3.62.44 \(\int \frac {-2+8 e^{4-60 x+225 x^2}-12 e^{8-120 x+450 x^2}+8 e^{12-180 x+675 x^2}-2 e^{16-240 x+900 x^2}+(10+e^{12-180 x+675 x^2} (720 x-5400 x^2)+e^{4-60 x+225 x^2} (-20+240 x-1800 x^2)+e^{16-240 x+900 x^2} (-240 x+1800 x^2)+e^{8-120 x+450 x^2} (10-720 x+5400 x^2)) \log (x)+(e^{8-120 x+450 x^2} (1200 x-9000 x^2)+e^{4-60 x+225 x^2} (-1200 x+9000 x^2)) \log ^2(x)-\log ^3(x)}{x \log ^3(x)} \, dx\)

Optimal. Leaf size=30 \[ \log \left (\frac {e^{\left (5-\frac {\left (-1+e^{(2-15 x)^2}\right )^2}{\log (x)}\right )^2}}{x}\right ) \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2 + 8*E^(4 - 60*x + 225*x^2) - 12*E^(8 - 120*x + 450*x^2) + 8*E^(12 - 180*x + 675*x^2) - 2*E^(16 - 240*x
+ 900*x^2) + (10 + E^(12 - 180*x + 675*x^2)*(720*x - 5400*x^2) + E^(4 - 60*x + 225*x^2)*(-20 + 240*x - 1800*x
^2) + E^(16 - 240*x + 900*x^2)*(-240*x + 1800*x^2) + E^(8 - 120*x + 450*x^2)*(10 - 720*x + 5400*x^2))*Log[x] +
(E^(8 - 120*x + 450*x^2)*(1200*x - 9000*x^2) + E^(4 - 60*x + 225*x^2)*(-1200*x + 9000*x^2))*Log[x]^2 - Log[x]
^3)/(x*Log[x]^3),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.38, size = 42, normalized size = 1.40 \begin {gather*} \frac {\left (-1+e^{(2-15 x)^2}\right )^4-10 \left (-1+e^{(2-15 x)^2}\right )^2 \log (x)-\log ^3(x)}{\log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 8*E^(4 - 60*x + 225*x^2) - 12*E^(8 - 120*x + 450*x^2) + 8*E^(12 - 180*x + 675*x^2) - 2*E^(16 -
240*x + 900*x^2) + (10 + E^(12 - 180*x + 675*x^2)*(720*x - 5400*x^2) + E^(4 - 60*x + 225*x^2)*(-20 + 240*x -
1800*x^2) + E^(16 - 240*x + 900*x^2)*(-240*x + 1800*x^2) + E^(8 - 120*x + 450*x^2)*(10 - 720*x + 5400*x^2))*Lo
g[x] + (E^(8 - 120*x + 450*x^2)*(1200*x - 9000*x^2) + E^(4 - 60*x + 225*x^2)*(-1200*x + 9000*x^2))*Log[x]^2 -
Log[x]^3)/(x*Log[x]^3),x]

[Out]

((-1 + E^(2 - 15*x)^2)^4 - 10*(-1 + E^(2 - 15*x)^2)^2*Log[x] - Log[x]^3)/Log[x]^2

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fricas [B]  time = 0.69, size = 94, normalized size = 3.13 \begin {gather*} -\frac {\log \relax (x)^{3} + 10 \, {\left (e^{\left (450 \, x^{2} - 120 \, x + 8\right )} - 2 \, e^{\left (225 \, x^{2} - 60 \, x + 4\right )} + 1\right )} \log \relax (x) - e^{\left (900 \, x^{2} - 240 \, x + 16\right )} + 4 \, e^{\left (675 \, x^{2} - 180 \, x + 12\right )} - 6 \, e^{\left (450 \, x^{2} - 120 \, x + 8\right )} + 4 \, e^{\left (225 \, x^{2} - 60 \, x + 4\right )} - 1}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+((-9000*x^2+1200*x)*exp(225*x^2-60*x+4)^2+(9000*x^2-1200*x)*exp(225*x^2-60*x+4))*log(x)^2
+((1800*x^2-240*x)*exp(225*x^2-60*x+4)^4+(-5400*x^2+720*x)*exp(225*x^2-60*x+4)^3+(5400*x^2-720*x+10)*exp(225*x
^2-60*x+4)^2+(-1800*x^2+240*x-20)*exp(225*x^2-60*x+4)+10)*log(x)-2*exp(225*x^2-60*x+4)^4+8*exp(225*x^2-60*x+4)
^3-12*exp(225*x^2-60*x+4)^2+8*exp(225*x^2-60*x+4)-2)/x/log(x)^3,x, algorithm="fricas")

[Out]

-(log(x)^3 + 10*(e^(450*x^2 - 120*x + 8) - 2*e^(225*x^2 - 60*x + 4) + 1)*log(x) - e^(900*x^2 - 240*x + 16) + 4
*e^(675*x^2 - 180*x + 12) - 6*e^(450*x^2 - 120*x + 8) + 4*e^(225*x^2 - 60*x + 4) - 1)/log(x)^2

