3.31.46 \(\int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} (1-6 x+9 x^2)+e^{\log ^2(x)} (-4 x+12 x^2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {5}{\frac {1}{2} (-1+x)+x+e^{-\log ^2(x)} x} \]

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Rubi [F]  time = 4.23, 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 {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30*E^(2*Log[x]^2) + E^Log[x]^2*(-20 + 40*Log[x]))/(4*x^2 + E^(2*Log[x]^2)*(1 - 6*x + 9*x^2) + E^Log[x]^2
*(-4*x + 12*x^2)),x]

[Out]

20*Defer[Int][E^Log[x]^2/((-1 + 3*x)*(-E^Log[x]^2 + 2*x + 3*E^Log[x]^2*x)^2), x] - 30*Defer[Int][E^Log[x]^2/((
-1 + 3*x)*(-E^Log[x]^2 + 2*x + 3*E^Log[x]^2*x)), x] + 40*Defer[Int][(E^Log[x]^2*Log[x])/(-E^Log[x]^2 + 2*x + 3
*E^Log[x]^2*x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {10 e^{\log ^2(x)} \left (-2-3 e^{\log ^2(x)}+4 \log (x)\right )}{\left (2 x+e^{\log ^2(x)} (-1+3 x)\right )^2} \, dx\\ &=10 \int \frac {e^{\log ^2(x)} \left (-2-3 e^{\log ^2(x)}+4 \log (x)\right )}{\left (2 x+e^{\log ^2(x)} (-1+3 x)\right )^2} \, dx\\ &=10 \int \left (-\frac {3 e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )}+\frac {2 e^{\log ^2(x)} (1-2 \log (x)+6 x \log (x))}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}\right ) \, dx\\ &=20 \int \frac {e^{\log ^2(x)} (1-2 \log (x)+6 x \log (x))}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx-30 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )} \, dx\\ &=20 \int \left (\frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}-\frac {2 e^{\log ^2(x)} \log (x)}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}+\frac {6 e^{\log ^2(x)} x \log (x)}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}\right ) \, dx-30 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )} \, dx\\ &=20 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx-30 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )} \, dx-40 \int \frac {e^{\log ^2(x)} \log (x)}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx+120 \int \frac {e^{\log ^2(x)} x \log (x)}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx\\ &=20 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx-30 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )} \, dx-40 \int \frac {e^{\log ^2(x)} \log (x)}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx+120 \int \left (\frac {e^{\log ^2(x)} \log (x)}{3 \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}+\frac {e^{\log ^2(x)} \log (x)}{3 (-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2}\right ) \, dx\\ &=20 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx-30 \int \frac {e^{\log ^2(x)}}{(-1+3 x) \left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )} \, dx+40 \int \frac {e^{\log ^2(x)} \log (x)}{\left (-e^{\log ^2(x)}+2 x+3 e^{\log ^2(x)} x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 26, normalized size = 1.13 \begin {gather*} \frac {10 e^{\log ^2(x)}}{2 x+e^{\log ^2(x)} (-1+3 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30*E^(2*Log[x]^2) + E^Log[x]^2*(-20 + 40*Log[x]))/(4*x^2 + E^(2*Log[x]^2)*(1 - 6*x + 9*x^2) + E^Lo
g[x]^2*(-4*x + 12*x^2)),x]

[Out]

(10*E^Log[x]^2)/(2*x + E^Log[x]^2*(-1 + 3*x))

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fricas [A]  time = 0.59, size = 24, normalized size = 1.04 \begin {gather*} \frac {10 \, e^{\left (\log \relax (x)^{2}\right )}}{{\left (3 \, x - 1\right )} e^{\left (\log \relax (x)^{2}\right )} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1)*exp(log(x)^2)^2+(12*x^2-4*x)*exp(l
og(x)^2)+4*x^2),x, algorithm="fricas")

[Out]

10*e^(log(x)^2)/((3*x - 1)*e^(log(x)^2) + 2*x)

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giac [A]  time = 0.28, size = 28, normalized size = 1.22 \begin {gather*} \frac {10 \, e^{\left (\log \relax (x)^{2}\right )}}{3 \, x e^{\left (\log \relax (x)^{2}\right )} + 2 \, x - e^{\left (\log \relax (x)^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1)*exp(log(x)^2)^2+(12*x^2-4*x)*exp(l
og(x)^2)+4*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 35, normalized size = 1.52

method	result	size
norman	\(\frac {20 x +30 \,{\mathrm e}^{\ln \relax (x )^{2}} x}{3 \,{\mathrm e}^{\ln \relax (x )^{2}} x -{\mathrm e}^{\ln \relax (x )^{2}}+2 x}\)	\(35\)
risch	\(\frac {10}{3 \left (x -\frac {1}{3}\right )}-\frac {20 x}{\left (3 x -1\right ) \left (3 \,{\mathrm e}^{\ln \relax (x )^{2}} x -{\mathrm e}^{\ln \relax (x )^{2}}+2 x \right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-30*exp(ln(x)^2)^2+(40*ln(x)-20)*exp(ln(x)^2))/((9*x^2-6*x+1)*exp(ln(x)^2)^2+(12*x^2-4*x)*exp(ln(x)^2)+4*
x^2),x,method=_RETURNVERBOSE)

[Out]

(20*x+30*exp(ln(x)^2)*x)/(3*exp(ln(x)^2)*x-exp(ln(x)^2)+2*x)

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maxima [A]  time = 0.92, size = 24, normalized size = 1.04 \begin {gather*} \frac {10 \, e^{\left (\log \relax (x)^{2}\right )}}{{\left (3 \, x - 1\right )} e^{\left (\log \relax (x)^{2}\right )} + 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1)*exp(log(x)^2)^2+(12*x^2-4*x)*exp(l
og(x)^2)+4*x^2),x, algorithm="maxima")

[Out]

10*e^(log(x)^2)/((3*x - 1)*e^(log(x)^2) + 2*x)

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mupad [B]  time = 2.00, size = 28, normalized size = 1.22 \begin {gather*} \frac {10\,{\mathrm {e}}^{{\ln \relax (x)}^2}}{2\,x-{\mathrm {e}}^{{\ln \relax (x)}^2}+3\,x\,{\mathrm {e}}^{{\ln \relax (x)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(30*exp(2*log(x)^2) - exp(log(x)^2)*(40*log(x) - 20))/(exp(2*log(x)^2)*(9*x^2 - 6*x + 1) - exp(log(x)^2)*
(4*x - 12*x^2) + 4*x^2),x)

[Out]

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

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sympy [A]  time = 0.34, size = 34, normalized size = 1.48 \begin {gather*} - \frac {20 x}{6 x^{2} - 2 x + \left (9 x^{2} - 6 x + 1\right ) e^{\log {\relax (x )}^{2}}} + \frac {30}{9 x - 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-30*exp(ln(x)**2)**2+(40*ln(x)-20)*exp(ln(x)**2))/((9*x**2-6*x+1)*exp(ln(x)**2)**2+(12*x**2-4*x)*ex
p(ln(x)**2)+4*x**2),x)

[Out]

-20*x/(6*x**2 - 2*x + (9*x**2 - 6*x + 1)*exp(log(x)**2)) + 30/(9*x - 3)

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