3.57.37 \(\int \frac {12+\log ^2(\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3})}{x \log ^2(\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3})} \, dx\)

Optimal. Leaf size=21 \[ \log (x)+\frac {4}{\log \left (\frac {(-16-e+\log (3))^2}{x^3}\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 1, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {14} \begin {gather*} \frac {4}{\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3]^2)/(x*Log[(256 + 32*E + E^2 + (-32 - 2*E
)*Log[3] + Log[3]^2)/x^3]^2),x]

[Out]

Log[x] + 4/Log[(16 + E - Log[3])^2/x^3]

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} &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \frac {12+x^2}{x^2} \, dx,x,\log \left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )\right )\right )\\ &=-\left (\frac {1}{3} \operatorname {Subst}\left (\int \left (1+\frac {12}{x^2}\right ) \, dx,x,\log \left (\frac {256+32 e+e^2+(-32-2 e) \log (3)+\log ^2(3)}{x^3}\right )\right )\right )\\ &=\log (x)+\frac {4}{\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} \log (x)+\frac {4}{\log \left (\frac {(16+e-\log (3))^2}{x^3}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + Log[(256 + 32*E + E^2 + (-32 - 2*E)*Log[3] + Log[3]^2)/x^3]^2)/(x*Log[(256 + 32*E + E^2 + (-32
- 2*E)*Log[3] + Log[3]^2)/x^3]^2),x]

[Out]

Log[x] + 4/Log[(16 + E - Log[3])^2/x^3]

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fricas [B]  time = 0.58, size = 68, normalized size = 3.24 \begin {gather*} -\frac {\log \left (-\frac {2 \, {\left (e + 16\right )} \log \relax (3) - \log \relax (3)^{2} - e^{2} - 32 \, e - 256}{x^{3}}\right )^{2} - 12}{3 \, \log \left (-\frac {2 \, {\left (e + 16\right )} \log \relax (3) - \log \relax (3)^{2} - e^{2} - 32 \, e - 256}{x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12)/x/log((log(3)^2+(-2*exp(1)-3
2)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2,x, algorithm="fricas")

[Out]

-1/3*(log(-(2*(e + 16)*log(3) - log(3)^2 - e^2 - 32*e - 256)/x^3)^2 - 12)/log(-(2*(e + 16)*log(3) - log(3)^2 -
e^2 - 32*e - 256)/x^3)

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giac [B]  time = 0.18, size = 69, normalized size = 3.29 \begin {gather*} \frac {\log \left (x^{3}\right ) \log \relax (x) - \log \left (-2 \, e \log \relax (3) + \log \relax (3)^{2} + e^{2} + 32 \, e - 32 \, \log \relax (3) + 256\right ) \log \relax (x) - 4}{\log \left (x^{3}\right ) - \log \left (-2 \, e \log \relax (3) + \log \relax (3)^{2} + e^{2} + 32 \, e - 32 \, \log \relax (3) + 256\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12)/x/log((log(3)^2+(-2*exp(1)-3
2)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2,x, algorithm="giac")

[Out]

(log(x^3)*log(x) - log(-2*e*log(3) + log(3)^2 + e^2 + 32*e - 32*log(3) + 256)*log(x) - 4)/(log(x^3) - log(-2*e
*log(3) + log(3)^2 + e^2 + 32*e - 32*log(3) + 256))

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maple [A]  time = 0.03, size = 34, normalized size = 1.62

method	result	size
risch	\(\frac {4}{\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}+\ln \relax (x )\)	\(34\)
norman	\(\frac {4}{\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}+\ln \relax (x )\)	\(36\)
derivativedivides	\(-\frac {\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}{3}+\frac {4}{\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}\)	\(64\)
default	\(-\frac {\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}{3}+\frac {4}{\ln \left (\frac {\ln \relax (3)^{2}+\left (-2 \,{\mathrm e}-32\right ) \ln \relax (3)+{\mathrm e}^{2}+32 \,{\mathrm e}+256}{x^{3}}\right )}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12)/x/ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+ex
p(1)^2+32*exp(1)+256)/x^3)^2,x,method=_RETURNVERBOSE)

[Out]

4/ln((ln(3)^2+(-2*exp(1)-32)*ln(3)+exp(2)+32*exp(1)+256)/x^3)+ln(x)

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maxima [A]  time = 0.46, size = 23, normalized size = 1.10 \begin {gather*} -\frac {4}{3 \, \log \relax (x) - 2 \, \log \left (-e + \log \relax (3) - 16\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log((log(3)^2+(-2*exp(1)-32)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2+12)/x/log((log(3)^2+(-2*exp(1)-3
2)*log(3)+exp(1)^2+32*exp(1)+256)/x^3)^2,x, algorithm="maxima")

[Out]

-4/(3*log(x) - 2*log(-e + log(3) - 16)) + log(x)

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mupad [B]  time = 3.72, size = 39, normalized size = 1.86 \begin {gather*} \frac {4}{\ln \left (\frac {1}{x^3}\right )+\ln \left (32\,\mathrm {e}+{\mathrm {e}}^2-32\,\ln \relax (3)-2\,\mathrm {e}\,\ln \relax (3)+{\ln \relax (3)}^2+256\right )}-\frac {\ln \left (\frac {1}{x^3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((32*exp(1) + exp(2) + log(3)^2 - log(3)*(2*exp(1) + 32) + 256)/x^3)^2 + 12)/(x*log((32*exp(1) + exp(2
) + log(3)^2 - log(3)*(2*exp(1) + 32) + 256)/x^3)^2),x)

[Out]

4/(log(1/x^3) + log(32*exp(1) + exp(2) - 32*log(3) - 2*exp(1)*log(3) + log(3)^2 + 256)) - log(1/x^3)/3

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sympy [A]  time = 0.12, size = 36, normalized size = 1.71 \begin {gather*} \log {\relax (x )} + \frac {4}{\log {\left (\frac {\left (-32 - 2 e\right ) \log {\relax (3 )} + \log {\relax (3 )}^{2} + e^{2} + 32 e + 256}{x^{3}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln((ln(3)**2+(-2*exp(1)-32)*ln(3)+exp(1)**2+32*exp(1)+256)/x**3)**2+12)/x/ln((ln(3)**2+(-2*exp(1)-3
2)*ln(3)+exp(1)**2+32*exp(1)+256)/x**3)**2,x)

[Out]

log(x) + 4/log(((-32 - 2*E)*log(3) + log(3)**2 + exp(2) + 32*E + 256)/x**3)

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