3.99.76 \(\int \frac {4 e^5 x+24 e^5 \log (3)-x \log (\frac {1}{25} (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)))}{x^3} \, dx\)

Optimal. Leaf size=33 \[ \frac {-e^5 \left (4+\frac {12 \log (3)}{x}\right )+\log \left (\frac {1}{25} (-\log (4)+\log (5))^2\right )}{x} \]

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {6, 37} \begin {gather*} -\frac {\left (x \left (4 e^5+\log (25)-2 \log \left (\log \left (\frac {5}{4}\right )\right )\right )+24 e^5 \log (3)\right )^2}{48 e^5 x^2 \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^5*x + 24*E^5*Log[3] - x*Log[(Log[4]^2 - 2*Log[4]*Log[5] + Log[5]^2)/25])/x^3,x]

[Out]

-1/48*(24*E^5*Log[3] + x*(4*E^5 + Log[25] - 2*Log[Log[5/4]]))^2/(E^5*x^2*Log[3])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {24 e^5 \log (3)+x \left (4 e^5-\log \left (\frac {1}{25} \left (\log ^2(4)-2 \log (4) \log (5)+\log ^2(5)\right )\right )\right )}{x^3} \, dx\\ &=-\frac {\left (24 e^5 \log (3)+x \left (4 e^5+\log (25)-2 \log \left (\log \left (\frac {5}{4}\right )\right )\right )\right )^2}{48 e^5 x^2 \log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 29, normalized size = 0.88 \begin {gather*} -\frac {2 e^5 (2 x+\log (729))+x \left (\log (25)-2 \log \left (\log \left (\frac {5}{4}\right )\right )\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^5*x + 24*E^5*Log[3] - x*Log[(Log[4]^2 - 2*Log[4]*Log[5] + Log[5]^2)/25])/x^3,x]

[Out]

-((2*E^5*(2*x + Log[729]) + x*(Log[25] - 2*Log[Log[5/4]]))/x^2)

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fricas [A]  time = 0.77, size = 40, normalized size = 1.21 \begin {gather*} -\frac {4 \, x e^{5} + 12 \, e^{5} \log \relax (3) - x \log \left (\frac {1}{25} \, \log \relax (5)^{2} - \frac {4}{25} \, \log \relax (5) \log \relax (2) + \frac {4}{25} \, \log \relax (2)^{2}\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5)*log(3)+4*x*exp(5))/x^3,x, algorith
m="fricas")

[Out]

-(4*x*e^5 + 12*e^5*log(3) - x*log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/25*log(2)^2))/x^2

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giac [A]  time = 0.27, size = 40, normalized size = 1.21 \begin {gather*} -\frac {4 \, x e^{5} + 12 \, e^{5} \log \relax (3) - x \log \left (\frac {1}{25} \, \log \relax (5)^{2} - \frac {4}{25} \, \log \relax (5) \log \relax (2) + \frac {4}{25} \, \log \relax (2)^{2}\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5)*log(3)+4*x*exp(5))/x^3,x, algorith
m="giac")

[Out]

-(4*x*e^5 + 12*e^5*log(3) - x*log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/25*log(2)^2))/x^2

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maple [A]  time = 0.07, size = 33, normalized size = 1.00

method	result	size
risch	\(\frac {\left (-2 \ln \relax (5)+2 \ln \left (\ln \relax (5)-2 \ln \relax (2)\right )-4 \,{\mathrm e}^{5}\right ) x -12 \,{\mathrm e}^{5} \ln \relax (3)}{x^{2}}\)	\(33\)
gosper	\(-\frac {12 \,{\mathrm e}^{5} \ln \relax (3)+4 x \,{\mathrm e}^{5}-x \ln \left (\frac {\ln \relax (5)^{2}}{25}-\frac {4 \ln \relax (2) \ln \relax (5)}{25}+\frac {4 \ln \relax (2)^{2}}{25}\right )}{x^{2}}\)	\(41\)
norman	\(\frac {\left (-2 \ln \relax (5)+\ln \left (\ln \relax (5)^{2}-4 \ln \relax (2) \ln \relax (5)+4 \ln \relax (2)^{2}\right )-4 \,{\mathrm e}^{5}\right ) x -12 \,{\mathrm e}^{5} \ln \relax (3)}{x^{2}}\)	\(41\)
default	\(-\frac {12 \,{\mathrm e}^{5} \ln \relax (3)}{x^{2}}-\frac {-\ln \left (\frac {\ln \relax (5)^{2}}{25}-\frac {4 \ln \relax (2) \ln \relax (5)}{25}+\frac {4 \ln \relax (2)^{2}}{25}\right )+4 \,{\mathrm e}^{5}}{x}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(1/25*ln(5)^2-4/25*ln(2)*ln(5)+4/25*ln(2)^2)+24*exp(5)*ln(3)+4*x*exp(5))/x^3,x,method=_RETURNVERBOSE
)

[Out]

((-2*ln(5)+2*ln(ln(5)-2*ln(2))-4*exp(5))*x-12*exp(5)*ln(3))/x^2

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maxima [A]  time = 0.36, size = 41, normalized size = 1.24 \begin {gather*} -\frac {x {\left (4 \, e^{5} - \log \left (\frac {1}{25} \, \log \relax (5)^{2} - \frac {4}{25} \, \log \relax (5) \log \relax (2) + \frac {4}{25} \, \log \relax (2)^{2}\right )\right )} + 12 \, e^{5} \log \relax (3)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(1/25*log(5)^2-4/25*log(2)*log(5)+4/25*log(2)^2)+24*exp(5)*log(3)+4*x*exp(5))/x^3,x, algorith
m="maxima")

[Out]

-(x*(4*e^5 - log(1/25*log(5)^2 - 4/25*log(5)*log(2) + 4/25*log(2)^2)) + 12*e^5*log(3))/x^2

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mupad [B]  time = 0.08, size = 39, normalized size = 1.18 \begin {gather*} \frac {\ln \left (\frac {4\,{\ln \relax (2)}^2}{25}-\frac {4\,\ln \relax (2)\,\ln \relax (5)}{25}+\frac {{\ln \relax (5)}^2}{25}\right )-4\,{\mathrm {e}}^5}{x}-\frac {12\,{\mathrm {e}}^5\,\ln \relax (3)}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((24*exp(5)*log(3) - x*log((4*log(2)^2)/25 - (4*log(2)*log(5))/25 + log(5)^2/25) + 4*x*exp(5))/x^3,x)

[Out]

(log((4*log(2)^2)/25 - (4*log(2)*log(5))/25 + log(5)^2/25) - 4*exp(5))/x - (12*exp(5)*log(3))/x^2

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sympy [A]  time = 0.24, size = 44, normalized size = 1.33 \begin {gather*} \frac {x \left (- 4 e^{5} - 2 \log {\relax (5 )} + \log {\left (- 4 \log {\relax (2 )} \log {\relax (5 )} + 4 \log {\relax (2 )}^{2} + \log {\relax (5 )}^{2} \right )}\right ) - 12 e^{5} \log {\relax (3 )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(1/25*ln(5)**2-4/25*ln(2)*ln(5)+4/25*ln(2)**2)+24*exp(5)*ln(3)+4*x*exp(5))/x**3,x)

[Out]

(x*(-4*exp(5) - 2*log(5) + log(-4*log(2)*log(5) + 4*log(2)**2 + log(5)**2)) - 12*exp(5)*log(3))/x**2

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