3.64.33 \(\int \frac {-153-5 x^2+48 \log (4)-4 \log ^2(4)}{3 x^2} \, dx\)

Optimal. Leaf size=28 \[ 2+\frac {3 \left (1-x^2+\left (-4+\frac {2}{3} (-x+\log (4))\right )^2\right )}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 14} \begin {gather*} \frac {153+4 \log ^2(4)-48 \log (4)}{3 x}-\frac {5 x}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-153 - 5*x^2 + 48*Log[4] - 4*Log[4]^2)/(3*x^2),x]

[Out]

(-5*x)/3 + (153 - 48*Log[4] + 4*Log[4]^2)/(3*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 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} &=\frac {1}{3} \int \frac {-153-5 x^2+48 \log (4)-4 \log ^2(4)}{x^2} \, dx\\ &=\frac {1}{3} \int \left (-5+\frac {-153+48 \log (4)-4 \log ^2(4)}{x^2}\right ) \, dx\\ &=-\frac {5 x}{3}+\frac {153-48 \log (4)+4 \log ^2(4)}{3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.89 \begin {gather*} \frac {1}{3} \left (-5 x-\frac {-153+48 \log (4)-4 \log ^2(4)}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-153 - 5*x^2 + 48*Log[4] - 4*Log[4]^2)/(3*x^2),x]

[Out]

(-5*x - (-153 + 48*Log[4] - 4*Log[4]^2)/x)/3

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fricas [A]  time = 0.55, size = 22, normalized size = 0.79 \begin {gather*} -\frac {5 \, x^{2} - 16 \, \log \relax (2)^{2} + 96 \, \log \relax (2) - 153}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-16*log(2)^2+96*log(2)-5*x^2-153)/x^2,x, algorithm="fricas")

[Out]

-1/3*(5*x^2 - 16*log(2)^2 + 96*log(2) - 153)/x

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giac [A]  time = 0.13, size = 21, normalized size = 0.75 \begin {gather*} -\frac {5}{3} \, x + \frac {16 \, \log \relax (2)^{2} - 96 \, \log \relax (2) + 153}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-16*log(2)^2+96*log(2)-5*x^2-153)/x^2,x, algorithm="giac")

[Out]

-5/3*x + 1/3*(16*log(2)^2 - 96*log(2) + 153)/x

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maple [A]  time = 0.06, size = 22, normalized size = 0.79

method	result	size
default	\(-\frac {5 x}{3}-\frac {-16 \ln \relax (2)^{2}+96 \ln \relax (2)-153}{3 x}\)	\(22\)
norman	\(\frac {-\frac {5 x^{2}}{3}+\frac {16 \ln \relax (2)^{2}}{3}-32 \ln \relax (2)+51}{x}\)	\(22\)
gosper	\(\frac {16 \ln \relax (2)^{2}-5 x^{2}-96 \ln \relax (2)+153}{3 x}\)	\(23\)
risch	\(-\frac {5 x}{3}+\frac {16 \ln \relax (2)^{2}}{3 x}-\frac {32 \ln \relax (2)}{x}+\frac {51}{x}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-16*ln(2)^2+96*ln(2)-5*x^2-153)/x^2,x,method=_RETURNVERBOSE)

[Out]

-5/3*x-1/3*(-16*ln(2)^2+96*ln(2)-153)/x

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maxima [A]  time = 0.36, size = 21, normalized size = 0.75 \begin {gather*} -\frac {5}{3} \, x + \frac {16 \, \log \relax (2)^{2} - 96 \, \log \relax (2) + 153}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-16*log(2)^2+96*log(2)-5*x^2-153)/x^2,x, algorithm="maxima")

[Out]

-5/3*x + 1/3*(16*log(2)^2 - 96*log(2) + 153)/x

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mupad [B]  time = 0.05, size = 20, normalized size = 0.71 \begin {gather*} \frac {\frac {16\,{\ln \relax (2)}^2}{3}-32\,\ln \relax (2)+51}{x}-\frac {5\,x}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((16*log(2)^2)/3 - 32*log(2) + (5*x^2)/3 + 51)/x^2,x)

[Out]

((16*log(2)^2)/3 - 32*log(2) + 51)/x - (5*x)/3

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sympy [A]  time = 0.11, size = 22, normalized size = 0.79 \begin {gather*} - \frac {5 x}{3} - \frac {-153 - 16 \log {\relax (2 )}^{2} + 96 \log {\relax (2 )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-16*ln(2)**2+96*ln(2)-5*x**2-153)/x**2,x)

[Out]

-5*x/3 - (-153 - 16*log(2)**2 + 96*log(2))/(3*x)

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