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giac [B]  time = 0.24, size = 98, normalized size = 3.27 \begin {gather*} -\frac {\log \relax (x)^{3} + 10 \, e^{\left (450 \, x^{2} - 120 \, x + 8\right )} \log \relax (x) - 20 \, e^{\left (225 \, x^{2} - 60 \, x + 4\right )} \log \relax (x) - e^{\left (900 \, x^{2} - 240 \, x + 16\right )} + 4 \, e^{\left (675 \, x^{2} - 180 \, x + 12\right )} - 6 \, e^{\left (450 \, x^{2} - 120 \, x + 8\right )} + 4 \, e^{\left (225 \, x^{2} - 60 \, x + 4\right )} + 10 \, \log \relax (x) - 1}{\log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+((-9000*x^2+1200*x)*exp(225*x^2-60*x+4)^2+(9000*x^2-1200*x)*exp(225*x^2-60*x+4))*log(x)^2
+((1800*x^2-240*x)*exp(225*x^2-60*x+4)^4+(-5400*x^2+720*x)*exp(225*x^2-60*x+4)^3+(5400*x^2-720*x+10)*exp(225*x
^2-60*x+4)^2+(-1800*x^2+240*x-20)*exp(225*x^2-60*x+4)+10)*log(x)-2*exp(225*x^2-60*x+4)^4+8*exp(225*x^2-60*x+4)
^3-12*exp(225*x^2-60*x+4)^2+8*exp(225*x^2-60*x+4)-2)/x/log(x)^3,x, algorithm="giac")

[Out]

-(log(x)^3 + 10*e^(450*x^2 - 120*x + 8)*log(x) - 20*e^(225*x^2 - 60*x + 4)*log(x) - e^(900*x^2 - 240*x + 16) +
4*e^(675*x^2 - 180*x + 12) - 6*e^(450*x^2 - 120*x + 8) + 4*e^(225*x^2 - 60*x + 4) + 10*log(x) - 1)/log(x)^2

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maple [B]  time = 0.08, size = 87, normalized size = 2.90

method	result	size
risch	\(-\ln \relax (x )+\frac {{\mathrm e}^{4 \left (15 x -2\right )^{2}}-4 \,{\mathrm e}^{3 \left (15 x -2\right )^{2}}-10 \ln \relax (x ) {\mathrm e}^{2 \left (15 x -2\right )^{2}}+6 \,{\mathrm e}^{2 \left (15 x -2\right )^{2}}+20 \ln \relax (x ) {\mathrm e}^{\left (15 x -2\right )^{2}}-4 \,{\mathrm e}^{\left (15 x -2\right )^{2}}-10 \ln \relax (x )+1}{\ln \relax (x )^{2}}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)^3+((-9000*x^2+1200*x)*exp(225*x^2-60*x+4)^2+(9000*x^2-1200*x)*exp(225*x^2-60*x+4))*ln(x)^2+((1800*
x^2-240*x)*exp(225*x^2-60*x+4)^4+(-5400*x^2+720*x)*exp(225*x^2-60*x+4)^3+(5400*x^2-720*x+10)*exp(225*x^2-60*x+
4)^2+(-1800*x^2+240*x-20)*exp(225*x^2-60*x+4)+10)*ln(x)-2*exp(225*x^2-60*x+4)^4+8*exp(225*x^2-60*x+4)^3-12*exp
(225*x^2-60*x+4)^2+8*exp(225*x^2-60*x+4)-2)/x/ln(x)^3,x,method=_RETURNVERBOSE)

[Out]

-ln(x)+(exp(4*(15*x-2)^2)-4*exp(3*(15*x-2)^2)-10*ln(x)*exp(2*(15*x-2)^2)+6*exp(2*(15*x-2)^2)+20*ln(x)*exp((15*
x-2)^2)-4*exp((15*x-2)^2)-10*ln(x)+1)/ln(x)^2

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maxima [B]  time = 0.42, size = 97, normalized size = 3.23 \begin {gather*} -\frac {{\left (2 \, {\left (5 \, e^{8} \log \relax (x) - 3 \, e^{8}\right )} e^{\left (450 \, x^{2} + 120 \, x\right )} - 4 \, {\left (5 \, e^{4} \log \relax (x) - e^{4}\right )} e^{\left (225 \, x^{2} + 180 \, x\right )} + 10 \, e^{\left (240 \, x\right )} \log \relax (x) - e^{\left (900 \, x^{2} + 16\right )} + 4 \, e^{\left (675 \, x^{2} + 60 \, x + 12\right )}\right )} e^{\left (-240 \, x\right )}}{\log \relax (x)^{2}} + \frac {1}{\log \relax (x)^{2}} - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)^3+((-9000*x^2+1200*x)*exp(225*x^2-60*x+4)^2+(9000*x^2-1200*x)*exp(225*x^2-60*x+4))*log(x)^2
+((1800*x^2-240*x)*exp(225*x^2-60*x+4)^4+(-5400*x^2+720*x)*exp(225*x^2-60*x+4)^3+(5400*x^2-720*x+10)*exp(225*x
^2-60*x+4)^2+(-1800*x^2+240*x-20)*exp(225*x^2-60*x+4)+10)*log(x)-2*exp(225*x^2-60*x+4)^4+8*exp(225*x^2-60*x+4)
^3-12*exp(225*x^2-60*x+4)^2+8*exp(225*x^2-60*x+4)-2)/x/log(x)^3,x, algorithm="maxima")

