3.31.62 \(\int \frac {e^{\frac {2 x+x^2+x \log (x)+(4 x+2 x^2+2 x \log (x)) \log (\frac {1}{4 x+x \log (x)})}{\log (\frac {1}{4 x+x \log (x)})}} (50+25 x+(35+5 x) \log (x)+5 \log ^2(x)+(60+40 x+(35+10 x) \log (x)+5 \log ^2(x)) \log (\frac {1}{4 x+x \log (x)})+(120+80 x+(70+20 x) \log (x)+10 \log ^2(x)) \log ^2(\frac {1}{4 x+x \log (x)}))}{(4+\log (x)) \log ^2(\frac {1}{4 x+x \log (x)})} \, dx\)

Optimal. Leaf size=29 \[ 5 e^{(2+x+\log (x)) \left (2 x+\frac {x}{\log \left (\frac {1}{x (4+\log (x))}\right )}\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 6.61, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (\frac {2 x+x^2+x \log (x)+\left (4 x+2 x^2+2 x \log (x)\right ) \log \left (\frac {1}{4 x+x \log (x)}\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) \left (50+25 x+(35+5 x) \log (x)+5 \log ^2(x)+\left (60+40 x+(35+10 x) \log (x)+5 \log ^2(x)\right ) \log \left (\frac {1}{4 x+x \log (x)}\right )+\left (120+80 x+(70+20 x) \log (x)+10 \log ^2(x)\right ) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )\right )}{(4+\log (x)) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*x + x^2 + x*Log[x] + (4*x + 2*x^2 + 2*x*Log[x])*Log[(4*x + x*Log[x])^(-1)])/Log[(4*x + x*Log[x])^(-
1)])*(50 + 25*x + (35 + 5*x)*Log[x] + 5*Log[x]^2 + (60 + 40*x + (35 + 10*x)*Log[x] + 5*Log[x]^2)*Log[(4*x + x*
Log[x])^(-1)] + (120 + 80*x + (70 + 20*x)*Log[x] + 10*Log[x]^2)*Log[(4*x + x*Log[x])^(-1)]^2))/((4 + Log[x])*L
og[(4*x + x*Log[x])^(-1)]^2),x]

[Out]

30*Defer[Int][E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)]), x] + 20*
Defer[Int][E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])*x, x] + 10*D
efer[Int][E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])*Log[x], x] +
50*Defer[Int][E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])/((4 + Log
[x])*Log[(4*x + x*Log[x])^(-1)]^2), x] + 25*Defer[Int][(E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1
)]))/Log[(4*x + x*Log[x])^(-1)])*x)/((4 + Log[x])*Log[(4*x + x*Log[x])^(-1)]^2), x] + 35*Defer[Int][(E^((x*(2
+ x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])*Log[x])/((4 + Log[x])*Log[(4*x +
x*Log[x])^(-1)]^2), x] + 5*Defer[Int][(E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x +
x*Log[x])^(-1)])*x*Log[x])/((4 + Log[x])*Log[(4*x + x*Log[x])^(-1)]^2), x] + 5*Defer[Int][(E^((x*(2 + x + Log[
x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])*Log[x]^2)/((4 + Log[x])*Log[(4*x + x*Log[x
])^(-1)]^2), x] + 15*Defer[Int][E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x]
)^(-1)])/Log[(4*x + x*Log[x])^(-1)], x] + 10*Defer[Int][(E^((x*(2 + x + Log[x])*(1 + 2*Log[(4*x + x*Log[x])^(-
1)]))/Log[(4*x + x*Log[x])^(-1)])*x)/Log[(4*x + x*Log[x])^(-1)], x] + 5*Defer[Int][(E^((x*(2 + x + Log[x])*(1
+ 2*Log[(4*x + x*Log[x])^(-1)]))/Log[(4*x + x*Log[x])^(-1)])*Log[x])/Log[(4*x + x*Log[x])^(-1)], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) \left (50+25 x+(35+5 x) \log (x)+5 \log ^2(x)+\left (60+40 x+(35+10 x) \log (x)+5 \log ^2(x)\right ) \log \left (\frac {1}{4 x+x \log (x)}\right )+\left (120+80 x+(70+20 x) \log (x)+10 \log ^2(x)\right ) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )\right )}{(4+\log (x)) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )} \, dx\\ &=\int \left (10 \exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (3+2 x+\log (x))+\frac {5 \exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (5+\log (x)) (2+x+\log (x))}{(4+\log (x)) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )}+\frac {5 \exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (3+2 x+\log (x))}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) \, dx\\ &=5 \int \frac {\exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (5+\log (x)) (2+x+\log (x))}{(4+\log (x)) \log ^2\left (\frac {1}{4 x+x \log (x)}\right )} \, dx+5 \int \frac {\exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (3+2 x+\log (x))}{\log \left (\frac {1}{4 x+x \log (x)}\right )} \, dx+10 \int \exp \left (\frac {x (2+x+\log (x)) \left (1+2 \log \left (\frac {1}{4 x+x \log (x)}\right )\right )}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right ) (3+2 x+\log (x)) \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 35, normalized size = 1.21 \begin {gather*} 5 e^{x \left (4+2 x+\frac {2+x+\log (x)}{\log \left (\frac {1}{4 x+x \log (x)}\right )}\right )} x^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*x + x^2 + x*Log[x] + (4*x + 2*x^2 + 2*x*Log[x])*Log[(4*x + x*Log[x])^(-1)])/Log[(4*x + x*Log[
x])^(-1)])*(50 + 25*x + (35 + 5*x)*Log[x] + 5*Log[x]^2 + (60 + 40*x + (35 + 10*x)*Log[x] + 5*Log[x]^2)*Log[(4*
x + x*Log[x])^(-1)] + (120 + 80*x + (70 + 20*x)*Log[x] + 10*Log[x]^2)*Log[(4*x + x*Log[x])^(-1)]^2))/((4 + Log
[x])*Log[(4*x + x*Log[x])^(-1)]^2),x]

