3.57.60 \(\int \frac {-2+2 x^4 \log ^2(\frac {1+200 \log (3)+8 \log (3) \log (5)}{40 \log (3)})}{x^3} \, dx\)

Optimal. Leaf size=28 \[ \left (-\frac {1}{x}+x \log \left (5+\frac {1}{5} \left (\frac {1}{8 \log (3)}+\log (5)\right )\right )\right )^2 \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 1, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {14} \begin {gather*} \frac {1}{x^2}+x^2 \log ^2\left (5+\frac {1}{40 \log (3)}+\frac {\log (5)}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + 2*x^4*Log[(1 + 200*Log[3] + 8*Log[3]*Log[5])/(40*Log[3])]^2)/x^3,x]

[Out]

x^(-2) + x^2*Log[5 + 1/(40*Log[3]) + Log[5]/5]^2

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} &=\int \left (-\frac {2}{x^3}+2 x \log ^2\left (5+\frac {1}{40 \log (3)}+\frac {\log (5)}{5}\right )\right ) \, dx\\ &=\frac {1}{x^2}+x^2 \log ^2\left (5+\frac {1}{40 \log (3)}+\frac {\log (5)}{5}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.96 \begin {gather*} \frac {1}{x^2}+x^2 \log ^2\left (5+\frac {1}{40 \log (3)}+\frac {\log (5)}{5}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + 2*x^4*Log[(1 + 200*Log[3] + 8*Log[3]*Log[5])/(40*Log[3])]^2)/x^3,x]

[Out]

x^(-2) + x^2*Log[5 + 1/(40*Log[3]) + Log[5]/5]^2

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fricas [A]  time = 0.61, size = 31, normalized size = 1.11 \begin {gather*} \frac {x^{4} \log \left (\frac {8 \, \log \relax (5) \log \relax (3) + 200 \, \log \relax (3) + 1}{40 \, \log \relax (3)}\right )^{2} + 1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4*log(1/40*(8*log(3)*log(5)+200*log(3)+1)/log(3))^2-2)/x^3,x, algorithm="fricas")

[Out]

(x^4*log(1/40*(8*log(5)*log(3) + 200*log(3) + 1)/log(3))^2 + 1)/x^2

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giac [A]  time = 0.13, size = 29, normalized size = 1.04 \begin {gather*} x^{2} \log \left (\frac {8 \, \log \relax (5) \log \relax (3) + 200 \, \log \relax (3) + 1}{40 \, \log \relax (3)}\right )^{2} + \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4*log(1/40*(8*log(3)*log(5)+200*log(3)+1)/log(3))^2-2)/x^3,x, algorithm="giac")

[Out]

x^2*log(1/40*(8*log(5)*log(3) + 200*log(3) + 1)/log(3))^2 + 1/x^2

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maple [A]  time = 0.10, size = 30, normalized size = 1.07

method	result	size
default	\(\ln \left (\frac {8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1}{40 \ln \relax (3)}\right )^{2} x^{2}+\frac {1}{x^{2}}\)	\(30\)
gosper	\(\frac {x^{4} \ln \left (\frac {8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1}{40 \ln \relax (3)}\right )^{2}+1}{x^{2}}\)	\(32\)
norman	\(\frac {1+\left (\ln \left (40\right )^{2}-2 \ln \left (40\right ) \ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right )+2 \ln \left (40\right ) \ln \left (\ln \relax (3)\right )+\ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right )^{2}-2 \ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right ) \ln \left (\ln \relax (3)\right )+\ln \left (\ln \relax (3)\right )^{2}\right ) x^{4}}{x^{2}}\)	\(78\)
risch	\(\ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right )^{2} x^{2}-6 \ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right ) \ln \relax (2) x^{2}-2 \ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right ) \ln \relax (5) x^{2}-2 \ln \left (8 \ln \relax (3) \ln \relax (5)+200 \ln \relax (3)+1\right ) \ln \left (\ln \relax (3)\right ) x^{2}+9 x^{2} \ln \relax (2)^{2}+6 x^{2} \ln \relax (2) \ln \relax (5)+6 \ln \relax (2) \ln \left (\ln \relax (3)\right ) x^{2}+x^{2} \ln \relax (5)^{2}+2 \ln \relax (5) \ln \left (\ln \relax (3)\right ) x^{2}+\ln \left (\ln \relax (3)\right )^{2} x^{2}+\frac {1}{x^{2}}\)	\(140\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4*ln(1/40*(8*ln(3)*ln(5)+200*ln(3)+1)/ln(3))^2-2)/x^3,x,method=_RETURNVERBOSE)

[Out]

ln(1/40*(8*ln(3)*ln(5)+200*ln(3)+1)/ln(3))^2*x^2+1/x^2

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maxima [A]  time = 0.36, size = 29, normalized size = 1.04 \begin {gather*} x^{2} \log \left (\frac {8 \, \log \relax (5) \log \relax (3) + 200 \, \log \relax (3) + 1}{40 \, \log \relax (3)}\right )^{2} + \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^4*log(1/40*(8*log(3)*log(5)+200*log(3)+1)/log(3))^2-2)/x^3,x, algorithm="maxima")

[Out]

x^2*log(1/40*(8*log(5)*log(3) + 200*log(3) + 1)/log(3))^2 + 1/x^2

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mupad [B]  time = 0.09, size = 47, normalized size = 1.68 \begin {gather*} x^2\,\left ({\ln \left (5\,\ln \relax (3)+\frac {\ln \relax (3)\,\ln \relax (5)}{5}+\frac {1}{40}\right )}^2-2\,\ln \left (5\,\ln \relax (3)+\frac {\ln \relax (3)\,\ln \relax (5)}{5}+\frac {1}{40}\right )\,\ln \left (\ln \relax (3)\right )+{\ln \left (\ln \relax (3)\right )}^2\right )+\frac {1}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^4*log((5*log(3) + (log(3)*log(5))/5 + 1/40)/log(3))^2 - 2)/x^3,x)

[Out]

x^2*(log(5*log(3) + (log(3)*log(5))/5 + 1/40)^2 - 2*log(5*log(3) + (log(3)*log(5))/5 + 1/40)*log(log(3)) + log
(log(3))^2) + 1/x^2

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sympy [B]  time = 0.12, size = 60, normalized size = 2.14 \begin {gather*} x^{2} \left (- 2 \log {\left (\frac {1}{40} + \frac {\log {\relax (3 )} \log {\relax (5 )}}{5} + 5 \log {\relax (3 )} \right )} \log {\left (\log {\relax (3 )} \right )} + \log {\left (\log {\relax (3 )} \right )}^{2} + \log {\left (\frac {1}{40} + \frac {\log {\relax (3 )} \log {\relax (5 )}}{5} + 5 \log {\relax (3 )} \right )}^{2}\right ) + \frac {1}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**4*ln(1/40*(8*ln(3)*ln(5)+200*ln(3)+1)/ln(3))**2-2)/x**3,x)

[Out]

x**2*(-2*log(1/40 + log(3)*log(5)/5 + 5*log(3))*log(log(3)) + log(log(3))**2 + log(1/40 + log(3)*log(5)/5 + 5*
log(3))**2) + x**(-2)

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