3.39.100 \(\int \frac {-9 e^{10} x+(6 e^{20}-6 e^{10} x) \log (x)-3 e^{10} x \log ^2(x)}{-((9 e^{20} x-18 e^{10} x^2+9 x^3) \log (\frac {5}{2}))-(6 e^{20} x-12 e^{10} x^2+6 x^3) \log (\frac {5}{2}) \log ^2(x)-(e^{20} x-2 e^{10} x^2+x^3) \log (\frac {5}{2}) \log ^4(x)} \, dx\)

Optimal. Leaf size=30 \[ -\frac {3 x}{\left (-x+\frac {x^2}{e^{10}}\right ) \log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )} \]

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Rubi [F]  time = 1.01, 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 {-9 e^{10} x+\left (6 e^{20}-6 e^{10} x\right ) \log (x)-3 e^{10} x \log ^2(x)}{-\left (\left (9 e^{20} x-18 e^{10} x^2+9 x^3\right ) \log \left (\frac {5}{2}\right )\right )-\left (6 e^{20} x-12 e^{10} x^2+6 x^3\right ) \log \left (\frac {5}{2}\right ) \log ^2(x)-\left (e^{20} x-2 e^{10} x^2+x^3\right ) \log \left (\frac {5}{2}\right ) \log ^4(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-9*E^10*x + (6*E^20 - 6*E^10*x)*Log[x] - 3*E^10*x*Log[x]^2)/(-((9*E^20*x - 18*E^10*x^2 + 9*x^3)*Log[5/2])
- (6*E^20*x - 12*E^10*x^2 + 6*x^3)*Log[5/2]*Log[x]^2 - (E^20*x - 2*E^10*x^2 + x^3)*Log[5/2]*Log[x]^4),x]

[Out]

3/(Log[5/2]*(3 + Log[x]^2)) - (6*Defer[Int][Log[x]/((E^10 - x)*(3 + Log[x]^2)^2), x])/Log[5/2] + (3*E^10*Defer
[Int][1/((E^10 - x)^2*(3 + Log[x]^2)), x])/Log[5/2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{10} \left (3 x-2 \left (e^{10}-x\right ) \log (x)+x \log ^2(x)\right )}{\left (e^{10}-x\right )^2 x \log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )^2} \, dx\\ &=\frac {\left (3 e^{10}\right ) \int \frac {3 x-2 \left (e^{10}-x\right ) \log (x)+x \log ^2(x)}{\left (e^{10}-x\right )^2 x \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}\\ &=\frac {\left (3 e^{10}\right ) \int \left (-\frac {2 \log (x)}{\left (e^{10}-x\right ) x \left (3+\log ^2(x)\right )^2}+\frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )}\right ) \, dx}{\log \left (\frac {5}{2}\right )}\\ &=\frac {\left (3 e^{10}\right ) \int \frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )} \, dx}{\log \left (\frac {5}{2}\right )}-\frac {\left (6 e^{10}\right ) \int \frac {\log (x)}{\left (e^{10}-x\right ) x \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}\\ &=\frac {\left (3 e^{10}\right ) \int \frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )} \, dx}{\log \left (\frac {5}{2}\right )}-\frac {\left (6 e^{10}\right ) \int \left (\frac {\log (x)}{e^{10} \left (e^{10}-x\right ) \left (3+\log ^2(x)\right )^2}+\frac {\log (x)}{e^{10} x \left (3+\log ^2(x)\right )^2}\right ) \, dx}{\log \left (\frac {5}{2}\right )}\\ &=-\frac {6 \int \frac {\log (x)}{\left (e^{10}-x\right ) \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}-\frac {6 \int \frac {\log (x)}{x \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}+\frac {\left (3 e^{10}\right ) \int \frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )} \, dx}{\log \left (\frac {5}{2}\right )}\\ &=-\frac {6 \int \frac {\log (x)}{\left (e^{10}-x\right ) \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}-\frac {6 \operatorname {Subst}\left (\int \frac {x}{\left (3+x^2\right )^2} \, dx,x,\log (x)\right )}{\log \left (\frac {5}{2}\right )}+\frac {\left (3 e^{10}\right ) \int \frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )} \, dx}{\log \left (\frac {5}{2}\right )}\\ &=\frac {3}{\log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )}-\frac {6 \int \frac {\log (x)}{\left (e^{10}-x\right ) \left (3+\log ^2(x)\right )^2} \, dx}{\log \left (\frac {5}{2}\right )}+\frac {\left (3 e^{10}\right ) \int \frac {1}{\left (e^{10}-x\right )^2 \left (3+\log ^2(x)\right )} \, dx}{\log \left (\frac {5}{2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 28, normalized size = 0.93 \begin {gather*} \frac {3 e^{10}}{\left (e^{10}-x\right ) \log \left (\frac {5}{2}\right ) \left (3+\log ^2(x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*E^10*x + (6*E^20 - 6*E^10*x)*Log[x] - 3*E^10*x*Log[x]^2)/(-((9*E^20*x - 18*E^10*x^2 + 9*x^3)*Log
[5/2]) - (6*E^20*x - 12*E^10*x^2 + 6*x^3)*Log[5/2]*Log[x]^2 - (E^20*x - 2*E^10*x^2 + x^3)*Log[5/2]*Log[x]^4),x
]

[Out]

(3*E^10)/((E^10 - x)*Log[5/2]*(3 + Log[x]^2))

