3.23.69 \(\int \frac {x^3+(-24+3 x) \log (\log (2))}{x^3} \, dx\)

Optimal. Leaf size=15 \[ -7+e+x-\frac {3 (-4+x) \log (\log (2))}{x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 1.20, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \begin {gather*} \frac {12 \log (\log (2))}{x^2}+x-\frac {3 \log (\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3 + (-24 + 3*x)*Log[Log[2]])/x^3,x]

[Out]

x + (12*Log[Log[2]])/x^2 - (3*Log[Log[2]])/x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {24 \log (\log (2))}{x^3}+\frac {3 \log (\log (2))}{x^2}\right ) \, dx\\ &=x+\frac {12 \log (\log (2))}{x^2}-\frac {3 \log (\log (2))}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 18, normalized size = 1.20 \begin {gather*} x+\frac {12 \log (\log (2))}{x^2}-\frac {3 \log (\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3 + (-24 + 3*x)*Log[Log[2]])/x^3,x]

[Out]

x + (12*Log[Log[2]])/x^2 - (3*Log[Log[2]])/x

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fricas [A]  time = 0.65, size = 16, normalized size = 1.07 \begin {gather*} \frac {x^{3} - 3 \, {\left (x - 4\right )} \log \left (\log \relax (2)\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-24)*log(log(2))+x^3)/x^3,x, algorithm="fricas")

[Out]

(x^3 - 3*(x - 4)*log(log(2)))/x^2

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giac [A]  time = 0.31, size = 18, normalized size = 1.20 \begin {gather*} x - \frac {3 \, {\left (x \log \left (\log \relax (2)\right ) - 4 \, \log \left (\log \relax (2)\right )\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-24)*log(log(2))+x^3)/x^3,x, algorithm="giac")

[Out]

x - 3*(x*log(log(2)) - 4*log(log(2)))/x^2

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maple [A]  time = 0.04, size = 19, normalized size = 1.27

method	result	size
default	\(x -\frac {3 \ln \left (\ln \relax (2)\right )}{x}+\frac {12 \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(19\)
risch	\(x +\frac {-3 x \ln \left (\ln \relax (2)\right )+12 \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(19\)
norman	\(\frac {x^{3}-3 x \ln \left (\ln \relax (2)\right )+12 \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(20\)
gosper	\(-\frac {-x^{3}+3 x \ln \left (\ln \relax (2)\right )-12 \ln \left (\ln \relax (2)\right )}{x^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x-24)*ln(ln(2))+x^3)/x^3,x,method=_RETURNVERBOSE)

[Out]

x-3*ln(ln(2))/x+12*ln(ln(2))/x^2

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maxima [A]  time = 0.44, size = 18, normalized size = 1.20 \begin {gather*} x - \frac {3 \, {\left (x \log \left (\log \relax (2)\right ) - 4 \, \log \left (\log \relax (2)\right )\right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-24)*log(log(2))+x^3)/x^3,x, algorithm="maxima")

[Out]

x - 3*(x*log(log(2)) - 4*log(log(2)))/x^2

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mupad [B]  time = 0.04, size = 18, normalized size = 1.20 \begin {gather*} x+\frac {12\,\ln \left (\ln \relax (2)\right )-3\,x\,\ln \left (\ln \relax (2)\right )}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(2))*(3*x - 24) + x^3)/x^3,x)

[Out]

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

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sympy [A]  time = 0.13, size = 19, normalized size = 1.27 \begin {gather*} x + \frac {- 3 x \log {\left (\log {\relax (2 )} \right )} + 12 \log {\left (\log {\relax (2 )} \right )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-24)*ln(ln(2))+x**3)/x**3,x)

[Out]

x + (-3*x*log(log(2)) + 12*log(log(2)))/x**2

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