3.25.67 \(\int \frac {1125 \log (2 x)-1125 \log ^2(2 x)}{8 x^3} \, dx\)

Optimal. Leaf size=13 \[ \frac {1125 \log ^2(2 x)}{16 x^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 14, 2304, 2305} \begin {gather*} \frac {1125 \log ^2(2 x)}{16 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1125*Log[2*x] - 1125*Log[2*x]^2)/(8*x^3),x]

[Out]

(1125*Log[2*x]^2)/(16*x^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 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]]

Rule 2304

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

Rule 2305

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {1125 \log (2 x)-1125 \log ^2(2 x)}{x^3} \, dx\\ &=\frac {1}{8} \int \left (\frac {1125 \log (2 x)}{x^3}-\frac {1125 \log ^2(2 x)}{x^3}\right ) \, dx\\ &=\frac {1125}{8} \int \frac {\log (2 x)}{x^3} \, dx-\frac {1125}{8} \int \frac {\log ^2(2 x)}{x^3} \, dx\\ &=-\frac {1125}{32 x^2}-\frac {1125 \log (2 x)}{16 x^2}+\frac {1125 \log ^2(2 x)}{16 x^2}-\frac {1125}{8} \int \frac {\log (2 x)}{x^3} \, dx\\ &=\frac {1125 \log ^2(2 x)}{16 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {1125 \log ^2(2 x)}{16 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1125*Log[2*x] - 1125*Log[2*x]^2)/(8*x^3),x]

[Out]

(1125*Log[2*x]^2)/(16*x^2)

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fricas [A]  time = 0.78, size = 11, normalized size = 0.85 \begin {gather*} \frac {1125 \, \log \left (2 \, x\right )^{2}}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1125*log(2*x)^2+1125*log(2*x))/x^3,x, algorithm="fricas")

[Out]

1125/16*log(2*x)^2/x^2

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giac [A]  time = 0.19, size = 11, normalized size = 0.85 \begin {gather*} \frac {1125 \, \log \left (2 \, x\right )^{2}}{16 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1125*log(2*x)^2+1125*log(2*x))/x^3,x, algorithm="giac")

[Out]

1125/16*log(2*x)^2/x^2

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maple [A]  time = 0.02, size = 12, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {1125 \ln \left (2 x \right )^{2}}{16 x^{2}}\)	\(12\)
default	\(\frac {1125 \ln \left (2 x \right )^{2}}{16 x^{2}}\)	\(12\)
norman	\(\frac {1125 \ln \left (2 x \right )^{2}}{16 x^{2}}\)	\(12\)
risch	\(\frac {1125 \ln \left (2 x \right )^{2}}{16 x^{2}}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*(-1125*ln(2*x)^2+1125*ln(2*x))/x^3,x,method=_RETURNVERBOSE)

[Out]

1125/16/x^2*ln(2*x)^2

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maxima [B]  time = 0.54, size = 35, normalized size = 2.69 \begin {gather*} \frac {1125 \, {\left (2 \, \log \left (2 \, x\right )^{2} + 2 \, \log \left (2 \, x\right ) + 1\right )}}{32 \, x^{2}} - \frac {1125 \, {\left (2 \, \log \left (2 \, x\right ) + 1\right )}}{32 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1125*log(2*x)^2+1125*log(2*x))/x^3,x, algorithm="maxima")

[Out]

1125/32*(2*log(2*x)^2 + 2*log(2*x) + 1)/x^2 - 1125/32*(2*log(2*x) + 1)/x^2

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mupad [B]  time = 1.44, size = 11, normalized size = 0.85 \begin {gather*} \frac {1125\,{\ln \left (2\,x\right )}^2}{16\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1125*log(2*x))/8 - (1125*log(2*x)^2)/8)/x^3,x)

[Out]

(1125*log(2*x)^2)/(16*x^2)

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sympy [A]  time = 0.11, size = 12, normalized size = 0.92 \begin {gather*} \frac {1125 \log {\left (2 x \right )}^{2}}{16 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*(-1125*ln(2*x)**2+1125*ln(2*x))/x**3,x)

[Out]

1125*log(2*x)**2/(16*x**2)

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