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fricas [A]  time = 0.63, size = 30, normalized size = 1.00 \begin {gather*} \frac {3 \, e^{10}}{{\left (x - e^{10}\right )} \log \left (\frac {2}{5}\right ) \log \relax (x)^{2} + 3 \, {\left (x - e^{10}\right )} \log \left (\frac {2}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp(5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x
^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12*x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x
^3)*log(2/5)),x, algorithm="fricas")

[Out]

3*e^10/((x - e^10)*log(2/5)*log(x)^2 + 3*(x - e^10)*log(2/5))

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giac [B]  time = 0.70, size = 65, normalized size = 2.17 \begin {gather*} -\frac {6 \, e^{10}}{x \log \relax (5) \log \relax (x)^{2} - e^{10} \log \relax (5) \log \relax (x)^{2} - x \log \relax (2) \log \relax (x)^{2} + e^{10} \log \relax (2) \log \relax (x)^{2} + 3 \, x \log \relax (5) - 3 \, e^{10} \log \relax (5) - 3 \, x \log \relax (2) + 3 \, e^{10} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp(5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x
^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12*x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x
^3)*log(2/5)),x, algorithm="giac")

[Out]

-6*e^10/(x*log(5)*log(x)^2 - e^10*log(5)*log(x)^2 - x*log(2)*log(x)^2 + e^10*log(2)*log(x)^2 + 3*x*log(5) - 3*
e^10*log(5) - 3*x*log(2) + 3*e^10*log(2))

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maple [A]  time = 0.30, size = 30, normalized size = 1.00

method	result	size
risch	\(-\frac {3 \,{\mathrm e}^{10}}{\left (\ln \relax (2)-\ln \relax (5)\right ) \left (\ln \relax (x )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\)	\(30\)
norman	\(-\frac {3 \,{\mathrm e}^{10}}{\left (\ln \relax (2)-\ln \relax (5)\right ) \left (\ln \relax (x )^{2}+3\right ) \left ({\mathrm e}^{10}-x \right )}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3*x*exp(5)^2*ln(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*ln(x)-9*x*exp(5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x^3)*ln(2
/5)*ln(x)^4+(6*x*exp(5)^4-12*x^2*exp(5)^2+6*x^3)*ln(2/5)*ln(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x^3)*ln(2/5))
,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.48, size = 54, normalized size = 1.80 \begin {gather*} -\frac {3 \, e^{10}}{{\left (x {\left (\log \relax (5) - \log \relax (2)\right )} - {\left (\log \relax (5) - \log \relax (2)\right )} e^{10}\right )} \log \relax (x)^{2} + 3 \, x {\left (\log \relax (5) - \log \relax (2)\right )} - 3 \, {\left (\log \relax (5) - \log \relax (2)\right )} e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp(5)^2*log(x)^2+(6*exp(5)^4-6*x*exp(5)^2)*log(x)-9*x*exp(5)^2)/((x*exp(5)^4-2*x^2*exp(5)^2+x
^3)*log(2/5)*log(x)^4+(6*x*exp(5)^4-12*x^2*exp(5)^2+6*x^3)*log(2/5)*log(x)^2+(9*x*exp(5)^4-18*x^2*exp(5)^2+9*x
^3)*log(2/5)),x, algorithm="maxima")

[Out]

-3*e^10/((x*(log(5) - log(2)) - (log(5) - log(2))*e^10)*log(x)^2 + 3*x*(log(5) - log(2)) - 3*(log(5) - log(2))
*e^10)

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mupad [B]  time = 2.75, size = 24, normalized size = 0.80 \begin {gather*} \frac {3\,{\mathrm {e}}^{10}}{\ln \left (\frac {2}{5}\right )\,\left (x-{\mathrm {e}}^{10}\right )\,\left ({\ln \relax (x)}^2+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*x*exp(10) - log(x)*(6*exp(20) - 6*x*exp(10)) + 3*x*exp(10)*log(x)^2)/(log(2/5)*(9*x*exp(20) - 18*x^2*e
xp(10) + 9*x^3) + log(2/5)*log(x)^4*(x*exp(20) - 2*x^2*exp(10) + x^3) + log(2/5)*log(x)^2*(6*x*exp(20) - 12*x^
2*exp(10) + 6*x^3)),x)

[Out]

(3*exp(10))/(log(2/5)*(x - exp(10))*(log(x)^2 + 3))

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sympy [B]  time = 0.19, size = 63, normalized size = 2.10 \begin {gather*} \frac {3 e^{10}}{- 3 x \log {\relax (5 )} + 3 x \log {\relax (2 )} + \left (- x \log {\relax (5 )} + x \log {\relax (2 )} - e^{10} \log {\relax (2 )} + e^{10} \log {\relax (5 )}\right ) \log {\relax (x )}^{2} - 3 e^{10} \log {\relax (2 )} + 3 e^{10} \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3*x*exp(5)**2*ln(x)**2+(6*exp(5)**4-6*x*exp(5)**2)*ln(x)-9*x*exp(5)**2)/((x*exp(5)**4-2*x**2*exp(5
)**2+x**3)*ln(2/5)*ln(x)**4+(6*x*exp(5)**4-12*x**2*exp(5)**2+6*x**3)*ln(2/5)*ln(x)**2+(9*x*exp(5)**4-18*x**2*e
xp(5)**2+9*x**3)*ln(2/5)),x)

[Out]

3*exp(10)/(-3*x*log(5) + 3*x*log(2) + (-x*log(5) + x*log(2) - exp(10)*log(2) + exp(10)*log(5))*log(x)**2 - 3*e
xp(10)*log(2) + 3*exp(10)*log(5))

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