3.99.96 \(\int \frac {-36 x+42 x^2-12 x^3+e^2 (-18+24 x-6 x^2)+(-72 x+90 x^2-24 x^3) \log (x)}{e^4 (9 x^2-12 x^3+4 x^4)+e^2 (36 x^3-48 x^4+16 x^5) \log (x)+(36 x^4-48 x^5+16 x^6) \log ^2(x)} \, dx\)

Optimal. Leaf size=33 \[ \frac {\frac {1}{x}+\frac {3-x}{3 x-2 x^2}}{e^2+2 x \log (x)} \]

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Rubi [F]  time = 2.37, 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 {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-36*x + 42*x^2 - 12*x^3 + E^2*(-18 + 24*x - 6*x^2) + (-72*x + 90*x^2 - 24*x^3)*Log[x])/(E^4*(9*x^2 - 12*x
^3 + 4*x^4) + E^2*(36*x^3 - 48*x^4 + 16*x^5)*Log[x] + (36*x^4 - 48*x^5 + 16*x^6)*Log[x]^2),x]

[Out]

2*E^2*Defer[Int][1/(x^2*(E^2 + 2*x*Log[x])^2), x] - ((12 - E^2)*Defer[Int][1/(x*(E^2 + 2*x*Log[x])^2), x])/3 +
(2*(3 - E^2)*Defer[Int][1/((-3 + 2*x)*(E^2 + 2*x*Log[x])^2), x])/3 - 4*Defer[Int][1/(x^2*(E^2 + 2*x*Log[x])),
x] - Defer[Int][1/(x*(E^2 + 2*x*Log[x])), x]/3 + 2*Defer[Int][1/((-3 + 2*x)^2*(E^2 + 2*x*Log[x])), x] + (2*De
fer[Int][1/((-3 + 2*x)*(E^2 + 2*x*Log[x])), x])/3

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {6 \left (-e^2 \left (3-4 x+x^2\right )-x \left (6-7 x+2 x^2\right )-x \left (12-15 x+4 x^2\right ) \log (x)\right )}{(3-2 x)^2 x^2 \left (e^2+2 x \log (x)\right )^2} \, dx\\ &=6 \int \frac {-e^2 \left (3-4 x+x^2\right )-x \left (6-7 x+2 x^2\right )-x \left (12-15 x+4 x^2\right ) \log (x)}{(3-2 x)^2 x^2 \left (e^2+2 x \log (x)\right )^2} \, dx\\ &=6 \int \left (\frac {\left (e^2-2 x\right ) (-2+x)}{2 x^2 (-3+2 x) \left (e^2+2 x \log (x)\right )^2}+\frac {-12+15 x-4 x^2}{2 x^2 (-3+2 x)^2 \left (e^2+2 x \log (x)\right )}\right ) \, dx\\ &=3 \int \frac {\left (e^2-2 x\right ) (-2+x)}{x^2 (-3+2 x) \left (e^2+2 x \log (x)\right )^2} \, dx+3 \int \frac {-12+15 x-4 x^2}{x^2 (-3+2 x)^2 \left (e^2+2 x \log (x)\right )} \, dx\\ &=3 \int \left (\frac {2 e^2}{3 x^2 \left (e^2+2 x \log (x)\right )^2}+\frac {-12+e^2}{9 x \left (e^2+2 x \log (x)\right )^2}-\frac {2 \left (-3+e^2\right )}{9 (-3+2 x) \left (e^2+2 x \log (x)\right )^2}\right ) \, dx+3 \int \left (-\frac {4}{3 x^2 \left (e^2+2 x \log (x)\right )}-\frac {1}{9 x \left (e^2+2 x \log (x)\right )}+\frac {2}{3 (-3+2 x)^2 \left (e^2+2 x \log (x)\right )}+\frac {2}{9 (-3+2 x) \left (e^2+2 x \log (x)\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{x \left (e^2+2 x \log (x)\right )} \, dx\right )+\frac {2}{3} \int \frac {1}{(-3+2 x) \left (e^2+2 x \log (x)\right )} \, dx+2 \int \frac {1}{(-3+2 x)^2 \left (e^2+2 x \log (x)\right )} \, dx-4 \int \frac {1}{x^2 \left (e^2+2 x \log (x)\right )} \, dx+\left (2 e^2\right ) \int \frac {1}{x^2 \left (e^2+2 x \log (x)\right )^2} \, dx+\frac {1}{3} \left (2 \left (3-e^2\right )\right ) \int \frac {1}{(-3+2 x) \left (e^2+2 x \log (x)\right )^2} \, dx+\frac {1}{3} \left (-12+e^2\right ) \int \frac {1}{x \left (e^2+2 x \log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 28, normalized size = 0.85 \begin {gather*} -\frac {3 (2-x)}{x (-3+2 x) \left (e^2+2 x \log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36*x + 42*x^2 - 12*x^3 + E^2*(-18 + 24*x - 6*x^2) + (-72*x + 90*x^2 - 24*x^3)*Log[x])/(E^4*(9*x^2
- 12*x^3 + 4*x^4) + E^2*(36*x^3 - 48*x^4 + 16*x^5)*Log[x] + (36*x^4 - 48*x^5 + 16*x^6)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.60, size = 35, normalized size = 1.06 \begin {gather*} \frac {3 \, {\left (x - 2\right )}}{{\left (2 \, x^{2} - 3 \, x\right )} e^{2} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42*x^2-36*x)/((16*x^6-48*x^5+36*x^4)*lo
g(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*log(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42*x^2-36*x)/((16*x^6-48*x^5+36*x^4)*lo
g(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*log(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.23, size = 26, normalized size = 0.79

