3.95.13 \(\int \frac {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+(24 e^{96+x} x^5+12 e^{96} x^4 \log (3)) \log (x)+(-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)) \log ^2(x)+(8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)) \log ^3(x)} \, dx\)

Optimal. Leaf size=27 \[ \frac {25}{\left (-\log (3)+2 x \left (-e^x+\frac {e^{48} x}{\log (x)}\right )\right )^2} \]

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Rubi [F]  time = 7.67, 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 {-100 e^{48} x \log (x)+200 e^{48} x \log ^2(x)+e^x (-100-100 x) \log ^3(x)}{-8 e^{144} x^6+\left (24 e^{96+x} x^5+12 e^{96} x^4 \log (3)\right ) \log (x)+\left (-24 e^{48+2 x} x^4-24 e^{48+x} x^3 \log (3)-6 e^{48} x^2 \log ^2(3)\right ) \log ^2(x)+\left (8 e^{3 x} x^3+12 e^{2 x} x^2 \log (3)+6 e^x x \log ^2(3)+\log ^3(3)\right ) \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-100*E^48*x*Log[x] + 200*E^48*x*Log[x]^2 + E^x*(-100 - 100*x)*Log[x]^3)/(-8*E^144*x^6 + (24*E^(96 + x)*x^
5 + 12*E^96*x^4*Log[3])*Log[x] + (-24*E^(48 + 2*x)*x^4 - 24*E^(48 + x)*x^3*Log[3] - 6*E^48*x^2*Log[3]^2)*Log[x
]^2 + (8*E^(3*x)*x^3 + 12*E^(2*x)*x^2*Log[3] + 6*E^x*x*Log[3]^2 + Log[3]^3)*Log[x]^3),x]

[Out]

