((ln(x)+2)*x)+4*ln((ln(x)+2)*x)*exp((5*ln((ln(x)+2)*x)
-1)/ln((ln(x)+2)*x))*x^2+20*ln((ln(x)+2)*x))/ln((ln(x)+2)*x)/exp((5*ln((ln(x)+2)*x)-1)/ln((ln(x)+2)*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (23) = 46\).

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx={\left ({\left (x^{3} + x^{2}\right )} e^{\left (\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} + 5\right )} e^{\left (-\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} \]

[In]

integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x
))-5*log(x)-15)/(x*log(x)+2*x)/log(x*log(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm
="fricas")

[Out]

((x^3 + x^2)*e^((5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x)) + 5)*e^(-(5*log(x*log(x) + 2*x) - 1)/log(x*lo
g(x) + 2*x))

Sympy [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + 5 e^{- \frac {5 \log {\left (x \log {\left (x \right )} + 2 x \right )} - 1}{\log {\left (x \log {\left (x \right )} + 2 x \right )}}} \]

[In]

integrate((((3*x**3+2*x**2)*ln(x)+6*x**3+4*x**2)*ln(x*ln(x)+2*x)**2*exp((5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x))
-5*ln(x)-15)/(x*ln(x)+2*x)/ln(x*ln(x)+2*x)**2/exp((5*ln(x*ln(x)+2*x)-1)/ln(x*ln(x)+2*x)),x)

[Out]

x**3 + x**2 + 5*exp(-(5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (23) = 46\).

Time = 0.40 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^{3} + x^{2} + \frac {5 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )} \log \left (x\right )}{e^{5} \log \left (x\right ) + 3 \, e^{5}} + \frac {15 \, e^{\left (\frac {1}{\log \left (x\right ) + \log \left (\log \left (x\right ) + 2\right )}\right )}}{e^{5} \log \left (x\right ) + 3 \, e^{5}} \]

[In]

integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x
))-5*log(x)-15)/(x*log(x)+2*x)/log(x*log(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm
="maxima")

[Out]

x^3 + x^2 + 5*e^(1/(log(x) + log(log(x) + 2)))*log(x)/(e^5*log(x) + 3*e^5) + 15*e^(1/(log(x) + log(log(x) + 2)
))/(e^5*log(x) + 3*e^5)

Giac [F]

\[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=\int { \frac {{\left ({\left (6 \, x^{3} + 4 \, x^{2} + {\left (3 \, x^{3} + 2 \, x^{2}\right )} \log \left (x\right )\right )} e^{\left (\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )} \log \left (x \log \left (x\right ) + 2 \, x\right )^{2} - 5 \, \log \left (x\right ) - 15\right )} e^{\left (-\frac {5 \, \log \left (x \log \left (x\right ) + 2 \, x\right ) - 1}{\log \left (x \log \left (x\right ) + 2 \, x\right )}\right )}}{{\left (x \log \left (x\right ) + 2 \, x\right )} \log \left (x \log \left (x\right ) + 2 \, x\right )^{2}} \,d x } \]

[In]

integrate((((3*x^3+2*x^2)*log(x)+6*x^3+4*x^2)*log(x*log(x)+2*x)^2*exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x
))-5*log(x)-15)/(x*log(x)+2*x)/log(x*log(x)+2*x)^2/exp((5*log(x*log(x)+2*x)-1)/log(x*log(x)+2*x)),x, algorithm
="giac")

[Out]

integrate(((6*x^3 + 4*x^2 + (3*x^3 + 2*x^2)*log(x))*e^((5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x))*log(x*
log(x) + 2*x)^2 - 5*log(x) - 15)*e^(-(5*log(x*log(x) + 2*x) - 1)/log(x*log(x) + 2*x))/((x*log(x) + 2*x)*log(x*
log(x) + 2*x)^2), x)

Mupad [B] (verification not implemented)

Time = 14.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (-15-5 \log (x)+e^{\frac {-1+5 \log (2 x+x \log (x))}{\log (2 x+x \log (x))}} \left (4 x^2+6 x^3+\left (2 x^2+3 x^3\right ) \log (x)\right ) \log ^2(2 x+x \log (x))\right )}{(2 x+x \log (x)) \log ^2(2 x+x \log (x))} \, dx=x^2+x^3+5\,{\mathrm {e}}^{\frac {1}{\ln \left (2\,x+x\,\ln \left (x\right )\right )}}\,{\mathrm {e}}^{-5} \]

[In]

int(-(exp(-(5*log(2*x + x*log(x)) - 1)/log(2*x + x*log(x)))*(5*log(x) - log(2*x + x*log(x))^2*exp((5*log(2*x +
x*log(x)) - 1)/log(2*x + x*log(x)))*(log(x)*(2*x^2 + 3*x^3) + 4*x^2 + 6*x^3) + 15))/(log(2*x + x*log(x))^2*(2
*x + x*log(x))),x)

[Out]

x^2 + x^3 + 5*exp(1/log(2*x + x*log(x)))*exp(-5)