3.100.39 \(\int \frac {(-20 x-20 x^2-5 x^3) \log (2 x)+\log ^{\frac {9}{2+x}}(2 x) (216+108 x-108 x \log (2 x) \log (\log (2 x)))}{(48 x+48 x^2+12 x^3) \log (2 x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{12} (-4-5 x)+\log ^{\frac {9}{2+x}}(2 x) \]

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Rubi [F]  time = 1.94, 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 {\left (-20 x-20 x^2-5 x^3\right ) \log (2 x)+\log ^{\frac {9}{2+x}}(2 x) (216+108 x-108 x \log (2 x) \log (\log (2 x)))}{\left (48 x+48 x^2+12 x^3\right ) \log (2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((-20*x - 20*x^2 - 5*x^3)*Log[2*x] + Log[2*x]^(9/(2 + x))*(216 + 108*x - 108*x*Log[2*x]*Log[Log[2*x]]))/((
48*x + 48*x^2 + 12*x^3)*Log[2*x]),x]

[Out]

(-5*x)/12 + (9*Defer[Int][Log[2*x]^((7 - x)/(2 + x))/x, x])/2 - (9*Defer[Int][Log[2*x]^((7 - x)/(2 + x))/(2 +
x), x])/2 - 9*Defer[Int][(Log[2*x]^(9/(2 + x))*Log[Log[2*x]])/(2 + x)^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (-20 x-20 x^2-5 x^3\right ) \log (2 x)+\log ^{\frac {9}{2+x}}(2 x) (216+108 x-108 x \log (2 x) \log (\log (2 x)))}{x \left (48+48 x+12 x^2\right ) \log (2 x)} \, dx\\ &=\int \frac {\left (-20 x-20 x^2-5 x^3\right ) \log (2 x)+\log ^{\frac {9}{2+x}}(2 x) (216+108 x-108 x \log (2 x) \log (\log (2 x)))}{12 x (2+x)^2 \log (2 x)} \, dx\\ &=\frac {1}{12} \int \frac {\left (-20 x-20 x^2-5 x^3\right ) \log (2 x)+\log ^{\frac {9}{2+x}}(2 x) (216+108 x-108 x \log (2 x) \log (\log (2 x)))}{x (2+x)^2 \log (2 x)} \, dx\\ &=\frac {1}{12} \int \left (-5+\frac {108 \log ^{\frac {7-x}{2+x}}(2 x) (2+x-x \log (2 x) \log (\log (2 x)))}{x (2+x)^2}\right ) \, dx\\ &=-\frac {5 x}{12}+9 \int \frac {\log ^{\frac {7-x}{2+x}}(2 x) (2+x-x \log (2 x) \log (\log (2 x)))}{x (2+x)^2} \, dx\\ &=-\frac {5 x}{12}+9 \int \left (\frac {\log ^{\frac {7-x}{2+x}}(2 x)}{x (2+x)}-\frac {\log ^{1+\frac {7-x}{2+x}}(2 x) \log (\log (2 x))}{(2+x)^2}\right ) \, dx\\ &=-\frac {5 x}{12}+9 \int \frac {\log ^{\frac {7-x}{2+x}}(2 x)}{x (2+x)} \, dx-9 \int \frac {\log ^{1+\frac {7-x}{2+x}}(2 x) \log (\log (2 x))}{(2+x)^2} \, dx\\ &=-\frac {5 x}{12}+9 \int \left (\frac {\log ^{\frac {7-x}{2+x}}(2 x)}{2 x}-\frac {\log ^{\frac {7-x}{2+x}}(2 x)}{2 (2+x)}\right ) \, dx-9 \int \frac {\log ^{\frac {9}{2+x}}(2 x) \log (\log (2 x))}{(2+x)^2} \, dx\\ &=-\frac {5 x}{12}+\frac {9}{2} \int \frac {\log ^{\frac {7-x}{2+x}}(2 x)}{x} \, dx-\frac {9}{2} \int \frac {\log ^{\frac {7-x}{2+x}}(2 x)}{2+x} \, dx-9 \int \frac {\log ^{\frac {9}{2+x}}(2 x) \log (\log (2 x))}{(2+x)^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 18, normalized size = 0.82 \begin {gather*} -\frac {5 x}{12}+\log ^{\frac {9}{2+x}}(2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-20*x - 20*x^2 - 5*x^3)*Log[2*x] + Log[2*x]^(9/(2 + x))*(216 + 108*x - 108*x*Log[2*x]*Log[Log[2*x]
]))/((48*x + 48*x^2 + 12*x^3)*Log[2*x]),x]

