3.52.61 \(\int \frac {150-24 x-12 x \log (x)}{e^{10} x-4 e^5 x^2+4 x^3+(-100 x^2+8 x^3+e^5 (50 x-4 x^2)) \log (x)+(625 x-100 x^2+4 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {6}{-e^5+2 x-(25-2 x) \log (x)} \]

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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 {150-24 x-12 x \log (x)}{e^{10} x-4 e^5 x^2+4 x^3+\left (-100 x^2+8 x^3+e^5 \left (50 x-4 x^2\right )\right ) \log (x)+\left (625 x-100 x^2+4 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(150 - 24*x - 12*x*Log[x])/(E^10*x - 4*E^5*x^2 + 4*x^3 + (-100*x^2 + 8*x^3 + E^5*(50*x - 4*x^2))*Log[x] +
(625*x - 100*x^2 + 4*x^3)*Log[x]^2),x]

[Out]

-12*Defer[Int][(E^5 - 2*x + 25*Log[x] - 2*x*Log[x])^(-2), x] + 150*Defer[Int][1/(x*(-E^5 + 2*x - 25*Log[x] + 2
*x*Log[x])^2), x] + 12*(25 - E^5)*Defer[Int][1/((-25 + 2*x)*(-E^5 + 2*x - 25*Log[x] + 2*x*Log[x])^2), x] - 12*
Defer[Int][1/((-25 + 2*x)*(-E^5 + 2*x - 25*Log[x] + 2*x*Log[x])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 (25-4 x-2 x \log (x))}{x \left (e^5-2 x+(25-2 x) \log (x)\right )^2} \, dx\\ &=6 \int \frac {25-4 x-2 x \log (x)}{x \left (e^5-2 x+(25-2 x) \log (x)\right )^2} \, dx\\ &=6 \int \left (\frac {625-2 \left (75-e^5\right ) x+4 x^2}{(25-2 x) x \left (e^5-2 x+25 \log (x)-2 x \log (x)\right )^2}-\frac {2}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )}\right ) \, dx\\ &=6 \int \frac {625-2 \left (75-e^5\right ) x+4 x^2}{(25-2 x) x \left (e^5-2 x+25 \log (x)-2 x \log (x)\right )^2} \, dx-12 \int \frac {1}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )} \, dx\\ &=6 \int \left (-\frac {2}{\left (e^5-2 x+25 \log (x)-2 x \log (x)\right )^2}+\frac {25}{x \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )^2}-\frac {2 \left (-25+e^5\right )}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )^2}\right ) \, dx-12 \int \frac {1}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )} \, dx\\ &=-\left (12 \int \frac {1}{\left (e^5-2 x+25 \log (x)-2 x \log (x)\right )^2} \, dx\right )-12 \int \frac {1}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )} \, dx+150 \int \frac {1}{x \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )^2} \, dx+\left (12 \left (25-e^5\right )\right ) \int \frac {1}{(-25+2 x) \left (-e^5+2 x-25 \log (x)+2 x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.35, size = 19, normalized size = 0.86 \begin {gather*} -\frac {6}{e^5-2 x+(25-2 x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(150 - 24*x - 12*x*Log[x])/(E^10*x - 4*E^5*x^2 + 4*x^3 + (-100*x^2 + 8*x^3 + E^5*(50*x - 4*x^2))*Log
[x] + (625*x - 100*x^2 + 4*x^3)*Log[x]^2),x]

[Out]

-6/(E^5 - 2*x + (25 - 2*x)*Log[x])

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fricas [A]  time = 0.46, size = 20, normalized size = 0.91 \begin {gather*} \frac {6}{{\left (2 \, x - 25\right )} \log \relax (x) + 2 \, x - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x*log(x)-24*x+150)/((4*x^3-100*x^2+625*x)*log(x)^2+((-4*x^2+50*x)*exp(5)+8*x^3-100*x^2)*log(x)+
x*exp(5)^2-4*x^2*exp(5)+4*x^3),x, algorithm="fricas")

[Out]

6/((2*x - 25)*log(x) + 2*x - e^5)

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giac [A]  time = 0.16, size = 21, normalized size = 0.95 \begin {gather*} \frac {6}{2 \, x \log \relax (x) + 2 \, x - e^{5} - 25 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x*log(x)-24*x+150)/((4*x^3-100*x^2+625*x)*log(x)^2+((-4*x^2+50*x)*exp(5)+8*x^3-100*x^2)*log(x)+
x*exp(5)^2-4*x^2*exp(5)+4*x^3),x, algorithm="giac")

[Out]

6/(2*x*log(x) + 2*x - e^5 - 25*log(x))

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maple [A]  time = 0.14, size = 20, normalized size = 0.91

method	result	size
norman	\(-\frac {6}{-2 x \ln \relax (x )+{\mathrm e}^{5}-2 x +25 \ln \relax (x )}\)	\(20\)
risch	\(-\frac {6}{-2 x \ln \relax (x )+{\mathrm e}^{5}-2 x +25 \ln \relax (x )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-12*x*ln(x)-24*x+150)/((4*x^3-100*x^2+625*x)*ln(x)^2+((-4*x^2+50*x)*exp(5)+8*x^3-100*x^2)*ln(x)+x*exp(5)^
2-4*x^2*exp(5)+4*x^3),x,method=_RETURNVERBOSE)

[Out]

-6/(-2*x*ln(x)+exp(5)-2*x+25*ln(x))

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maxima [A]  time = 0.39, size = 20, normalized size = 0.91 \begin {gather*} \frac {6}{{\left (2 \, x - 25\right )} \log \relax (x) + 2 \, x - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x*log(x)-24*x+150)/((4*x^3-100*x^2+625*x)*log(x)^2+((-4*x^2+50*x)*exp(5)+8*x^3-100*x^2)*log(x)+
x*exp(5)^2-4*x^2*exp(5)+4*x^3),x, algorithm="maxima")

[Out]

6/((2*x - 25)*log(x) + 2*x - e^5)

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mupad [B]  time = 3.83, size = 36, normalized size = 1.64 \begin {gather*} \frac {6\,{\mathrm {e}}^{-5}\,\left (2\,x-25\,\ln \relax (x)+2\,x\,\ln \relax (x)\right )}{2\,x-{\mathrm {e}}^5-25\,\ln \relax (x)+2\,x\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x + 12*x*log(x) - 150)/(x*exp(10) + log(x)*(exp(5)*(50*x - 4*x^2) - 100*x^2 + 8*x^3) + log(x)^2*(625*
x - 100*x^2 + 4*x^3) - 4*x^2*exp(5) + 4*x^3),x)

[Out]

(6*exp(-5)*(2*x - 25*log(x) + 2*x*log(x)))/(2*x - exp(5) - 25*log(x) + 2*x*log(x))

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sympy [A]  time = 0.20, size = 15, normalized size = 0.68 \begin {gather*} \frac {6}{2 x + \left (2 x - 25\right ) \log {\relax (x )} - e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-12*x*ln(x)-24*x+150)/((4*x**3-100*x**2+625*x)*ln(x)**2+((-4*x**2+50*x)*exp(5)+8*x**3-100*x**2)*ln(
x)+x*exp(5)**2-4*x**2*exp(5)+4*x**3),x)

[Out]

6/(2*x + (2*x - 25)*log(x) - exp(5))

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