3.6.100 \(\int \frac {-8 e+8 x^2+(-8 e+8 x^2) \log (x)+(32 x+16 x^2 \log (x)) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {8}{3} x \left (-\frac {e}{x}+x\right ) \log (4+2 x \log (x)) \]

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Rubi [F]  time = 0.73, 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 {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{6+3 x \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-8*E + 8*x^2 + (-8*E + 8*x^2)*Log[x] + (32*x + 16*x^2*Log[x])*Log[4 + 2*x*Log[x]])/(6 + 3*x*Log[x]),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{3 (2+x \log (x))} \, dx\\ &=\frac {1}{3} \int \frac {-8 e+8 x^2+\left (-8 e+8 x^2\right ) \log (x)+\left (32 x+16 x^2 \log (x)\right ) \log (4+2 x \log (x))}{2+x \log (x)} \, dx\\ &=\frac {1}{3} \int \left (-\frac {8 \left (e-x^2\right ) (1+\log (x))}{2+x \log (x)}+16 x \log (2 (2+x \log (x)))\right ) \, dx\\ &=-\left (\frac {8}{3} \int \frac {\left (e-x^2\right ) (1+\log (x))}{2+x \log (x)} \, dx\right )+\frac {16}{3} \int x \log (2 (2+x \log (x))) \, dx\\ &=\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \left (\frac {e-x^2}{x}-\frac {(-2+x) \left (-e+x^2\right )}{x (2+x \log (x))}\right ) \, dx-\frac {16}{3} \int \frac {x^2 (1+\log (x))}{2 (2+x \log (x))} \, dx\\ &=\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \frac {e-x^2}{x} \, dx+\frac {8}{3} \int \frac {(-2+x) \left (-e+x^2\right )}{x (2+x \log (x))} \, dx-\frac {8}{3} \int \frac {x^2 (1+\log (x))}{2+x \log (x)} \, dx\\ &=\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \left (\frac {e}{x}-x\right ) \, dx-\frac {8}{3} \int \left (x+\frac {(-2+x) x}{2+x \log (x)}\right ) \, dx+\frac {8}{3} \int \left (-\frac {e}{2+x \log (x)}+\frac {2 e}{x (2+x \log (x))}-\frac {2 x}{2+x \log (x)}+\frac {x^2}{2+x \log (x)}\right ) \, dx\\ &=-\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {8}{3} \int \frac {(-2+x) x}{2+x \log (x)} \, dx+\frac {8}{3} \int \frac {x^2}{2+x \log (x)} \, dx-\frac {16}{3} \int \frac {x}{2+x \log (x)} \, dx-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx\\ &=-\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))+\frac {8}{3} \int \frac {x^2}{2+x \log (x)} \, dx-\frac {8}{3} \int \left (-\frac {2 x}{2+x \log (x)}+\frac {x^2}{2+x \log (x)}\right ) \, dx-\frac {16}{3} \int \frac {x}{2+x \log (x)} \, dx-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx\\ &=-\frac {8}{3} e \log (x)+\frac {8}{3} x^2 \log (2 (2+x \log (x)))-\frac {1}{3} (8 e) \int \frac {1}{2+x \log (x)} \, dx+\frac {1}{3} (16 e) \int \frac {1}{x (2+x \log (x))} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 30, normalized size = 1.43 \begin {gather*} 8 \left (-\frac {1}{3} e \log (2+x \log (x))+\frac {1}{3} x^2 \log (4+2 x \log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8*E + 8*x^2 + (-8*E + 8*x^2)*Log[x] + (32*x + 16*x^2*Log[x])*Log[4 + 2*x*Log[x]])/(6 + 3*x*Log[x])
,x]

[Out]

8*(-1/3*(E*Log[2 + x*Log[x]]) + (x^2*Log[4 + 2*x*Log[x]])/3)

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fricas [A]  time = 0.54, size = 18, normalized size = 0.86 \begin {gather*} \frac {8}{3} \, {\left (x^{2} - e\right )} \log \left (2 \, x \log \relax (x) + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2*log(x)+32*x)*log(2*x*log(x)+4)+(-8*exp(1)+8*x^2)*log(x)-8*exp(1)+8*x^2)/(3*x*log(x)+6),x, a
lgorithm="fricas")

