3.40.54 \(\int \frac {e^5 (-4+4 x^2)+e^5 (-4+3 x-4 x^2) \log (\frac {12-9 x+12 x^2}{4 x})+(8 x-6 x^2+8 x^3) \log ^2(\frac {12-9 x+12 x^2}{4 x})}{(20-15 x+20 x^2) \log ^2(\frac {12-9 x+12 x^2}{4 x})} \, dx\)

Optimal. Leaf size=25 \[ \frac {1}{5} x \left (x-\frac {e^5}{\log \left (3 \left (-\frac {3}{4}+\frac {1}{x}+x\right )\right )}\right ) \]

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Rubi [F]  time = 0.98, 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 {e^5 \left (-4+4 x^2\right )+e^5 \left (-4+3 x-4 x^2\right ) \log \left (\frac {12-9 x+12 x^2}{4 x}\right )+\left (8 x-6 x^2+8 x^3\right ) \log ^2\left (\frac {12-9 x+12 x^2}{4 x}\right )}{\left (20-15 x+20 x^2\right ) \log ^2\left (\frac {12-9 x+12 x^2}{4 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^5*(-4 + 4*x^2) + E^5*(-4 + 3*x - 4*x^2)*Log[(12 - 9*x + 12*x^2)/(4*x)] + (8*x - 6*x^2 + 8*x^3)*Log[(12
- 9*x + 12*x^2)/(4*x)]^2)/((20 - 15*x + 20*x^2)*Log[(12 - 9*x + 12*x^2)/(4*x)]^2),x]

[Out]

x^2/5 + (E^5*Defer[Int][Log[-9/4 + 3/x + 3*x]^(-2), x])/5 - (((64*I)/5)*E^5*Defer[Int][1/((3 + I*Sqrt[55] - 8*
x)*Log[-9/4 + 3/x + 3*x]^2), x])/Sqrt[55] + (3*(55 - (3*I)*Sqrt[55])*E^5*Defer[Int][1/((-3 - I*Sqrt[55] + 8*x)
*Log[-9/4 + 3/x + 3*x]^2), x])/275 - (((64*I)/5)*E^5*Defer[Int][1/((-3 + I*Sqrt[55] + 8*x)*Log[-9/4 + 3/x + 3*
x]^2), x])/Sqrt[55] + (3*(55 + (3*I)*Sqrt[55])*E^5*Defer[Int][1/((-3 + I*Sqrt[55] + 8*x)*Log[-9/4 + 3/x + 3*x]
^2), x])/275 - (E^5*Defer[Int][Log[-9/4 + 3/x + 3*x]^(-1), x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{5} \left (2 x+\frac {4 e^5 \left (-1+x^2\right )}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}-\frac {e^5}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx\\ &=\frac {1}{5} \int \left (2 x+\frac {4 e^5 \left (-1+x^2\right )}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}-\frac {e^5}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx\\ &=\frac {x^2}{5}-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{5} \left (4 e^5\right ) \int \frac {-1+x^2}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx\\ &=\frac {x^2}{5}-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{5} \left (4 e^5\right ) \int \left (\frac {1}{4 \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}+\frac {-8+3 x}{4 \left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx\\ &=\frac {x^2}{5}+\frac {1}{5} e^5 \int \frac {1}{\log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{5} e^5 \int \frac {-8+3 x}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx\\ &=\frac {x^2}{5}+\frac {1}{5} e^5 \int \left (-\frac {8}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}+\frac {3 x}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx+\frac {1}{5} e^5 \int \frac {1}{\log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx\\ &=\frac {x^2}{5}+\frac {1}{5} e^5 \int \frac {1}{\log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{5} \left (3 e^5\right ) \int \frac {x}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} \left (8 e^5\right ) \int \frac {1}{\left (4-3 x+4 x^2\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx\\ &=\frac {x^2}{5}+\frac {1}{5} e^5 \int \frac {1}{\log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{5} \left (3 e^5\right ) \int \left (\frac {1-\frac {3 i}{\sqrt {55}}}{\left (-3-i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}+\frac {1+\frac {3 i}{\sqrt {55}}}{\left (-3+i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx-\frac {1}{5} \left (8 e^5\right ) \int \left (\frac {8 i}{\sqrt {55} \left (3+i \sqrt {55}-8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}+\frac {8 i}{\sqrt {55} \left (-3+i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \, dx\\ &=\frac {x^2}{5}+\frac {1}{5} e^5 \int \frac {1}{\log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {1}{5} e^5 \int \frac {1}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx-\frac {\left (64 i e^5\right ) \int \frac {1}{\left (3+i \sqrt {55}-8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx}{5 \sqrt {55}}-\frac {\left (64 i e^5\right ) \int \frac {1}{\left (-3+i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx}{5 \sqrt {55}}+\frac {1}{275} \left (3 \left (55-3 i \sqrt {55}\right ) e^5\right ) \int \frac {1}{\left (-3-i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx+\frac {1}{275} \left (3 \left (55+3 i \sqrt {55}\right ) e^5\right ) \int \frac {1}{\left (-3+i \sqrt {55}+8 x\right ) \log ^2\left (-\frac {9}{4}+\frac {3}{x}+3 x\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 29, normalized size = 1.16 \begin {gather*} \frac {1}{5} \left (x^2-\frac {e^5 x}{\log \left (-\frac {9}{4}+\frac {3}{x}+3 x\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-4 + 4*x^2) + E^5*(-4 + 3*x - 4*x^2)*Log[(12 - 9*x + 12*x^2)/(4*x)] + (8*x - 6*x^2 + 8*x^3)*Lo
g[(12 - 9*x + 12*x^2)/(4*x)]^2)/((20 - 15*x + 20*x^2)*Log[(12 - 9*x + 12*x^2)/(4*x)]^2),x]

