3.5.79 \(\int \frac {4 x+(96-16 x) \log (2)}{9 x^3 \log (2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {16 \left (-3+x+x^2-\frac {x}{4 \log (2)}\right )}{9 x^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 186, 37} \begin {gather*} -\frac {(x (1-4 \log (2))+24 \log (2))^2}{108 x^2 \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + (96 - 16*x)*Log[2])/(9*x^3*Log[2]),x]

[Out]

-1/108*(x*(1 - 4*Log[2]) + 24*Log[2])^2/(x^2*Log[2]^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {4 x+(96-16 x) \log (2)}{x^3} \, dx}{9 \log (2)}\\ &=\frac {\int \frac {4 x (1-4 \log (2))+96 \log (2)}{x^3} \, dx}{9 \log (2)}\\ &=-\frac {(x (1-4 \log (2))+24 \log (2))^2}{108 x^2 \log ^2(2)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {4 (-12 \log (2)+x (-1+\log (16)))}{9 x^2 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + (96 - 16*x)*Log[2])/(9*x^3*Log[2]),x]

[Out]

(4*(-12*Log[2] + x*(-1 + Log[16])))/(9*x^2*Log[2])

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fricas [A]  time = 0.59, size = 20, normalized size = 0.91 \begin {gather*} \frac {4 \, {\left (4 \, {\left (x - 3\right )} \log \relax (2) - x\right )}}{9 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-16*x+96)*log(2)+4*x)/x^3/log(2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 22, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (4 \, x \log \relax (2) - x - 12 \, \log \relax (2)\right )}}{9 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-16*x+96)*log(2)+4*x)/x^3/log(2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 20, normalized size = 0.91

method	result	size
norman	\(\frac {-\frac {16}{3}+\frac {4 \left (4 \ln \relax (2)-1\right ) x}{9 \ln \relax (2)}}{x^{2}}\)	\(20\)
gosper	\(\frac {\frac {16 x \ln \relax (2)}{9}-\frac {16 \ln \relax (2)}{3}-\frac {4 x}{9}}{x^{2} \ln \relax (2)}\)	\(23\)
risch	\(\frac {\left (16 \ln \relax (2)-4\right ) x -48 \ln \relax (2)}{9 \ln \relax (2) x^{2}}\)	\(23\)
default	\(\frac {-\frac {48 \ln \relax (2)}{x^{2}}-\frac {4 \left (-4 \ln \relax (2)+1\right )}{x}}{9 \ln \relax (2)}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((-16*x+96)*ln(2)+4*x)/x^3/ln(2),x,method=_RETURNVERBOSE)

[Out]

(-16/3+4/9*(4*ln(2)-1)/ln(2)*x)/x^2

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maxima [A]  time = 0.76, size = 22, normalized size = 1.00 \begin {gather*} \frac {4 \, {\left (x {\left (4 \, \log \relax (2) - 1\right )} - 12 \, \log \relax (2)\right )}}{9 \, x^{2} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-16*x+96)*log(2)+4*x)/x^3/log(2),x, algorithm="maxima")

[Out]

4/9*(x*(4*log(2) - 1) - 12*log(2))/(x^2*log(2))

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mupad [B]  time = 0.43, size = 23, normalized size = 1.05 \begin {gather*} -\frac {48\,\ln \relax (2)-x\,\left (16\,\ln \relax (2)-4\right )}{9\,x^2\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x)/9 - (log(2)*(16*x - 96))/9)/(x^3*log(2)),x)

[Out]

-(48*log(2) - x*(16*log(2) - 4))/(9*x^2*log(2))

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sympy [A]  time = 0.13, size = 20, normalized size = 0.91 \begin {gather*} \frac {x \left (-4 + 16 \log {\relax (2 )}\right ) - 48 \log {\relax (2 )}}{9 x^{2} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((-16*x+96)*ln(2)+4*x)/x**3/ln(2),x)

[Out]

(x*(-4 + 16*log(2)) - 48*log(2))/(9*x**2*log(2))

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