100*E^48*Defer[Int][(x*Log[x])/(2*E^48*x^2 - 2*E^x*x*Log[x] - Log[3]*Log[x])^3, x] - 100*E^48*Defer[Int][(x*Lo
g[x]^2)/(2*E^48*x^2 - 2*E^x*x*Log[x] - Log[3]*Log[x])^3, x] + 100*E^48*Defer[Int][(x^2*Log[x]^2)/(2*E^48*x^2 -
2*E^x*x*Log[x] - Log[3]*Log[x])^3, x] - 50*Log[3]*Defer[Int][Log[x]^3/(x*(2*E^48*x^2 - 2*E^x*x*Log[x] - Log[3
]*Log[x])^3), x] - 50*Defer[Int][Log[x]^2/(x*(2*E^48*x^2 - 2*E^x*x*Log[x] - Log[3]*Log[x])^2), x] + 50*Log[3]*
Defer[Int][Log[x]^3/(-2*E^48*x^2 + 2*E^x*x*Log[x] + Log[3]*Log[x])^3, x] - 50*Defer[Int][Log[x]^2/(-2*E^48*x^2
+ 2*E^x*x*Log[x] + Log[3]*Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100 \log (x) \left (e^{48} x-2 e^{48} x \log (x)+e^x (1+x) \log ^2(x)\right )}{\left (2 e^{48} x^2-\left (2 e^x x+\log (3)\right ) \log (x)\right )^3} \, dx\\ &=100 \int \frac {\log (x) \left (e^{48} x-2 e^{48} x \log (x)+e^x (1+x) \log ^2(x)\right )}{\left (2 e^{48} x^2-\left (2 e^x x+\log (3)\right ) \log (x)\right )^3} \, dx\\ &=100 \int \left (-\frac {(1+x) \log ^2(x)}{2 x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^2}+\frac {\log (x) \left (2 e^{48} x^2-2 e^{48} x^2 \log (x)+2 e^{48} x^3 \log (x)-\log (3) \log ^2(x)-x \log (3) \log ^2(x)\right )}{2 x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3}\right ) \, dx\\ &=-\left (50 \int \frac {(1+x) \log ^2(x)}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^2} \, dx\right )+50 \int \frac {\log (x) \left (2 e^{48} x^2-2 e^{48} x^2 \log (x)+2 e^{48} x^3 \log (x)-\log (3) \log ^2(x)-x \log (3) \log ^2(x)\right )}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3} \, dx\\ &=50 \int \left (\frac {2 e^{48} x \log (x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3}-\frac {2 e^{48} x \log ^2(x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3}+\frac {2 e^{48} x^2 \log ^2(x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3}-\frac {\log (3) \log ^3(x)}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3}+\frac {\log (3) \log ^3(x)}{\left (-2 e^{48} x^2+2 e^x x \log (x)+\log (3) \log (x)\right )^3}\right ) \, dx-50 \int \left (\frac {\log ^2(x)}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^2}+\frac {\log ^2(x)}{\left (-2 e^{48} x^2+2 e^x x \log (x)+\log (3) \log (x)\right )^2}\right ) \, dx\\ &=-\left (50 \int \frac {\log ^2(x)}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^2} \, dx\right )-50 \int \frac {\log ^2(x)}{\left (-2 e^{48} x^2+2 e^x x \log (x)+\log (3) \log (x)\right )^2} \, dx+\left (100 e^{48}\right ) \int \frac {x \log (x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3} \, dx-\left (100 e^{48}\right ) \int \frac {x \log ^2(x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3} \, dx+\left (100 e^{48}\right ) \int \frac {x^2 \log ^2(x)}{\left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3} \, dx-(50 \log (3)) \int \frac {\log ^3(x)}{x \left (2 e^{48} x^2-2 e^x x \log (x)-\log (3) \log (x)\right )^3} \, dx+(50 \log (3)) \int \frac {\log ^3(x)}{\left (-2 e^{48} x^2+2 e^x x \log (x)+\log (3) \log (x)\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.62, size = 29, normalized size = 1.07 \begin {gather*} \frac {25 \log ^2(x)}{\left (-2 e^{48} x^2+\left (2 e^x x+\log (3)\right ) \log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100*E^48*x*Log[x] + 200*E^48*x*Log[x]^2 + E^x*(-100 - 100*x)*Log[x]^3)/(-8*E^144*x^6 + (24*E^(96 +
x)*x^5 + 12*E^96*x^4*Log[3])*Log[x] + (-24*E^(48 + 2*x)*x^4 - 24*E^(48 + x)*x^3*Log[3] - 6*E^48*x^2*Log[3]^2)
*Log[x]^2 + (8*E^(3*x)*x^3 + 12*E^(2*x)*x^2*Log[3] + 6*E^x*x*Log[3]^2 + Log[3]^3)*Log[x]^3),x]

[Out]

(25*Log[x]^2)/(-2*E^48*x^2 + (2*E^x*x + Log[3])*Log[x])^2

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fricas [B]  time = 0.75, size = 73, normalized size = 2.70 \begin {gather*} \frac {25 \, e^{192} \log \relax (x)^{2}}{4 \, x^{4} e^{288} + {\left (4 \, x^{2} e^{\left (2 \, x + 192\right )} + 4 \, x e^{\left (x + 192\right )} \log \relax (3) + e^{192} \log \relax (3)^{2}\right )} \log \relax (x)^{2} - 4 \, {\left (2 \, x^{3} e^{\left (x + 240\right )} + x^{2} e^{240} \log \relax (3)\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(48)*log(x))/((8*x^3*exp(x)^3+12*x^2*l
og(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log(3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6
*x^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*log(3))*log(x)-8*x^6*exp(48)^3),x, a
lgorithm="fricas")

[Out]

25*e^192*log(x)^2/(4*x^4*e^288 + (4*x^2*e^(2*x + 192) + 4*x*e^(x + 192)*log(3) + e^192*log(3)^2)*log(x)^2 - 4*
(2*x^3*e^(x + 240) + x^2*e^240*log(3))*log(x))