[Out]

5*E^(x*(4 + 2*x + (2 + x + Log[x])/Log[(4*x + x*Log[x])^(-1)]))*x^(2*x)

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 52, normalized size = 1.79 \begin {gather*} 5 \, e^{\left (\frac {x^{2} + x \log \relax (x) + 2 \, {\left (x^{2} + x \log \relax (x) + 2 \, x\right )} \log \left (\frac {1}{x \log \relax (x) + 4 \, x}\right ) + 2 \, x}{\log \left (\frac {1}{x \log \relax (x) + 4 \, x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*log(x)^2+(20*x+70)*log(x)+80*x+120)*log(1/(x*log(x)+4*x))^2+(5*log(x)^2+(10*x+35)*log(x)+40*x+6
0)*log(1/(x*log(x)+4*x))+5*log(x)^2+(5*x+35)*log(x)+25*x+50)*exp(((2*x*log(x)+2*x^2+4*x)*log(1/(x*log(x)+4*x))
+x*log(x)+x^2+2*x)/log(1/(x*log(x)+4*x)))/(log(x)+4)/log(1/(x*log(x)+4*x))^2,x, algorithm="fricas")

[Out]

5*e^((x^2 + x*log(x) + 2*(x^2 + x*log(x) + 2*x)*log(1/(x*log(x) + 4*x)) + 2*x)/log(1/(x*log(x) + 4*x)))

________________________________________________________________________________________

giac [B]  time = 67.65, size = 63, normalized size = 2.17 \begin {gather*} 5 \, e^{\left (2 \, x^{2} + 2 \, x \log \relax (x) + 4 \, x - \frac {x^{2}}{\log \left (x \log \relax (x) + 4 \, x\right )} - \frac {x \log \relax (x)}{\log \left (x \log \relax (x) + 4 \, x\right )} - \frac {2 \, x}{\log \left (x \log \relax (x) + 4 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*log(x)^2+(20*x+70)*log(x)+80*x+120)*log(1/(x*log(x)+4*x))^2+(5*log(x)^2+(10*x+35)*log(x)+40*x+6
0)*log(1/(x*log(x)+4*x))+5*log(x)^2+(5*x+35)*log(x)+25*x+50)*exp(((2*x*log(x)+2*x^2+4*x)*log(1/(x*log(x)+4*x))
+x*log(x)+x^2+2*x)/log(1/(x*log(x)+4*x)))/(log(x)+4)/log(1/(x*log(x)+4*x))^2,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.77, size = 255, normalized size = 8.79

method	result	size
risch	\(5 \,{\mathrm e}^{\frac {2 x \left (x +\ln \relax (x )+2\right ) \left (i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+4}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )+4}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{3}+2 \ln \relax (x )+2 \ln \left (\ln \relax (x )+4\right )-1\right )}{i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )+4}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )-i \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (x )+4}\right ) \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{2}+i \pi \mathrm {csgn}\left (\frac {i}{x \left (\ln \relax (x )+4\right )}\right )^{3}+2 \ln \relax (x )+2 \ln \left (\ln \relax (x )+4\right )}}\)	\(255\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*ln(x)^2+(20*x+70)*ln(x)+80*x+120)*ln(1/(x*ln(x)+4*x))^2+(5*ln(x)^2+(10*x+35)*ln(x)+40*x+60)*ln(1/(x*l
n(x)+4*x))+5*ln(x)^2+(5*x+35)*ln(x)+25*x+50)*exp(((2*x*ln(x)+2*x^2+4*x)*ln(1/(x*ln(x)+4*x))+x*ln(x)+x^2+2*x)/l
n(1/(x*ln(x)+4*x)))/(ln(x)+4)/ln(1/(x*ln(x)+4*x))^2,x,method=_RETURNVERBOSE)

