3.23.60 \(\int \frac {25 x^4 \log ^4(\frac {4}{3})}{-32 e^{50}+5 x^5 \log ^4(\frac {4}{3})} \, dx\)

Optimal. Leaf size=22 \[ \log \left (5 \left (-2 e^{50}+\frac {5}{16} x^5 \log ^4\left (\frac {4}{3}\right )\right )\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 18, normalized size of antiderivative = 0.82, number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 260} \begin {gather*} \log \left (32 e^{50}-5 x^5 \log ^4\left (\frac {4}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*x^4*Log[4/3]^4)/(-32*E^50 + 5*x^5*Log[4/3]^4),x]

[Out]

Log[32*E^50 - 5*x^5*Log[4/3]^4]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 260

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (25 \log ^4\left (\frac {4}{3}\right )\right ) \int \frac {x^4}{-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )} \, dx\\ &=\log \left (32 e^{50}-5 x^5 \log ^4\left (\frac {4}{3}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.82 \begin {gather*} \log \left (-32 e^{50}+5 x^5 \log ^4\left (\frac {4}{3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^4*Log[4/3]^4)/(-32*E^50 + 5*x^5*Log[4/3]^4),x]

[Out]

Log[-32*E^50 + 5*x^5*Log[4/3]^4]

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fricas [A]  time = 1.03, size = 15, normalized size = 0.68 \begin {gather*} \log \left (5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm="fricas")

[Out]

log(5*x^5*log(3/4)^4 - 32*e^50)

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giac [A]  time = 0.21, size = 16, normalized size = 0.73 \begin {gather*} \log \left ({\left | 5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm="giac")

[Out]

log(abs(5*x^5*log(3/4)^4 - 32*e^50))

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maple [A]  time = 0.47, size = 15, normalized size = 0.68

method	result	size
meijerg	\(\ln \left (1-\frac {5 x^{5} \ln \left (\frac {3}{4}\right )^{4} {\mathrm e}^{-50}}{32}\right )\)	\(15\)
derivativedivides	\(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\)	\(18\)
default	\(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\)	\(18\)
norman	\(\ln \left (5 x^{5} \ln \left (\frac {3}{4}\right )^{4}-32 \,{\mathrm e}^{50}\right )\)	\(18\)
risch	\(\frac {16 \ln \left (\left (80 \ln \relax (2)^{4}-160 \ln \relax (3) \ln \relax (2)^{3}+120 \ln \relax (3)^{2} \ln \relax (2)^{2}-40 \ln \relax (2) \ln \relax (3)^{3}+5 \ln \relax (3)^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \relax (2)^{4}}{\ln \relax (3)^{4}-8 \ln \relax (2) \ln \relax (3)^{3}+24 \ln \relax (3)^{2} \ln \relax (2)^{2}-32 \ln \relax (3) \ln \relax (2)^{3}+16 \ln \relax (2)^{4}}-\frac {32 \ln \left (\left (80 \ln \relax (2)^{4}-160 \ln \relax (3) \ln \relax (2)^{3}+120 \ln \relax (3)^{2} \ln \relax (2)^{2}-40 \ln \relax (2) \ln \relax (3)^{3}+5 \ln \relax (3)^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \relax (2)^{3} \ln \relax (3)}{\ln \relax (3)^{4}-8 \ln \relax (2) \ln \relax (3)^{3}+24 \ln \relax (3)^{2} \ln \relax (2)^{2}-32 \ln \relax (3) \ln \relax (2)^{3}+16 \ln \relax (2)^{4}}+\frac {24 \ln \left (\left (80 \ln \relax (2)^{4}-160 \ln \relax (3) \ln \relax (2)^{3}+120 \ln \relax (3)^{2} \ln \relax (2)^{2}-40 \ln \relax (2) \ln \relax (3)^{3}+5 \ln \relax (3)^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \relax (2)^{2} \ln \relax (3)^{2}}{\ln \relax (3)^{4}-8 \ln \relax (2) \ln \relax (3)^{3}+24 \ln \relax (3)^{2} \ln \relax (2)^{2}-32 \ln \relax (3) \ln \relax (2)^{3}+16 \ln \relax (2)^{4}}-\frac {8 \ln \left (\left (80 \ln \relax (2)^{4}-160 \ln \relax (3) \ln \relax (2)^{3}+120 \ln \relax (3)^{2} \ln \relax (2)^{2}-40 \ln \relax (2) \ln \relax (3)^{3}+5 \ln \relax (3)^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \relax (2) \ln \relax (3)^{3}}{\ln \relax (3)^{4}-8 \ln \relax (2) \ln \relax (3)^{3}+24 \ln \relax (3)^{2} \ln \relax (2)^{2}-32 \ln \relax (3) \ln \relax (2)^{3}+16 \ln \relax (2)^{4}}+\frac {\ln \left (\left (80 \ln \relax (2)^{4}-160 \ln \relax (3) \ln \relax (2)^{3}+120 \ln \relax (3)^{2} \ln \relax (2)^{2}-40 \ln \relax (2) \ln \relax (3)^{3}+5 \ln \relax (3)^{4}\right ) x^{5}-32 \,{\mathrm e}^{50}\right ) \ln \relax (3)^{4}}{\ln \relax (3)^{4}-8 \ln \relax (2) \ln \relax (3)^{3}+24 \ln \relax (3)^{2} \ln \relax (2)^{2}-32 \ln \relax (3) \ln \relax (2)^{3}+16 \ln \relax (2)^{4}}\)	\(479\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(25*x^4*ln(3/4)^4/(5*x^5*ln(3/4)^4-32*exp(25)^2),x,method=_RETURNVERBOSE)

