3.13.39 \(\int \frac {-6 x+(2-12 x) \log (\frac {1}{3} (8-48 x))+(x-6 x^2+e^{x^2} (-x+6 x^2+2 x^3-12 x^4)) \log ^2(\frac {1}{3} (8-48 x))}{(-x^3+6 x^4) \log ^2(\frac {1}{3} (8-48 x))} \, dx\)

Optimal. Leaf size=28 \[ \frac {1}{x}-\frac {e^{x^2}}{x}+\frac {1}{x^2 \log \left (\frac {8}{3}-16 x\right )} \]

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Rubi [F]  time = 1.98, 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 {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{\left (-x^3+6 x^4\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-6*x + (2 - 12*x)*Log[(8 - 48*x)/3] + (x - 6*x^2 + E^x^2*(-x + 6*x^2 + 2*x^3 - 12*x^4))*Log[(8 - 48*x)/3]
^2)/((-x^3 + 6*x^4)*Log[(8 - 48*x)/3]^2),x]

[Out]

x^(-1) - E^x^2/x + 36/Log[8/3 - 16*x] + 6*Defer[Int][1/(x^2*Log[8/3 - 16*x]^2), x] + 36*Defer[Int][1/(x*Log[8/
3 - 16*x]^2), x] - 2*Defer[Int][1/(x^3*Log[8/3 - 16*x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-6 x+(2-12 x) \log \left (\frac {1}{3} (8-48 x)\right )+\left (x-6 x^2+e^{x^2} \left (-x+6 x^2+2 x^3-12 x^4\right )\right ) \log ^2\left (\frac {1}{3} (8-48 x)\right )}{x^3 (-1+6 x) \log ^2\left (\frac {1}{3} (8-48 x)\right )} \, dx\\ &=\int \frac {x \left (-1+e^{x^2} \left (1-2 x^2\right )\right )-\frac {6 x}{(-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )}-\frac {2}{\log \left (\frac {8}{3}-16 x\right )}}{x^3} \, dx\\ &=\int \left (-\frac {e^{x^2} \left (-1+2 x^2\right )}{x^2}+\frac {-6 x+2 \log \left (\frac {8}{3}-16 x\right )-12 x \log \left (\frac {8}{3}-16 x\right )+x \log ^2\left (\frac {8}{3}-16 x\right )-6 x^2 \log ^2\left (\frac {8}{3}-16 x\right )}{x^3 (-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )}\right ) \, dx\\ &=-\int \frac {e^{x^2} \left (-1+2 x^2\right )}{x^2} \, dx+\int \frac {-6 x+2 \log \left (\frac {8}{3}-16 x\right )-12 x \log \left (\frac {8}{3}-16 x\right )+x \log ^2\left (\frac {8}{3}-16 x\right )-6 x^2 \log ^2\left (\frac {8}{3}-16 x\right )}{x^3 (-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )} \, dx\\ &=-\frac {e^{x^2}}{x}+\int \left (-\frac {1}{x^2}-\frac {6}{x^2 (-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )}-\frac {2}{x^3 \log \left (\frac {8}{3}-16 x\right )}\right ) \, dx\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx-6 \int \frac {1}{x^2 (-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )} \, dx\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx-6 \int \left (-\frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )}-\frac {6}{x \log ^2\left (\frac {8}{3}-16 x\right )}+\frac {36}{(-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )}\right ) \, dx\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )} \, dx+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )} \, dx-216 \int \frac {1}{(-1+6 x) \log ^2\left (\frac {8}{3}-16 x\right )} \, dx\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )} \, dx+\frac {27}{2} \operatorname {Subst}\left (\int -\frac {8}{3 x \log ^2(x)} \, dx,x,\frac {8}{3}-16 x\right )+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )} \, dx\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )} \, dx+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )} \, dx-36 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\frac {8}{3}-16 x\right )\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )} \, dx+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )} \, dx-36 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {8}{3}-16 x\right )\right )\\ &=\frac {1}{x}-\frac {e^{x^2}}{x}+\frac {36}{\log \left (\frac {8}{3}-16 x\right )}-2 \int \frac {1}{x^3 \log \left (\frac {8}{3}-16 x\right )} \, dx+6 \int \frac {1}{x^2 \log ^2\left (\frac {8}{3}-16 x\right )} \, dx+36 \int \frac {1}{x \log ^2\left (\frac {8}{3}-16 x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 24, normalized size = 0.86 \begin {gather*} \frac {x-e^{x^2} x+\frac {1}{\log \left (\frac {8}{3}-16 x\right )}}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6*x + (2 - 12*x)*Log[(8 - 48*x)/3] + (x - 6*x^2 + E^x^2*(-x + 6*x^2 + 2*x^3 - 12*x^4))*Log[(8 - 48
*x)/3]^2)/((-x^3 + 6*x^4)*Log[(8 - 48*x)/3]^2),x]