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giac [B]  time = 80.35, size = 71, normalized size = 2.63 \begin {gather*} \frac {25 \, \log \relax (x)^{2}}{4 \, x^{4} e^{96} - 8 \, x^{3} e^{\left (x + 48\right )} \log \relax (x) - 4 \, x^{2} e^{48} \log \relax (3) \log \relax (x) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} + 4 \, x e^{x} \log \relax (3) \log \relax (x)^{2} + \log \relax (3)^{2} \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(48)*log(x))/((8*x^3*exp(x)^3+12*x^2*l
og(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log(3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6
*x^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*log(3))*log(x)-8*x^6*exp(48)^3),x, a
lgorithm="giac")

[Out]

25*log(x)^2/(4*x^4*e^96 - 8*x^3*e^(x + 48)*log(x) - 4*x^2*e^48*log(3)*log(x) + 4*x^2*e^(2*x)*log(x)^2 + 4*x*e^
x*log(3)*log(x)^2 + log(3)^2*log(x)^2)

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maple [B]  time = 0.09, size = 86, normalized size = 3.19

method	result	size
risch	\(\frac {25}{\left (2 \,{\mathrm e}^{x} x +\ln \relax (3)\right )^{2}}+\frac {100 \left (-{\mathrm e}^{48} x^{2}+2 x \,{\mathrm e}^{x} \ln \relax (x )+\ln \relax (3) \ln \relax (x )\right ) {\mathrm e}^{48} x^{2}}{\left (4 \,{\mathrm e}^{2 x} x^{2}+4 x \ln \relax (3) {\mathrm e}^{x}+\ln \relax (3)^{2}\right ) \left (2 x \,{\mathrm e}^{x} \ln \relax (x )-2 \,{\mathrm e}^{48} x^{2}+\ln \relax (3) \ln \relax (x )\right )^{2}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-100*x-100)*exp(x)*ln(x)^3+200*x*exp(48)*ln(x)^2-100*x*exp(48)*ln(x))/((8*x^3*exp(x)^3+12*x^2*ln(3)*exp(
x)^2+6*x*ln(3)^2*exp(x)+ln(3)^3)*ln(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*ln(3)*exp(x)-6*x^2*exp(48)*l
n(3)^2)*ln(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*ln(3))*ln(x)-8*x^6*exp(48)^3),x,method=_RETURNVERBOS
E)

[Out]

25/(2*exp(x)*x+ln(3))^2+100*(-exp(48)*x^2+2*x*exp(x)*ln(x)+ln(3)*ln(x))*exp(48)*x^2/(4*exp(2*x)*x^2+4*x*ln(3)*
exp(x)+ln(3)^2)/(2*x*exp(x)*ln(x)-2*exp(48)*x^2+ln(3)*ln(x))^2

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maxima [B]  time = 0.85, size = 72, normalized size = 2.67 \begin {gather*} \frac {25 \, \log \relax (x)^{2}}{4 \, x^{4} e^{96} - 4 \, x^{2} e^{48} \log \relax (3) \log \relax (x) + 4 \, x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} + \log \relax (3)^{2} \log \relax (x)^{2} - 4 \, {\left (2 \, x^{3} e^{48} \log \relax (x) - x \log \relax (3) \log \relax (x)^{2}\right )} e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x-100)*exp(x)*log(x)^3+200*x*exp(48)*log(x)^2-100*x*exp(48)*log(x))/((8*x^3*exp(x)^3+12*x^2*l
og(3)*exp(x)^2+6*x*log(3)^2*exp(x)+log(3)^3)*log(x)^3+(-24*x^4*exp(48)*exp(x)^2-24*x^3*exp(48)*log(3)*exp(x)-6
*x^2*exp(48)*log(3)^2)*log(x)^2+(24*x^5*exp(48)^2*exp(x)+12*x^4*exp(48)^2*log(3))*log(x)-8*x^6*exp(48)^3),x, a
lgorithm="maxima")

[Out]

