3.11.2 \(\int \frac {125-125 \log (\frac {x}{e})+(-175+250 x) \log ^2(\frac {x}{e})}{150 x^2+(420 x^2-300 x^3) \log (\frac {x}{e})+(294 x^2-420 x^3+150 x^4) \log ^2(\frac {x}{e})} \, dx\)

Optimal. Leaf size=26 \[ \frac {5}{2 x \left (\frac {21}{5}-3 x+\frac {3}{\log \left (\frac {x}{e}\right )}\right )} \]

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Rubi [F]  time = 1.61, 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 {125-125 \log \left (\frac {x}{e}\right )+(-175+250 x) \log ^2\left (\frac {x}{e}\right )}{150 x^2+\left (420 x^2-300 x^3\right ) \log \left (\frac {x}{e}\right )+\left (294 x^2-420 x^3+150 x^4\right ) \log ^2\left (\frac {x}{e}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(125 - 125*Log[x/E] + (-175 + 250*x)*Log[x/E]^2)/(150*x^2 + (420*x^2 - 300*x^3)*Log[x/E] + (294*x^2 - 420*
x^3 + 150*x^4)*Log[x/E]^2),x]

[Out]

25/(6*(7 - 5*x)*x) + (125*Defer[Int][1/(x^2*(2 - 5*x - 7*Log[x] + 5*x*Log[x])^2), x])/6 + (3125*Defer[Int][1/(
x*(2 - 5*x - 7*Log[x] + 5*x*Log[x])^2), x])/294 + (15625*Defer[Int][1/((-7 + 5*x)^2*(2 - 5*x - 7*Log[x] + 5*x*
Log[x])^2), x])/42 - (15625*Defer[Int][1/((-7 + 5*x)*(2 - 5*x - 7*Log[x] + 5*x*Log[x])^2), x])/294 - (125*Defe
r[Int][1/(x^2*(2 - 5*x - 7*Log[x] + 5*x*Log[x])), x])/42 + (625*Defer[Int][1/(x*(2 - 5*x - 7*Log[x] + 5*x*Log[
x])), x])/294 + (3125*Defer[Int][1/((-7 + 5*x)^2*(2 - 5*x - 7*Log[x] + 5*x*Log[x])), x])/21 - (3125*Defer[Int]
[1/((-7 + 5*x)*(2 - 5*x - 7*Log[x] + 5*x*Log[x])), x])/294

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 \left (3+10 x+(9-20 x) \log (x)+(-7+10 x) \log ^2(x)\right )}{6 x^2 (2-5 x+(-7+5 x) \log (x))^2} \, dx\\ &=\frac {25}{6} \int \frac {3+10 x+(9-20 x) \log (x)+(-7+10 x) \log ^2(x)}{x^2 (2-5 x+(-7+5 x) \log (x))^2} \, dx\\ &=\frac {25}{6} \int \left (\frac {-7+10 x}{x^2 (-7+5 x)^2}+\frac {5 \left (49-45 x+25 x^2\right )}{x^2 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))^2}+\frac {5 (-7+15 x)}{x^2 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))}\right ) \, dx\\ &=\frac {25}{6} \int \frac {-7+10 x}{x^2 (-7+5 x)^2} \, dx+\frac {125}{6} \int \frac {49-45 x+25 x^2}{x^2 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))^2} \, dx+\frac {125}{6} \int \frac {-7+15 x}{x^2 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))} \, dx\\ &=\frac {25}{6 (7-5 x) x}+\frac {125}{6} \int \left (\frac {1}{x^2 (2-5 x-7 \log (x)+5 x \log (x))^2}+\frac {25}{49 x (2-5 x-7 \log (x)+5 x \log (x))^2}+\frac {125}{7 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))^2}-\frac {125}{49 (-7+5 x) (2-5 x-7 \log (x)+5 x \log (x))^2}\right ) \, dx+\frac {125}{6} \int \left (-\frac {1}{7 x^2 (2-5 x-7 \log (x)+5 x \log (x))}+\frac {5}{49 x (2-5 x-7 \log (x)+5 x \log (x))}+\frac {50}{7 (-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))}-\frac {25}{49 (-7+5 x) (2-5 x-7 \log (x)+5 x \log (x))}\right ) \, dx\\ &=\frac {25}{6 (7-5 x) x}+\frac {625}{294} \int \frac {1}{x (2-5 x-7 \log (x)+5 x \log (x))} \, dx-\frac {125}{42} \int \frac {1}{x^2 (2-5 x-7 \log (x)+5 x \log (x))} \, dx+\frac {3125}{294} \int \frac {1}{x (2-5 x-7 \log (x)+5 x \log (x))^2} \, dx-\frac {3125}{294} \int \frac {1}{(-7+5 x) (2-5 x-7 \log (x)+5 x \log (x))} \, dx+\frac {125}{6} \int \frac {1}{x^2 (2-5 x-7 \log (x)+5 x \log (x))^2} \, dx-\frac {15625}{294} \int \frac {1}{(-7+5 x) (2-5 x-7 \log (x)+5 x \log (x))^2} \, dx+\frac {3125}{21} \int \frac {1}{(-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))} \, dx+\frac {15625}{42} \int \frac {1}{(-7+5 x)^2 (2-5 x-7 \log (x)+5 x \log (x))^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.49, size = 28, normalized size = 1.08 \begin {gather*} \frac {25 (1-\log (x))}{6 x (2-5 x+(-7+5 x) \log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(125 - 125*Log[x/E] + (-175 + 250*x)*Log[x/E]^2)/(150*x^2 + (420*x^2 - 300*x^3)*Log[x/E] + (294*x^2
- 420*x^3 + 150*x^4)*Log[x/E]^2),x]