[Out]

ln(1-5/32*x^5*ln(3/4)^4*exp(-50))

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maxima [A]  time = 0.37, size = 15, normalized size = 0.68 \begin {gather*} \log \left (5 \, x^{5} \log \left (\frac {3}{4}\right )^{4} - 32 \, e^{50}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25*x^4*log(3/4)^4/(5*x^5*log(3/4)^4-32*exp(25)^2),x, algorithm="maxima")

[Out]

log(5*x^5*log(3/4)^4 - 32*e^50)

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mupad [B]  time = 0.10, size = 15, normalized size = 0.68 \begin {gather*} \ln \left (5\,x^5\,{\ln \left (\frac {3}{4}\right )}^4-32\,{\mathrm {e}}^{50}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x^4*log(3/4)^4)/(32*exp(50) - 5*x^5*log(3/4)^4),x)

[Out]

log(5*x^5*log(3/4)^4 - 32*exp(50))

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sympy [B]  time = 2.82, size = 112, normalized size = 5.09 \begin {gather*} \frac {\left (- 800 \log {\relax (2 )}^{3} \log {\relax (3 )} - 200 \log {\relax (2 )} \log {\relax (3 )}^{3} + 25 \log {\relax (3 )}^{4} + 400 \log {\relax (2 )}^{4} + 600 \log {\relax (2 )}^{2} \log {\relax (3 )}^{2}\right ) \log {\left (x^{5} \left (- 160 \log {\relax (2 )}^{3} \log {\relax (3 )} - 40 \log {\relax (2 )} \log {\relax (3 )}^{3} + 5 \log {\relax (3 )}^{4} + 80 \log {\relax (2 )}^{4} + 120 \log {\relax (2 )}^{2} \log {\relax (3 )}^{2}\right ) - 32 e^{50} \right )}}{25 \left (- \log {\relax (3 )} + 2 \log {\relax (2 )}\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(25*x**4*ln(3/4)**4/(5*x**5*ln(3/4)**4-32*exp(25)**2),x)

[Out]

(-800*log(2)**3*log(3) - 200*log(2)*log(3)**3 + 25*log(3)**4 + 400*log(2)**4 + 600*log(2)**2*log(3)**2)*log(x*
*5*(-160*log(2)**3*log(3) - 40*log(2)*log(3)**3 + 5*log(3)**4 + 80*log(2)**4 + 120*log(2)**2*log(3)**2) - 32*e
xp(50))/(25*(-log(3) + 2*log(2))**4)

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