3.40.62 \(\int \frac {800 e^4 x+(100 e^2-800 e^4 x) \log (x)-100 e^2 \log ^2(x)}{1000 e^6 x^3+300 e^4 x^2 \log (x)+30 e^2 x \log ^2(x)+\log ^3(x)} \, dx\)

Optimal. Leaf size=20 \[ \left (5-\frac {x}{x+\frac {\log (x)}{10 e^2}}\right )^2 \]

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Rubi [F]  time = 0.49, 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 {800 e^4 x+\left (100 e^2-800 e^4 x\right ) \log (x)-100 e^2 \log ^2(x)}{1000 e^6 x^3+300 e^4 x^2 \log (x)+30 e^2 x \log ^2(x)+\log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(800*E^4*x + (100*E^2 - 800*E^4*x)*Log[x] - 100*E^2*Log[x]^2)/(1000*E^6*x^3 + 300*E^4*x^2*Log[x] + 30*E^2*
x*Log[x]^2 + Log[x]^3),x]

[Out]

100*E^2*Defer[Int][(-10*E^2*x - Log[x])^(-1), x] - 200*E^4*Defer[Int][x/(10*E^2*x + Log[x])^3, x] - 2000*E^6*D
efer[Int][x^2/(10*E^2*x + Log[x])^3, x] + 100*E^2*Defer[Int][(10*E^2*x + Log[x])^(-2), x] + 1200*E^4*Defer[Int
][x/(10*E^2*x + Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100 e^2 (1-\log (x)) \left (8 e^2 x+\log (x)\right )}{\left (10 e^2 x+\log (x)\right )^3} \, dx\\ &=\left (100 e^2\right ) \int \frac {(1-\log (x)) \left (8 e^2 x+\log (x)\right )}{\left (10 e^2 x+\log (x)\right )^3} \, dx\\ &=\left (100 e^2\right ) \int \left (\frac {1}{-10 e^2 x-\log (x)}-\frac {2 e^2 x \left (1+10 e^2 x\right )}{\left (10 e^2 x+\log (x)\right )^3}+\frac {1+12 e^2 x}{\left (10 e^2 x+\log (x)\right )^2}\right ) \, dx\\ &=\left (100 e^2\right ) \int \frac {1}{-10 e^2 x-\log (x)} \, dx+\left (100 e^2\right ) \int \frac {1+12 e^2 x}{\left (10 e^2 x+\log (x)\right )^2} \, dx-\left (200 e^4\right ) \int \frac {x \left (1+10 e^2 x\right )}{\left (10 e^2 x+\log (x)\right )^3} \, dx\\ &=\left (100 e^2\right ) \int \frac {1}{-10 e^2 x-\log (x)} \, dx+\left (100 e^2\right ) \int \left (\frac {1}{\left (10 e^2 x+\log (x)\right )^2}+\frac {12 e^2 x}{\left (10 e^2 x+\log (x)\right )^2}\right ) \, dx-\left (200 e^4\right ) \int \left (\frac {x}{\left (10 e^2 x+\log (x)\right )^3}+\frac {10 e^2 x^2}{\left (10 e^2 x+\log (x)\right )^3}\right ) \, dx\\ &=\left (100 e^2\right ) \int \frac {1}{-10 e^2 x-\log (x)} \, dx+\left (100 e^2\right ) \int \frac {1}{\left (10 e^2 x+\log (x)\right )^2} \, dx-\left (200 e^4\right ) \int \frac {x}{\left (10 e^2 x+\log (x)\right )^3} \, dx+\left (1200 e^4\right ) \int \frac {x}{\left (10 e^2 x+\log (x)\right )^2} \, dx-\left (2000 e^6\right ) \int \frac {x^2}{\left (10 e^2 x+\log (x)\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 26, normalized size = 1.30 \begin {gather*} -\frac {100 e^2 x \left (9 e^2 x+\log (x)\right )}{\left (10 e^2 x+\log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(800*E^4*x + (100*E^2 - 800*E^4*x)*Log[x] - 100*E^2*Log[x]^2)/(1000*E^6*x^3 + 300*E^4*x^2*Log[x] + 3
0*E^2*x*Log[x]^2 + Log[x]^3),x]

[Out]

(-100*E^2*x*(9*E^2*x + Log[x]))/(10*E^2*x + Log[x])^2

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fricas [A]  time = 0.69, size = 37, normalized size = 1.85 \begin {gather*} -\frac {100 \, {\left (9 \, x^{2} e^{4} + x e^{2} \log \relax (x)\right )}}{100 \, x^{2} e^{4} + 20 \, x e^{2} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*exp(2)*log(x)^2+(-800*x*exp(2)^2+100*exp(2))*log(x)+800*x*exp(2)^2)/(log(x)^3+30*x*exp(2)*log(
x)^2+300*x^2*exp(2)^2*log(x)+1000*x^3*exp(2)^3),x, algorithm="fricas")