[Out]

8/3*(x^2 - e)*log(2*x*log(x) + 4)

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giac [A]  time = 0.37, size = 25, normalized size = 1.19 \begin {gather*} \frac {8}{3} \, x^{2} \log \left (2 \, x \log \relax (x) + 4\right ) - \frac {8}{3} \, e \log \left (x \log \relax (x) + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2*log(x)+32*x)*log(2*x*log(x)+4)+(-8*exp(1)+8*x^2)*log(x)-8*exp(1)+8*x^2)/(3*x*log(x)+6),x, a
lgorithm="giac")

[Out]

8/3*x^2*log(2*x*log(x) + 4) - 8/3*e*log(x*log(x) + 2)

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maple [A]  time = 0.17, size = 27, normalized size = 1.29

method	result	size
norman	\(-\frac {8 \,{\mathrm e} \ln \left (2 x \ln \relax (x )+4\right )}{3}+\frac {8 x^{2} \ln \left (2 x \ln \relax (x )+4\right )}{3}\)	\(27\)
risch	\(\frac {8 x^{2} \ln \left (2 x \ln \relax (x )+4\right )}{3}-\frac {8 \,{\mathrm e} \ln \relax (x )}{3}-\frac {8 \,{\mathrm e} \ln \left (\frac {2}{x}+\ln \relax (x )\right )}{3}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((16*x^2*ln(x)+32*x)*ln(2*x*ln(x)+4)+(-8*exp(1)+8*x^2)*ln(x)-8*exp(1)+8*x^2)/(3*x*ln(x)+6),x,method=_RETUR
NVERBOSE)

[Out]

-8/3*exp(1)*ln(2*x*ln(x)+4)+8/3*x^2*ln(2*x*ln(x)+4)

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maxima [B]  time = 0.74, size = 41, normalized size = 1.95 \begin {gather*} \frac {8}{3} \, x^{2} \log \relax (2) + \frac {8}{3} \, x^{2} \log \left (x \log \relax (x) + 2\right ) - \frac {8}{3} \, e \log \relax (x) - \frac {8}{3} \, e \log \left (\frac {x \log \relax (x) + 2}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x^2*log(x)+32*x)*log(2*x*log(x)+4)+(-8*exp(1)+8*x^2)*log(x)-8*exp(1)+8*x^2)/(3*x*log(x)+6),x, a
lgorithm="maxima")

[Out]

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

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mupad [B]  time = 0.87, size = 35, normalized size = 1.67 \begin {gather*} \frac {8\,x^2\,\ln \left (2\,x\,\ln \relax (x)+4\right )}{3}-\frac {8\,\mathrm {e}\,\ln \relax (x)}{3}-\frac {8\,\mathrm {e}\,\ln \left (\frac {x\,\ln \relax (x)+2}{x}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*exp(1) - log(2*x*log(x) + 4)*(32*x + 16*x^2*log(x)) + log(x)*(8*exp(1) - 8*x^2) - 8*x^2)/(3*x*log(x) +
6),x)

[Out]

(8*x^2*log(2*x*log(x) + 4))/3 - (8*exp(1)*log(x))/3 - (8*exp(1)*log((x*log(x) + 2)/x))/3

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sympy [B]  time = 0.41, size = 41, normalized size = 1.95 \begin {gather*} \frac {8 x^{2} \log {\left (2 x \log {\relax (x )} + 4 \right )}}{3} - \frac {8 e \log {\relax (x )}}{3} - \frac {8 e \log {\left (\log {\relax (x )} + \frac {2}{x} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((16*x**2*ln(x)+32*x)*ln(2*x*ln(x)+4)+(-8*exp(1)+8*x**2)*ln(x)-8*exp(1)+8*x**2)/(3*x*ln(x)+6),x)

[Out]

8*x**2*log(2*x*log(x) + 4)/3 - 8*E*log(x)/3 - 8*E*log(log(x) + 2/x)/3

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