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

Optimal. Leaf size=24 \[ -25+5 e^{-5+\frac {1}{\log (x (2+\log (x)))}}+x \left (x+x^2\right ) \]

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Rubi [A]  time = 3.04, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 5, integrand size = 120, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {2561, 6742, 6688, 43, 6706} \begin {gather*} x^3+x^2+5 e^{\frac {1}{\log (x (\log (x)+2))}-5} \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2561

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 6688

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

Rule 6706

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 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 {-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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 28, normalized size = 1.17 \begin {gather*} \frac {5 e^{\frac {1}{\log (x (2+\log (x)))}}+e^5 x^2 (1+x)}{e^5} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [B]  time = 0.58, size = 64, normalized size = 2.67 \begin {gather*} {\left ({\left (x^{3} + x^{2}\right )} e^{\left (\frac {5 \, \log \left (x \log \relax (x) + 2 \, x\right ) - 1}{\log \left (x \log \relax (x) + 2 \, x\right )}\right )} + 5\right )} e^{\left (-\frac {5 \, \log \left (x \log \relax (x) + 2 \, x\right ) - 1}{\log \left (x \log \relax (x) + 2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left ({\left (6 \, x^{3} + 4 \, x^{2} + {\left (3 \, x^{3} + 2 \, x^{2}\right )} \log \relax (x)\right )} e^{\left (\frac {5 \, \log \left (x \log \relax (x) + 2 \, x\right ) - 1}{\log \left (x \log \relax (x) + 2 \, x\right )}\right )} \log \left (x \log \relax (x) + 2 \, x\right )^{2} - 5 \, \log \relax (x) - 15\right )} e^{\left (-\frac {5 \, \log \left (x \log \relax (x) + 2 \, x\right ) - 1}{\log \left (x \log \relax (x) + 2 \, x\right )}\right )}}{{\left (x \log \relax (x) + 2 \, x\right )} \log \left (x \log \relax (x) + 2 \, x\right )^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 0.44, size = 208, normalized size = 8.67

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

Verification of antiderivative is not currently implemented for this CAS.

[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]

x^3+x^2+5*exp(-(-5*I*Pi*csgn(I*x*(ln(x)+2))^3+5*I*Pi*csgn(I*x*(ln(x)+2))^2*csgn(I*x)+5*I*Pi*csgn(I*x*(ln(x)+2)
)^2*csgn(I*(ln(x)+2))-5*I*Pi*csgn(I*x*(ln(x)+2))*csgn(I*x)*csgn(I*(ln(x)+2))+10*ln(x)+10*ln(ln(x)+2)-2)/(-I*Pi
*csgn(I*x*(ln(x)+2))^3+I*Pi*csgn(I*x*(ln(x)+2))^2*csgn(I*x)+I*Pi*csgn(I*x*(ln(x)+2))^2*csgn(I*(ln(x)+2))-I*Pi*
csgn(I*x*(ln(x)+2))*csgn(I*x)*csgn(I*(ln(x)+2))+2*ln(x)+2*ln(ln(x)+2)))

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maxima [B]  time = 0.57, size = 59, normalized size = 2.46 \begin {gather*} x^{3} + x^{2} + \frac {5 \, e^{\left (\frac {1}{\log \relax (x) + \log \left (\log \relax (x) + 2\right )}\right )} \log \relax (x)}{e^{5} \log \relax (x) + 3 \, e^{5}} + \frac {15 \, e^{\left (\frac {1}{\log \relax (x) + \log \left (\log \relax (x) + 2\right )}\right )}}{e^{5} \log \relax (x) + 3 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 7.44, size = 23, normalized size = 0.96 \begin {gather*} x^2+x^3+5\,{\mathrm {e}}^{\frac {1}{\ln \left (2\,x+x\,\ln \relax (x)\right )}}\,{\mathrm {e}}^{-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.60, size = 32, normalized size = 1.33 \begin {gather*} x^{3} + x^{2} + 5 e^{- \frac {5 \log {\left (x \log {\relax (x )} + 2 x \right )} - 1}{\log {\left (x \log {\relax (x )} + 2 x \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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