[Out]

(x - E^x^2*x + Log[8/3 - 16*x]^(-1))/x^2

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fricas [A]  time = 0.69, size = 32, normalized size = 1.14 \begin {gather*} -\frac {{\left (x e^{\left (x^{2}\right )} - x\right )} \log \left (-16 \, x + \frac {8}{3}\right ) - 1}{x^{2} \log \left (-16 \, x + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-12*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^
3)/log(-16*x+8/3)^2,x, algorithm="fricas")

[Out]

-((x*e^(x^2) - x)*log(-16*x + 8/3) - 1)/(x^2*log(-16*x + 8/3))

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giac [A]  time = 0.39, size = 36, normalized size = 1.29 \begin {gather*} -\frac {x e^{\left (x^{2}\right )} \log \left (-16 \, x + \frac {8}{3}\right ) - x \log \left (-16 \, x + \frac {8}{3}\right ) - 1}{x^{2} \log \left (-16 \, x + \frac {8}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-12*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^
3)/log(-16*x+8/3)^2,x, algorithm="giac")

[Out]

-(x*e^(x^2)*log(-16*x + 8/3) - x*log(-16*x + 8/3) - 1)/(x^2*log(-16*x + 8/3))

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maple [A]  time = 0.22, size = 25, normalized size = 0.89

method	result	size
risch	\(-\frac {{\mathrm e}^{x^{2}}-1}{x}+\frac {1}{\ln \left (-16 x +\frac {8}{3}\right ) x^{2}}\)	\(25\)
default	\(\frac {1+\left (3 \ln \relax (2)-\ln \relax (3)\right ) x +x \ln \left (1-6 x \right )}{x^{2} \left (\ln \left (\frac {8}{3}\right )+\ln \left (1-6 x \right )\right )}-\frac {{\mathrm e}^{x^{2}}}{x}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*ln(-16*x+8/3)^2+(-12*x+2)*ln(-16*x+8/3)-6*x)/(6*x^4-x^3)/ln(-1
6*x+8/3)^2,x,method=_RETURNVERBOSE)

[Out]

-(exp(x^2)-1)/x+1/ln(-16*x+8/3)/x^2

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maxima [C]  time = 0.60, size = 80, normalized size = 2.86 \begin {gather*} -\frac {{\left (i \, \pi - \log \relax (3) + 3 \, \log \relax (2)\right )} x e^{\left (x^{2}\right )} + {\left (-i \, \pi + \log \relax (3) - 3 \, \log \relax (2)\right )} x + {\left (x e^{\left (x^{2}\right )} - x\right )} \log \left (6 \, x - 1\right ) - 1}{{\left (i \, \pi - \log \relax (3) + 3 \, \log \relax (2)\right )} x^{2} + x^{2} \log \left (6 \, x - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4+2*x^3+6*x^2-x)*exp(x^2)-6*x^2+x)*log(-16*x+8/3)^2+(-12*x+2)*log(-16*x+8/3)-6*x)/(6*x^4-x^
3)/log(-16*x+8/3)^2,x, algorithm="maxima")

[Out]

-((I*pi - log(3) + 3*log(2))*x*e^(x^2) + (-I*pi + log(3) - 3*log(2))*x + (x*e^(x^2) - x)*log(6*x - 1) - 1)/((I
*pi - log(3) + 3*log(2))*x^2 + x^2*log(6*x - 1))

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mupad [B]  time = 0.98, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{x^2\,\ln \left (\frac {8}{3}-16\,x\right )}-\frac {{\mathrm {e}}^{x^2}}{x}+\frac {1}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + log(8/3 - 16*x)*(12*x - 2) + log(8/3 - 16*x)^2*(exp(x^2)*(x - 6*x^2 - 2*x^3 + 12*x^4) - x + 6*x^2))
/(log(8/3 - 16*x)^2*(x^3 - 6*x^4)),x)

[Out]

1/(x^2*log(8/3 - 16*x)) - exp(x^2)/x + 1/x

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sympy [A]  time = 0.33, size = 22, normalized size = 0.79 \begin {gather*} - \frac {e^{x^{2}}}{x} + \frac {1}{x} + \frac {1}{x^{2} \log {\left (\frac {8}{3} - 16 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x**4+2*x**3+6*x**2-x)*exp(x**2)-6*x**2+x)*ln(-16*x+8/3)**2+(-12*x+2)*ln(-16*x+8/3)-6*x)/(6*x*
*4-x**3)/ln(-16*x+8/3)**2,x)

[Out]

-exp(x**2)/x + 1/x + 1/(x**2*log(8/3 - 16*x))

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