[Out]

(25*(1 - Log[x]))/(6*x*(2 - 5*x + (-7 + 5*x)*Log[x]))

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fricas [A]  time = 0.63, size = 28, normalized size = 1.08 \begin {gather*} -\frac {25 \, \log \left (x e^{\left (-1\right )}\right )}{6 \, {\left ({\left (5 \, x^{2} - 7 \, x\right )} \log \left (x e^{\left (-1\right )}\right ) - 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-420*x^3+294*x^2)*log(x/exp(1))^2+(-300
*x^3+420*x^2)*log(x/exp(1))+150*x^2),x, algorithm="fricas")

[Out]

-25/6*log(x*e^(-1))/((5*x^2 - 7*x)*log(x*e^(-1)) - 5*x)

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giac [B]  time = 0.43, size = 52, normalized size = 2.00 \begin {gather*} -\frac {125}{6 \, {\left (25 \, x^{3} \log \relax (x) - 25 \, x^{3} - 70 \, x^{2} \log \relax (x) + 45 \, x^{2} + 49 \, x \log \relax (x) - 14 \, x\right )}} - \frac {125}{42 \, {\left (5 \, x - 7\right )}} + \frac {25}{42 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-420*x^3+294*x^2)*log(x/exp(1))^2+(-300
*x^3+420*x^2)*log(x/exp(1))+150*x^2),x, algorithm="giac")

[Out]

-125/6/(25*x^3*log(x) - 25*x^3 - 70*x^2*log(x) + 45*x^2 + 49*x*log(x) - 14*x) - 125/42/(5*x - 7) + 25/42/x

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maple [A]  time = 0.05, size = 36, normalized size = 1.38

method	result	size
norman	\(-\frac {25 \ln \left ({\mathrm e}^{-1} x \right )}{6 x \left (5 \ln \left ({\mathrm e}^{-1} x \right ) x -7 \ln \left ({\mathrm e}^{-1} x \right )-5\right )}\)	\(36\)
risch	\(-\frac {25}{6 x \left (5 x -7\right )}-\frac {125}{6 x \left (5 x -7\right ) \left (5 \ln \left ({\mathrm e}^{-1} x \right ) x -7 \ln \left ({\mathrm e}^{-1} x \right )-5\right )}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((250*x-175)*ln(x/exp(1))^2-125*ln(x/exp(1))+125)/((150*x^4-420*x^3+294*x^2)*ln(x/exp(1))^2+(-300*x^3+420*
x^2)*ln(x/exp(1))+150*x^2),x,method=_RETURNVERBOSE)

[Out]

-25/6*ln(x/exp(1))/x/(5*ln(x/exp(1))*x-7*ln(x/exp(1))-5)

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maxima [A]  time = 0.38, size = 30, normalized size = 1.15 \begin {gather*} \frac {25 \, {\left (\log \relax (x) - 1\right )}}{6 \, {\left (5 \, x^{2} - {\left (5 \, x^{2} - 7 \, x\right )} \log \relax (x) - 2 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x-175)*log(x/exp(1))^2-125*log(x/exp(1))+125)/((150*x^4-420*x^3+294*x^2)*log(x/exp(1))^2+(-300
*x^3+420*x^2)*log(x/exp(1))+150*x^2),x, algorithm="maxima")

[Out]

25/6*(log(x) - 1)/(5*x^2 - (5*x^2 - 7*x)*log(x) - 2*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\left (250\,x-175\right )\,{\ln \left (x\,{\mathrm {e}}^{-1}\right )}^2-125\,\ln \left (x\,{\mathrm {e}}^{-1}\right )+125}{{\ln \left (x\,{\mathrm {e}}^{-1}\right )}^2\,\left (150\,x^4-420\,x^3+294\,x^2\right )+\ln \left (x\,{\mathrm {e}}^{-1}\right )\,\left (420\,x^2-300\,x^3\right )+150\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x*exp(-1))^2*(250*x - 175) - 125*log(x*exp(-1)) + 125)/(log(x*exp(-1))^2*(294*x^2 - 420*x^3 + 150*x^4
) + log(x*exp(-1))*(420*x^2 - 300*x^3) + 150*x^2),x)

[Out]

int((log(x*exp(-1))^2*(250*x - 175) - 125*log(x*exp(-1)) + 125)/(log(x*exp(-1))^2*(294*x^2 - 420*x^3 + 150*x^4
) + log(x*exp(-1))*(420*x^2 - 300*x^3) + 150*x^2), x)

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sympy [B]  time = 0.22, size = 41, normalized size = 1.58 \begin {gather*} - \frac {125}{- 150 x^{2} + 210 x + \left (150 x^{3} - 420 x^{2} + 294 x\right ) \log {\left (\frac {x}{e} \right )}} - \frac {25}{30 x^{2} - 42 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((250*x-175)*ln(x/exp(1))**2-125*ln(x/exp(1))+125)/((150*x**4-420*x**3+294*x**2)*ln(x/exp(1))**2+(-3
00*x**3+420*x**2)*ln(x/exp(1))+150*x**2),x)

[Out]

-125/(-150*x**2 + 210*x + (150*x**3 - 420*x**2 + 294*x)*log(x*exp(-1))) - 25/(30*x**2 - 42*x)

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