[Out]

(x^2 - (E^5*x)/Log[-9/4 + 3/x + 3*x])/5

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fricas [B]  time = 1.06, size = 46, normalized size = 1.84 \begin {gather*} \frac {x^{2} \log \left (\frac {3 \, {\left (4 \, x^{2} - 3 \, x + 4\right )}}{4 \, x}\right ) - x e^{5}}{5 \, \log \left (\frac {3 \, {\left (4 \, x^{2} - 3 \, x + 4\right )}}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-6*x^2+8*x)*log(1/4*(12*x^2-9*x+12)/x)^2+(-4*x^2+3*x-4)*exp(5)*log(1/4*(12*x^2-9*x+12)/x)+(4*
x^2-4)*exp(5))/(20*x^2-15*x+20)/log(1/4*(12*x^2-9*x+12)/x)^2,x, algorithm="fricas")

[Out]

1/5*(x^2*log(3/4*(4*x^2 - 3*x + 4)/x) - x*e^5)/log(3/4*(4*x^2 - 3*x + 4)/x)

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giac [B]  time = 0.33, size = 46, normalized size = 1.84 \begin {gather*} \frac {x^{2} \log \left (\frac {3 \, {\left (4 \, x^{2} - 3 \, x + 4\right )}}{4 \, x}\right ) - x e^{5}}{5 \, \log \left (\frac {3 \, {\left (4 \, x^{2} - 3 \, x + 4\right )}}{4 \, x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-6*x^2+8*x)*log(1/4*(12*x^2-9*x+12)/x)^2+(-4*x^2+3*x-4)*exp(5)*log(1/4*(12*x^2-9*x+12)/x)+(4*
x^2-4)*exp(5))/(20*x^2-15*x+20)/log(1/4*(12*x^2-9*x+12)/x)^2,x, algorithm="giac")

[Out]

1/5*(x^2*log(3/4*(4*x^2 - 3*x + 4)/x) - x*e^5)/log(3/4*(4*x^2 - 3*x + 4)/x)