method	result	size
risch	\(\frac {3 x -6}{x \left (2 x -3\right ) \left (2 x \ln \relax (x )+{\mathrm e}^{2}\right )}\)	\(26\)
norman	\(\frac {3 x -6}{x \left (2 x -3\right ) \left (2 x \ln \relax (x )+{\mathrm e}^{2}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-24*x^3+90*x^2-72*x)*ln(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42*x^2-36*x)/((16*x^6-48*x^5+36*x^4)*ln(x)^2+(
16*x^5-48*x^4+36*x^3)*exp(2)*ln(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x,method=_RETURNVERBOSE)

[Out]

3*(x-2)/x/(2*x-3)/(2*x*ln(x)+exp(2))

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maxima [A]  time = 0.40, size = 35, normalized size = 1.06 \begin {gather*} \frac {3 \, {\left (x - 2\right )}}{2 \, x^{2} e^{2} - 3 \, x e^{2} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42*x^2-36*x)/((16*x^6-48*x^5+36*x^4)*lo
g(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*log(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 6.14, size = 25, normalized size = 0.76 \begin {gather*} \frac {3\,\left (x-2\right )}{x\,\left (2\,x-3\right )\,\left ({\mathrm {e}}^2+2\,x\,\ln \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(36*x + exp(2)*(6*x^2 - 24*x + 18) - 42*x^2 + 12*x^3 + log(x)*(72*x - 90*x^2 + 24*x^3))/(log(x)^2*(36*x^4
- 48*x^5 + 16*x^6) + exp(4)*(9*x^2 - 12*x^3 + 4*x^4) + exp(2)*log(x)*(36*x^3 - 48*x^4 + 16*x^5)),x)

[Out]

(3*(x - 2))/(x*(2*x - 3)*(exp(2) + 2*x*log(x)))

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sympy [A]  time = 0.19, size = 32, normalized size = 0.97 \begin {gather*} \frac {3 x - 6}{2 x^{2} e^{2} - 3 x e^{2} + \left (4 x^{3} - 6 x^{2}\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-24*x**3+90*x**2-72*x)*ln(x)+(-6*x**2+24*x-18)*exp(2)-12*x**3+42*x**2-36*x)/((16*x**6-48*x**5+36*x
**4)*ln(x)**2+(16*x**5-48*x**4+36*x**3)*exp(2)*ln(x)+(4*x**4-12*x**3+9*x**2)*exp(2)**2),x)

[Out]

(3*x - 6)/(2*x**2*exp(2) - 3*x*exp(2) + (4*x**3 - 6*x**2)*log(x))

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