3.74.1 \(\int \frac {60-3 x \log ^2(4 x)}{20 x \log ^2(4 x)} \, dx\)

Optimal. Leaf size=21 \[ -5-\frac {3}{4} \left (e+\frac {x}{5}+\frac {4}{\log (4 x)}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 14, normalized size of antiderivative = 0.67, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {12, 6688, 2302, 30} \begin {gather*} -\frac {3 x}{20}-\frac {3}{\log (4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(60 - 3*x*Log[4*x]^2)/(20*x*Log[4*x]^2),x]

[Out]

(-3*x)/20 - 3/Log[4*x]

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{20} \int \frac {60-3 x \log ^2(4 x)}{x \log ^2(4 x)} \, dx\\ &=\frac {1}{20} \int \left (-3+\frac {60}{x \log ^2(4 x)}\right ) \, dx\\ &=-\frac {3 x}{20}+3 \int \frac {1}{x \log ^2(4 x)} \, dx\\ &=-\frac {3 x}{20}+3 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (4 x)\right )\\ &=-\frac {3 x}{20}-\frac {3}{\log (4 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 14, normalized size = 0.67 \begin {gather*} -\frac {3 x}{20}-\frac {3}{\log (4 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(60 - 3*x*Log[4*x]^2)/(20*x*Log[4*x]^2),x]

[Out]

(-3*x)/20 - 3/Log[4*x]

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fricas [A]  time = 0.58, size = 16, normalized size = 0.76 \begin {gather*} -\frac {3 \, {\left (x \log \left (4 \, x\right ) + 20\right )}}{20 \, \log \left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*(-3*x*log(4*x)^2+60)/x/log(4*x)^2,x, algorithm="fricas")

[Out]

-3/20*(x*log(4*x) + 20)/log(4*x)

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giac [A]  time = 0.13, size = 12, normalized size = 0.57 \begin {gather*} -\frac {3}{20} \, x - \frac {3}{\log \left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*(-3*x*log(4*x)^2+60)/x/log(4*x)^2,x, algorithm="giac")

[Out]

-3/20*x - 3/log(4*x)

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maple [A]  time = 0.02, size = 13, normalized size = 0.62

method	result	size
derivativedivides	\(-\frac {3 x}{20}-\frac {3}{\ln \left (4 x \right )}\)	\(13\)
default	\(-\frac {3 x}{20}-\frac {3}{\ln \left (4 x \right )}\)	\(13\)
risch	\(-\frac {3 x}{20}-\frac {3}{\ln \left (4 x \right )}\)	\(13\)
norman	\(\frac {-3-\frac {3 x \ln \left (4 x \right )}{20}}{\ln \left (4 x \right )}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/20*(-3*x*ln(4*x)^2+60)/x/ln(4*x)^2,x,method=_RETURNVERBOSE)

[Out]

-3/20*x-3/ln(4*x)

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maxima [A]  time = 0.35, size = 12, normalized size = 0.57 \begin {gather*} -\frac {3}{20} \, x - \frac {3}{\log \left (4 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*(-3*x*log(4*x)^2+60)/x/log(4*x)^2,x, algorithm="maxima")

[Out]

-3/20*x - 3/log(4*x)

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mupad [B]  time = 4.37, size = 12, normalized size = 0.57 \begin {gather*} -\frac {3\,x}{20}-\frac {3}{\ln \left (4\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((3*x*log(4*x)^2)/20 - 3)/(x*log(4*x)^2),x)

[Out]

- (3*x)/20 - 3/log(4*x)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.57 \begin {gather*} - \frac {3 x}{20} - \frac {3}{\log {\left (4 x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/20*(-3*x*ln(4*x)**2+60)/x/ln(4*x)**2,x)

[Out]

-3*x/20 - 3/log(4*x)

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