[Out]

-(2*(5*e^8*log(x) - 3*e^8)*e^(450*x^2 + 120*x) - 4*(5*e^4*log(x) - e^4)*e^(225*x^2 + 180*x) + 10*e^(240*x)*log
(x) - e^(900*x^2 + 16) + 4*e^(675*x^2 + 60*x + 12))*e^(-240*x)/log(x)^2 + 1/log(x)^2 - log(x)

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mupad [B]  time = 4.67, size = 101, normalized size = 3.37 \begin {gather*} \frac {{\mathrm {e}}^{900\,x^2-240\,x+16}}{{\ln \relax (x)}^2}-\frac {5}{\ln \relax (x)}-\frac {5\,\ln \relax (x)-1}{{\ln \relax (x)}^2}-\frac {4\,{\mathrm {e}}^{675\,x^2-180\,x+12}}{{\ln \relax (x)}^2}-\ln \relax (x)+\frac {{\mathrm {e}}^{225\,x^2-60\,x+4}\,\left (20\,\ln \relax (x)-4\right )}{{\ln \relax (x)}^2}-\frac {{\mathrm {e}}^{450\,x^2-120\,x+8}\,\left (10\,\ln \relax (x)-6\right )}{{\ln \relax (x)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*exp(450*x^2 - 120*x + 8) - 8*exp(225*x^2 - 60*x + 4) - 8*exp(675*x^2 - 180*x + 12) + 2*exp(900*x^2 -
240*x + 16) - log(x)*(exp(450*x^2 - 120*x + 8)*(5400*x^2 - 720*x + 10) - exp(900*x^2 - 240*x + 16)*(240*x - 18
00*x^2) - exp(225*x^2 - 60*x + 4)*(1800*x^2 - 240*x + 20) + exp(675*x^2 - 180*x + 12)*(720*x - 5400*x^2) + 10)
+ log(x)^2*(exp(225*x^2 - 60*x + 4)*(1200*x - 9000*x^2) - exp(450*x^2 - 120*x + 8)*(1200*x - 9000*x^2)) + log
(x)^3 + 2)/(x*log(x)^3),x)

[Out]

exp(900*x^2 - 240*x + 16)/log(x)^2 - 5/log(x) - (5*log(x) - 1)/log(x)^2 - (4*exp(675*x^2 - 180*x + 12))/log(x)
^2 - log(x) + (exp(225*x^2 - 60*x + 4)*(20*log(x) - 4))/log(x)^2 - (exp(450*x^2 - 120*x + 8)*(10*log(x) - 6))/
log(x)^2

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sympy [B]  time = 0.54, size = 105, normalized size = 3.50 \begin {gather*} \frac {1 - 10 \log {\relax (x )}}{\log {\relax (x )}^{2}} + \frac {\left (- 10 \log {\relax (x )}^{7} + 6 \log {\relax (x )}^{6}\right ) e^{450 x^{2} - 120 x + 8} + \left (20 \log {\relax (x )}^{7} - 4 \log {\relax (x )}^{6}\right ) e^{225 x^{2} - 60 x + 4} - 4 e^{675 x^{2} - 180 x + 12} \log {\relax (x )}^{6} + e^{900 x^{2} - 240 x + 16} \log {\relax (x )}^{6}}{\log {\relax (x )}^{8}} - \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)**3+((-9000*x**2+1200*x)*exp(225*x**2-60*x+4)**2+(9000*x**2-1200*x)*exp(225*x**2-60*x+4))*ln(
x)**2+((1800*x**2-240*x)*exp(225*x**2-60*x+4)**4+(-5400*x**2+720*x)*exp(225*x**2-60*x+4)**3+(5400*x**2-720*x+1
0)*exp(225*x**2-60*x+4)**2+(-1800*x**2+240*x-20)*exp(225*x**2-60*x+4)+10)*ln(x)-2*exp(225*x**2-60*x+4)**4+8*ex
p(225*x**2-60*x+4)**3-12*exp(225*x**2-60*x+4)**2+8*exp(225*x**2-60*x+4)-2)/x/ln(x)**3,x)

[Out]

(1 - 10*log(x))/log(x)**2 + ((-10*log(x)**7 + 6*log(x)**6)*exp(450*x**2 - 120*x + 8) + (20*log(x)**7 - 4*log(x
)**6)*exp(225*x**2 - 60*x + 4) - 4*exp(675*x**2 - 180*x + 12)*log(x)**6 + exp(900*x**2 - 240*x + 16)*log(x)**6
)/log(x)**8 - log(x)

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