[Out]

-100*(9*x^2*e^4 + x*e^2*log(x))/(100*x^2*e^4 + 20*x*e^2*log(x) + log(x)^2)

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giac [A]  time = 0.43, size = 37, normalized size = 1.85 \begin {gather*} -\frac {100 \, {\left (9 \, x^{2} e^{4} + x e^{2} \log \relax (x)\right )}}{100 \, x^{2} e^{4} + 20 \, x e^{2} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*exp(2)*log(x)^2+(-800*x*exp(2)^2+100*exp(2))*log(x)+800*x*exp(2)^2)/(log(x)^3+30*x*exp(2)*log(
x)^2+300*x^2*exp(2)^2*log(x)+1000*x^3*exp(2)^3),x, algorithm="giac")

[Out]

-100*(9*x^2*e^4 + x*e^2*log(x))/(100*x^2*e^4 + 20*x*e^2*log(x) + log(x)^2)

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maple [A]  time = 0.08, size = 24, normalized size = 1.20

method	result	size
risch	\(-\frac {100 \left (9 \,{\mathrm e}^{2} x +\ln \relax (x )\right ) {\mathrm e}^{2} x}{\left (10 \,{\mathrm e}^{2} x +\ln \relax (x )\right )^{2}}\)	\(24\)
norman	\(\frac {-900 x^{2} {\mathrm e}^{4}-100 x \,{\mathrm e}^{2} \ln \relax (x )}{\left (10 \,{\mathrm e}^{2} x +\ln \relax (x )\right )^{2}}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-100*exp(2)*ln(x)^2+(-800*x*exp(2)^2+100*exp(2))*ln(x)+800*x*exp(2)^2)/(ln(x)^3+30*x*exp(2)*ln(x)^2+300*x
^2*exp(2)^2*ln(x)+1000*x^3*exp(2)^3),x,method=_RETURNVERBOSE)

[Out]

-100*(9*exp(2)*x+ln(x))*exp(2)*x/(10*exp(2)*x+ln(x))^2

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maxima [A]  time = 2.00, size = 37, normalized size = 1.85 \begin {gather*} -\frac {100 \, {\left (9 \, x^{2} e^{4} + x e^{2} \log \relax (x)\right )}}{100 \, x^{2} e^{4} + 20 \, x e^{2} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*exp(2)*log(x)^2+(-800*x*exp(2)^2+100*exp(2))*log(x)+800*x*exp(2)^2)/(log(x)^3+30*x*exp(2)*log(
x)^2+300*x^2*exp(2)^2*log(x)+1000*x^3*exp(2)^3),x, algorithm="maxima")

[Out]

-100*(9*x^2*e^4 + x*e^2*log(x))/(100*x^2*e^4 + 20*x*e^2*log(x) + log(x)^2)

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mupad [B]  time = 3.03, size = 23, normalized size = 1.15 \begin {gather*} -\frac {100\,x\,{\mathrm {e}}^2\,\left (\ln \relax (x)+9\,x\,{\mathrm {e}}^2\right )}{{\left (\ln \relax (x)+10\,x\,{\mathrm {e}}^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((800*x*exp(4) - 100*exp(2)*log(x)^2 + log(x)*(100*exp(2) - 800*x*exp(4)))/(log(x)^3 + 1000*x^3*exp(6) + 30
*x*exp(2)*log(x)^2 + 300*x^2*exp(4)*log(x)),x)

[Out]

-(100*x*exp(2)*(log(x) + 9*x*exp(2)))/(log(x) + 10*x*exp(2))^2

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sympy [B]  time = 0.13, size = 42, normalized size = 2.10 \begin {gather*} \frac {- 900 x^{2} e^{4} - 100 x e^{2} \log {\relax (x )}}{100 x^{2} e^{4} + 20 x e^{2} \log {\relax (x )} + \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-100*exp(2)*ln(x)**2+(-800*x*exp(2)**2+100*exp(2))*ln(x)+800*x*exp(2)**2)/(ln(x)**3+30*x*exp(2)*ln(
x)**2+300*x**2*exp(2)**2*ln(x)+1000*x**3*exp(2)**3),x)

[Out]

(-900*x**2*exp(4) - 100*x*exp(2)*log(x))/(100*x**2*exp(4) + 20*x*exp(2)*log(x) + log(x)**2)

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