[Out]

(-5*x)/12 + Log[2*x]^(9/(2 + x))

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fricas [A]  time = 0.51, size = 16, normalized size = 0.73 \begin {gather*} -\frac {5}{12} \, x + \log \left (2 \, x\right )^{\frac {9}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-108*x*log(2*x)*log(log(2*x))+108*x+216)*exp(9*log(log(2*x))/(2+x))+(-5*x^3-20*x^2-20*x)*log(2*x))
/(12*x^3+48*x^2+48*x)/log(2*x),x, algorithm="fricas")

[Out]

-5/12*x + log(2*x)^(9/(x + 2))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-108*x*log(2*x)*log(log(2*x))+108*x+216)*exp(9*log(log(2*x))/(2+x))+(-5*x^3-20*x^2-20*x)*log(2*x))
/(12*x^3+48*x^2+48*x)/log(2*x),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Simplification assuming sageVARx near 0S
implification assuming sageVARx near 0undef

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maple [A]  time = 0.08, size = 17, normalized size = 0.77

method	result	size
risch	\(-\frac {5 x}{12}+\ln \left (2 x \right )^{\frac {9}{2+x}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-108*x*ln(2*x)*ln(ln(2*x))+108*x+216)*exp(9*ln(ln(2*x))/(2+x))+(-5*x^3-20*x^2-20*x)*ln(2*x))/(12*x^3+48*
x^2+48*x)/ln(2*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.55, size = 36, normalized size = 1.64 \begin {gather*} {\left (\log \relax (2) + \log \relax (x)\right )}^{\frac {9}{x + 2}} - \frac {5 \, {\left (x^{2} + 2 \, x - 4\right )}}{12 \, {\left (x + 2\right )}} - \frac {5}{3 \, {\left (x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-108*x*log(2*x)*log(log(2*x))+108*x+216)*exp(9*log(log(2*x))/(2+x))+(-5*x^3-20*x^2-20*x)*log(2*x))
/(12*x^3+48*x^2+48*x)/log(2*x),x, algorithm="maxima")

[Out]

(log(2) + log(x))^(9/(x + 2)) - 5/12*(x^2 + 2*x - 4)/(x + 2) - 5/3/(x + 2)

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mupad [B]  time = 7.78, size = 16, normalized size = 0.73 \begin {gather*} {\ln \left (2\,x\right )}^{\frac {9}{x+2}}-\frac {5\,x}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((9*log(log(2*x)))/(x + 2))*(108*x - 108*x*log(2*x)*log(log(2*x)) + 216) - log(2*x)*(20*x + 20*x^2 + 5
*x^3))/(log(2*x)*(48*x + 48*x^2 + 12*x^3)),x)

[Out]

log(2*x)^(9/(x + 2)) - (5*x)/12

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sympy [A]  time = 0.72, size = 17, normalized size = 0.77 \begin {gather*} - \frac {5 x}{12} + e^{\frac {9 \log {\left (\log {\left (2 x \right )} \right )}}{x + 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-108*x*ln(2*x)*ln(ln(2*x))+108*x+216)*exp(9*ln(ln(2*x))/(2+x))+(-5*x**3-20*x**2-20*x)*ln(2*x))/(12
*x**3+48*x**2+48*x)/ln(2*x),x)

[Out]

-5*x/12 + exp(9*log(log(2*x))/(x + 2))

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