3.55.2 \(\int \frac {1+e^{2 x} (-12+24 x-6 x^2)-\log (x)}{4 x^2+e^{2 x} (-96 x+24 x^2)+e^{4 x} (576-288 x+36 x^2)+(e^{2 x} (96-24 x)-8 x) \log (x)+4 \log ^2(x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {x}{4 \left (-3 e^{2 x} (4-x)+x-\log (x)\right )} \]

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Rubi [F]  time = 7.03, 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 {1+e^{2 x} \left (-12+24 x-6 x^2\right )-\log (x)}{4 x^2+e^{2 x} \left (-96 x+24 x^2\right )+e^{4 x} \left (576-288 x+36 x^2\right )+\left (e^{2 x} (96-24 x)-8 x\right ) \log (x)+4 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(1 + E^(2*x)*(-12 + 24*x - 6*x^2) - Log[x])/(4*x^2 + E^(2*x)*(-96*x + 24*x^2) + E^(4*x)*(576 - 288*x + 36*
x^2) + (E^(2*x)*(96 - 24*x) - 8*x)*Log[x] + 4*Log[x]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1-6 e^{2 x} \left (2-4 x+x^2\right )-\log (x)}{4 \left (3 e^{2 x} (-4+x)+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {1-6 e^{2 x} \left (2-4 x+x^2\right )-\log (x)}{\left (3 e^{2 x} (-4+x)+x-\log (x)\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {2 \left (2-4 x+x^2\right )}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )}+\frac {-4+5 x-8 x^2+2 x^3+7 x \log (x)-2 x^2 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx\\ &=\frac {1}{4} \int \frac {-4+5 x-8 x^2+2 x^3+7 x \log (x)-2 x^2 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\frac {1}{2} \int \frac {2-4 x+x^2}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )} \, dx\\ &=\frac {1}{4} \int \left (-\frac {4}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {5 x}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}-\frac {8 x^2}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {2 x^3}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {7 x \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}-\frac {2 x^2 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx-\frac {1}{2} \int \left (\frac {2}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )}+\frac {x}{-12 e^{2 x}+x+3 e^{2 x} x-\log (x)}\right ) \, dx\\ &=\frac {1}{2} \int \frac {x^3}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\frac {1}{2} \int \frac {x}{-12 e^{2 x}+x+3 e^{2 x} x-\log (x)} \, dx-\frac {1}{2} \int \frac {x^2 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+\frac {5}{4} \int \frac {x}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+\frac {7}{4} \int \frac {x \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-2 \int \frac {x^2}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \left (\frac {16}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {64}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {4 x}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {x^2}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx-\frac {1}{2} \int \frac {x}{-12 e^{2 x}+x+3 e^{2 x} x-\log (x)} \, dx-\frac {1}{2} \int \left (\frac {4 \log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {16 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {x \log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx+\frac {5}{4} \int \left (\frac {1}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {4}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx+\frac {7}{4} \int \left (\frac {\log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {4 \log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx-2 \int \left (\frac {4}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {16}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}+\frac {x}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2}\right ) \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )} \, dx\\ &=\frac {1}{2} \int \frac {x^2}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\frac {1}{2} \int \frac {x}{-12 e^{2 x}+x+3 e^{2 x} x-\log (x)} \, dx-\frac {1}{2} \int \frac {x \log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+\frac {5}{4} \int \frac {1}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+\frac {7}{4} \int \frac {\log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-2 \int \frac {\log (x)}{\left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+5 \int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx+7 \int \frac {\log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-8 \int \frac {\log (x)}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )^2} \, dx-\int \frac {1}{(-4+x) \left (-12 e^{2 x}+x+3 e^{2 x} x-\log (x)\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.10, size = 23, normalized size = 0.92 \begin {gather*} \frac {x}{4 \left (3 e^{2 x} (-4+x)+x-\log (x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^(2*x)*(-12 + 24*x - 6*x^2) - Log[x])/(4*x^2 + E^(2*x)*(-96*x + 24*x^2) + E^(4*x)*(576 - 288*x
+ 36*x^2) + (E^(2*x)*(96 - 24*x) - 8*x)*Log[x] + 4*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.61, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{4 \, {\left (3 \, {\left (x - 4\right )} e^{\left (2 \, x\right )} + x - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*exp(2*x)-8*x)*log(x)+(36*x^2-288*x+576
)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 24, normalized size = 0.96 \begin {gather*} \frac {x}{4 \, {\left (3 \, x e^{\left (2 \, x\right )} + x - 12 \, e^{\left (2 \, x\right )} - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*exp(2*x)-8*x)*log(x)+(36*x^2-288*x+576
)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x, algorithm="giac")

[Out]

1/4*x/(3*x*e^(2*x) + x - 12*e^(2*x) - log(x))

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maple [A]  time = 0.03, size = 25, normalized size = 1.00

method	result	size
risch	\(\frac {x}{12 x \,{\mathrm e}^{2 x}+4 x -48 \,{\mathrm e}^{2 x}-4 \ln \relax (x )}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*ln(x)^2+((-24*x+96)*exp(2*x)-8*x)*ln(x)+(36*x^2-288*x+576)*exp(2*x
)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x,method=_RETURNVERBOSE)

[Out]

1/4*x/(3*x*exp(2*x)+x-12*exp(2*x)-ln(x))

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maxima [A]  time = 0.61, size = 20, normalized size = 0.80 \begin {gather*} \frac {x}{4 \, {\left (3 \, {\left (x - 4\right )} e^{\left (2 \, x\right )} + x - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)+(-6*x^2+24*x-12)*exp(2*x)+1)/(4*log(x)^2+((-24*x+96)*exp(2*x)-8*x)*log(x)+(36*x^2-288*x+576
)*exp(2*x)^2+(24*x^2-96*x)*exp(2*x)+4*x^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \relax (x)+{\mathrm {e}}^{2\,x}\,\left (6\,x^2-24\,x+12\right )-1}{{\mathrm {e}}^{4\,x}\,\left (36\,x^2-288\,x+576\right )-{\mathrm {e}}^{2\,x}\,\left (96\,x-24\,x^2\right )+4\,{\ln \relax (x)}^2-\ln \relax (x)\,\left (8\,x+{\mathrm {e}}^{2\,x}\,\left (24\,x-96\right )\right )+4\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x) + exp(2*x)*(6*x^2 - 24*x + 12) - 1)/(exp(4*x)*(36*x^2 - 288*x + 576) - exp(2*x)*(96*x - 24*x^2) +
4*log(x)^2 - log(x)*(8*x + exp(2*x)*(24*x - 96)) + 4*x^2),x)

[Out]

int(-(log(x) + exp(2*x)*(6*x^2 - 24*x + 12) - 1)/(exp(4*x)*(36*x^2 - 288*x + 576) - exp(2*x)*(96*x - 24*x^2) +
4*log(x)^2 - log(x)*(8*x + exp(2*x)*(24*x - 96)) + 4*x^2), x)

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sympy [A]  time = 0.39, size = 19, normalized size = 0.76 \begin {gather*} \frac {x}{4 x + \left (12 x - 48\right ) e^{2 x} - 4 \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)+(-6*x**2+24*x-12)*exp(2*x)+1)/(4*ln(x)**2+((-24*x+96)*exp(2*x)-8*x)*ln(x)+(36*x**2-288*x+576
)*exp(2*x)**2+(24*x**2-96*x)*exp(2*x)+4*x**2),x)

[Out]

x/(4*x + (12*x - 48)*exp(2*x) - 4*log(x))

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