[Out]

5*exp(2*x*(x+ln(x)+2)*(I*Pi*csgn(I/x)*csgn(I/(ln(x)+4))*csgn(I/x/(ln(x)+4))-I*Pi*csgn(I/x)*csgn(I/x/(ln(x)+4))
^2-I*Pi*csgn(I/(ln(x)+4))*csgn(I/x/(ln(x)+4))^2+I*Pi*csgn(I/x/(ln(x)+4))^3+2*ln(x)+2*ln(ln(x)+4)-1)/(I*Pi*csgn
(I/x)*csgn(I/(ln(x)+4))*csgn(I/x/(ln(x)+4))-I*Pi*csgn(I/x)*csgn(I/x/(ln(x)+4))^2-I*Pi*csgn(I/(ln(x)+4))*csgn(I
/x/(ln(x)+4))^2+I*Pi*csgn(I/x/(ln(x)+4))^3+2*ln(x)+2*ln(ln(x)+4)))

________________________________________________________________________________________

maxima [B]  time = 0.76, size = 60, normalized size = 2.07 \begin {gather*} 5 \, e^{\left (2 \, x^{2} + 2 \, x \log \relax (x) + 4 \, x - \frac {x^{2}}{\log \relax (x) + \log \left (\log \relax (x) + 4\right )} - \frac {x \log \relax (x)}{\log \relax (x) + \log \left (\log \relax (x) + 4\right )} - \frac {2 \, x}{\log \relax (x) + \log \left (\log \relax (x) + 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*log(x)^2+(20*x+70)*log(x)+80*x+120)*log(1/(x*log(x)+4*x))^2+(5*log(x)^2+(10*x+35)*log(x)+40*x+6
0)*log(1/(x*log(x)+4*x))+5*log(x)^2+(5*x+35)*log(x)+25*x+50)*exp(((2*x*log(x)+2*x^2+4*x)*log(1/(x*log(x)+4*x))
+x*log(x)+x^2+2*x)/log(1/(x*log(x)+4*x)))/(log(x)+4)/log(1/(x*log(x)+4*x))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.22, size = 69, normalized size = 2.38 \begin {gather*} 5\,x^{2\,x}\,x^{\frac {x}{\ln \left (\frac {1}{4\,x+x\,\ln \relax (x)}\right )}}\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^{\frac {2\,x}{\ln \left (\frac {1}{4\,x+x\,\ln \relax (x)}\right )}}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^{\frac {x^2}{\ln \left (\frac {1}{4\,x+x\,\ln \relax (x)}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x + log(1/(4*x + x*log(x)))*(4*x + 2*x*log(x) + 2*x^2) + x*log(x) + x^2)/log(1/(4*x + x*log(x))))*
(25*x + 5*log(x)^2 + log(1/(4*x + x*log(x)))*(40*x + 5*log(x)^2 + log(x)*(10*x + 35) + 60) + log(x)*(5*x + 35)
+ log(1/(4*x + x*log(x)))^2*(80*x + 10*log(x)^2 + log(x)*(20*x + 70) + 120) + 50))/(log(1/(4*x + x*log(x)))^2
*(log(x) + 4)),x)

[Out]

5*x^(2*x)*x^(x/log(1/(4*x + x*log(x))))*exp(4*x)*exp((2*x)/log(1/(4*x + x*log(x))))*exp(2*x^2)*exp(x^2/log(1/(
4*x + x*log(x))))

________________________________________________________________________________________

sympy [B]  time = 1.27, size = 53, normalized size = 1.83 \begin {gather*} 5 e^{\frac {x^{2} + x \log {\relax (x )} + 2 x + \left (2 x^{2} + 2 x \log {\relax (x )} + 4 x\right ) \log {\left (\frac {1}{x \log {\relax (x )} + 4 x} \right )}}{\log {\left (\frac {1}{x \log {\relax (x )} + 4 x} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*ln(x)**2+(20*x+70)*ln(x)+80*x+120)*ln(1/(x*ln(x)+4*x))**2+(5*ln(x)**2+(10*x+35)*ln(x)+40*x+60)*
ln(1/(x*ln(x)+4*x))+5*ln(x)**2+(5*x+35)*ln(x)+25*x+50)*exp(((2*x*ln(x)+2*x**2+4*x)*ln(1/(x*ln(x)+4*x))+x*ln(x)
+x**2+2*x)/ln(1/(x*ln(x)+4*x)))/(ln(x)+4)/ln(1/(x*ln(x)+4*x))**2,x)

[Out]

5*exp((x**2 + x*log(x) + 2*x + (2*x**2 + 2*x*log(x) + 4*x)*log(1/(x*log(x) + 4*x)))/log(1/(x*log(x) + 4*x)))

________________________________________________________________________________________