25*log(x)^2/(4*x^4*e^96 - 4*x^2*e^48*log(3)*log(x) + 4*x^2*e^(2*x)*log(x)^2 + log(3)^2*log(x)^2 - 4*(2*x^3*e^4
8*log(x) - x*log(3)*log(x)^2)*e^x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left (100\,x+100\right )\,{\ln \relax (x)}^3-200\,x\,{\mathrm {e}}^{48}\,{\ln \relax (x)}^2+100\,x\,{\mathrm {e}}^{48}\,\ln \relax (x)}{\ln \relax (x)\,\left (12\,x^4\,{\mathrm {e}}^{96}\,\ln \relax (3)+24\,x^5\,{\mathrm {e}}^{96}\,{\mathrm {e}}^x\right )-{\ln \relax (x)}^2\,\left (6\,x^2\,{\mathrm {e}}^{48}\,{\ln \relax (3)}^2+24\,x^4\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{48}+24\,x^3\,{\mathrm {e}}^{48}\,{\mathrm {e}}^x\,\ln \relax (3)\right )-8\,x^6\,{\mathrm {e}}^{144}+{\ln \relax (x)}^3\,\left (8\,x^3\,{\mathrm {e}}^{3\,x}+{\ln \relax (3)}^3+6\,x\,{\mathrm {e}}^x\,{\ln \relax (3)}^2+12\,x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (3)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(100*x*exp(48)*log(x) - 200*x*exp(48)*log(x)^2 + exp(x)*log(x)^3*(100*x + 100))/(log(x)*(12*x^4*exp(96)*l
og(3) + 24*x^5*exp(96)*exp(x)) - log(x)^2*(6*x^2*exp(48)*log(3)^2 + 24*x^4*exp(2*x)*exp(48) + 24*x^3*exp(48)*e
xp(x)*log(3)) - 8*x^6*exp(144) + log(x)^3*(8*x^3*exp(3*x) + log(3)^3 + 6*x*exp(x)*log(3)^2 + 12*x^2*exp(2*x)*l
og(3))),x)

[Out]

int(-(100*x*exp(48)*log(x) - 200*x*exp(48)*log(x)^2 + exp(x)*log(x)^3*(100*x + 100))/(log(x)*(12*x^4*exp(96)*l
og(3) + 24*x^5*exp(96)*exp(x)) - log(x)^2*(6*x^2*exp(48)*log(3)^2 + 24*x^4*exp(2*x)*exp(48) + 24*x^3*exp(48)*e
xp(x)*log(3)) - 8*x^6*exp(144) + log(x)^3*(8*x^3*exp(3*x) + log(3)^3 + 6*x*exp(x)*log(3)^2 + 12*x^2*exp(2*x)*l
og(3))), x)

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sympy [B]  time = 0.54, size = 82, normalized size = 3.04 \begin {gather*} \frac {25 \log {\relax (x )}^{2}}{4 x^{4} e^{96} + 4 x^{2} e^{2 x} \log {\relax (x )}^{2} - 4 x^{2} e^{48} \log {\relax (3 )} \log {\relax (x )} + \left (- 8 x^{3} e^{48} \log {\relax (x )} + 4 x \log {\relax (3 )} \log {\relax (x )}^{2}\right ) e^{x} + \log {\relax (3 )}^{2} \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-100*x-100)*exp(x)*ln(x)**3+200*x*exp(48)*ln(x)**2-100*x*exp(48)*ln(x))/((8*x**3*exp(x)**3+12*x**2
*ln(3)*exp(x)**2+6*x*ln(3)**2*exp(x)+ln(3)**3)*ln(x)**3+(-24*x**4*exp(48)*exp(x)**2-24*x**3*exp(48)*ln(3)*exp(
x)-6*x**2*exp(48)*ln(3)**2)*ln(x)**2+(24*x**5*exp(48)**2*exp(x)+12*x**4*exp(48)**2*ln(3))*ln(x)-8*x**6*exp(48)
**3),x)

[Out]

25*log(x)**2/(4*x**4*exp(96) + 4*x**2*exp(2*x)*log(x)**2 - 4*x**2*exp(48)*log(3)*log(x) + (-8*x**3*exp(48)*log
(x) + 4*x*log(3)*log(x)**2)*exp(x) + log(3)**2*log(x)**2)

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