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maple [A]  time = 0.25, size = 30, normalized size = 1.20

method	result	size
risch	\(\frac {x^{2}}{5}-\frac {x \,{\mathrm e}^{5}}{5 \ln \left (\frac {12 x^{2}-9 x +12}{4 x}\right )}\)	\(30\)
norman	\(\frac {-\frac {x \,{\mathrm e}^{5}}{5}+\frac {x^{2} \ln \left (\frac {12 x^{2}-9 x +12}{4 x}\right )}{5}}{\ln \left (\frac {12 x^{2}-9 x +12}{4 x}\right )}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^3-6*x^2+8*x)*ln(1/4*(12*x^2-9*x+12)/x)^2+(-4*x^2+3*x-4)*exp(5)*ln(1/4*(12*x^2-9*x+12)/x)+(4*x^2-4)*e
xp(5))/(20*x^2-15*x+20)/ln(1/4*(12*x^2-9*x+12)/x)^2,x,method=_RETURNVERBOSE)

[Out]

1/5*x^2-1/5*x*exp(5)/ln(1/4*(12*x^2-9*x+12)/x)

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maxima [B]  time = 0.67, size = 65, normalized size = 2.60 \begin {gather*} \frac {x^{2} {\left (\log \relax (3) - 2 \, \log \relax (2)\right )} + x^{2} \log \left (4 \, x^{2} - 3 \, x + 4\right ) - x^{2} \log \relax (x) - x e^{5}}{5 \, {\left (\log \relax (3) - 2 \, \log \relax (2) + \log \left (4 \, x^{2} - 3 \, x + 4\right ) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^3-6*x^2+8*x)*log(1/4*(12*x^2-9*x+12)/x)^2+(-4*x^2+3*x-4)*exp(5)*log(1/4*(12*x^2-9*x+12)/x)+(4*
x^2-4)*exp(5))/(20*x^2-15*x+20)/log(1/4*(12*x^2-9*x+12)/x)^2,x, algorithm="maxima")

[Out]

1/5*(x^2*(log(3) - 2*log(2)) + x^2*log(4*x^2 - 3*x + 4) - x^2*log(x) - x*e^5)/(log(3) - 2*log(2) + log(4*x^2 -
3*x + 4) - log(x))

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mupad [B]  time = 2.54, size = 95, normalized size = 3.80 \begin {gather*} \frac {\frac {3\,{\mathrm {e}}^5}{2}-4\,x\,{\mathrm {e}}^5}{10\,x^2-10}-\frac {x\,{\mathrm {e}}^5}{5}-\frac {\frac {x\,{\mathrm {e}}^5}{5}-\frac {x\,\ln \left (\frac {3\,x^2-\frac {9\,x}{4}+3}{x}\right )\,{\mathrm {e}}^5\,\left (4\,x^2-3\,x+4\right )}{20\,\left (x^2-1\right )}}{\ln \left (\frac {3\,x^2-\frac {9\,x}{4}+3}{x}\right )}+\frac {x^2}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*(4*x^2 - 4) + log((3*x^2 - (9*x)/4 + 3)/x)^2*(8*x - 6*x^2 + 8*x^3) - log((3*x^2 - (9*x)/4 + 3)/x)*
exp(5)*(4*x^2 - 3*x + 4))/(log((3*x^2 - (9*x)/4 + 3)/x)^2*(20*x^2 - 15*x + 20)),x)

[Out]

((3*exp(5))/2 - 4*x*exp(5))/(10*x^2 - 10) - (x*exp(5))/5 - ((x*exp(5))/5 - (x*log((3*x^2 - (9*x)/4 + 3)/x)*exp
(5)*(4*x^2 - 3*x + 4))/(20*(x^2 - 1)))/log((3*x^2 - (9*x)/4 + 3)/x) + x^2/5

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sympy [A]  time = 0.18, size = 26, normalized size = 1.04 \begin {gather*} \frac {x^{2}}{5} - \frac {x e^{5}}{5 \log {\left (\frac {3 x^{2} - \frac {9 x}{4} + 3}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**3-6*x**2+8*x)*ln(1/4*(12*x**2-9*x+12)/x)**2+(-4*x**2+3*x-4)*exp(5)*ln(1/4*(12*x**2-9*x+12)/x)
+(4*x**2-4)*exp(5))/(20*x**2-15*x+20)/ln(1/4*(12*x**2-9*x+12)/x)**2,x)

[Out]

x**2/5 - x*exp(5)/(5*log((3*x**2 - 9*x/4 + 3)/x))

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