3.80.14 \(\int \frac {2 x^3-4 x^2 \log (24)+2 x \log ^2(24)+\frac {1}{81} e^{20} x^4 (-3 x+4 \log (24))}{x^3-2 x^2 \log (24)+x \log ^2(24)} \, dx\)

Optimal. Leaf size=28 \[ 4+2 x+\frac {e^{4 \left (5-\log \left (\frac {3}{x}\right )\right )}}{-x+\log (24)} \]

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Rubi [B]  time = 0.16, antiderivative size = 58, normalized size of antiderivative = 2.07, number of steps used = 5, number of rules used = 4, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1594, 27, 1586, 1850} \begin {gather*} -\frac {1}{81} e^{20} x^3-\frac {1}{81} e^{20} x^2 \log (24)-\frac {e^{20} \log ^4(24)}{81 (x-\log (24))}+\frac {1}{81} x \left (162-e^{20} \log ^2(24)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^3 - 4*x^2*Log[24] + 2*x*Log[24]^2 + (E^20*x^4*(-3*x + 4*Log[24]))/81)/(x^3 - 2*x^2*Log[24] + x*Log[24
]^2),x]

[Out]

-1/81*(E^20*x^3) - (E^20*x^2*Log[24])/81 - (E^20*Log[24]^4)/(81*(x - Log[24])) + (x*(162 - E^20*Log[24]^2))/81

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x^3-4 x^2 \log (24)+2 x \log ^2(24)+\frac {1}{81} e^{20} x^4 (-3 x+4 \log (24))}{x \left (x^2-2 x \log (24)+\log ^2(24)\right )} \, dx\\ &=\int \frac {2 x^3-4 x^2 \log (24)+2 x \log ^2(24)+\frac {1}{81} e^{20} x^4 (-3 x+4 \log (24))}{x (x-\log (24))^2} \, dx\\ &=\int \frac {2 x^2-\frac {e^{20} x^4}{27}-4 x \log (24)+\frac {4}{81} e^{20} x^3 \log (24)+2 \log ^2(24)}{(x-\log (24))^2} \, dx\\ &=\int \left (-\frac {1}{27} e^{20} x^2-\frac {2}{81} e^{20} x \log (24)+\frac {e^{20} \log ^4(24)}{81 (x-\log (24))^2}+\frac {1}{81} \left (162-e^{20} \log ^2(24)\right )\right ) \, dx\\ &=-\frac {1}{81} e^{20} x^3-\frac {1}{81} e^{20} x^2 \log (24)-\frac {e^{20} \log ^4(24)}{81 (x-\log (24))}+\frac {1}{81} x \left (162-e^{20} \log ^2(24)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 56, normalized size = 2.00 \begin {gather*} \frac {162 x^2-e^{20} x^4+162 \log ^2(24)-4 e^{20} \log ^4(24)+4 x \log (24) \left (-81+e^{20} \log ^2(24)\right )}{81 (x-\log (24))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^3 - 4*x^2*Log[24] + 2*x*Log[24]^2 + (E^20*x^4*(-3*x + 4*Log[24]))/81)/(x^3 - 2*x^2*Log[24] + x*
Log[24]^2),x]

[Out]

(162*x^2 - E^20*x^4 + 162*Log[24]^2 - 4*E^20*Log[24]^4 + 4*x*Log[24]*(-81 + E^20*Log[24]^2))/(81*(x - Log[24])
)

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fricas [A]  time = 0.86, size = 43, normalized size = 1.54 \begin {gather*} -\frac {x^{4} e^{20} - x e^{20} \log \left (24\right )^{3} + e^{20} \log \left (24\right )^{4} - 162 \, x^{2} + 162 \, x \log \left (24\right )}{81 \, {\left (x - \log \left (24\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(24)-3*x)*exp(-4*log(3/x)+20)+2*x*log(24)^2-4*x^2*log(24)+2*x^3)/(x*log(24)^2-2*x^2*log(24)+x
^3),x, algorithm="fricas")

[Out]

-1/81*(x^4*e^20 - x*e^20*log(24)^3 + e^20*log(24)^4 - 162*x^2 + 162*x*log(24))/(x - log(24))

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giac [A]  time = 0.20, size = 45, normalized size = 1.61 \begin {gather*} -\frac {1}{81} \, x^{3} e^{20} - \frac {1}{81} \, x^{2} e^{20} \log \left (24\right ) - \frac {1}{81} \, x e^{20} \log \left (24\right )^{2} - \frac {e^{20} \log \left (24\right )^{4}}{81 \, {\left (x - \log \left (24\right )\right )}} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(24)-3*x)*exp(-4*log(3/x)+20)+2*x*log(24)^2-4*x^2*log(24)+2*x^3)/(x*log(24)^2-2*x^2*log(24)+x
^3),x, algorithm="giac")

[Out]

-1/81*x^3*e^20 - 1/81*x^2*e^20*log(24) - 1/81*x*e^20*log(24)^2 - 1/81*e^20*log(24)^4/(x - log(24)) + 2*x

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maple [A]  time = 0.07, size = 29, normalized size = 1.04

method	result	size
norman	\(\frac {-2 x^{2}+\frac {{\mathrm e}^{20} x^{4}}{81}+2 \ln \left (24\right )^{2}}{\ln \left (24\right )-x}\)	\(29\)
risch	\(-\frac {\ln \relax (2)^{2} {\mathrm e}^{20} x}{9}-\frac {2 \ln \relax (2) \ln \relax (3) {\mathrm e}^{20} x}{27}-\frac {\ln \relax (2) {\mathrm e}^{20} x^{2}}{27}-\frac {\ln \relax (3)^{2} {\mathrm e}^{20} x}{81}-\frac {\ln \relax (3) {\mathrm e}^{20} x^{2}}{81}-\frac {{\mathrm e}^{20} x^{3}}{81}+2 x +\frac {\ln \relax (2)^{4} {\mathrm e}^{20}}{3 \ln \relax (2)+\ln \relax (3)-x}+\frac {4 \ln \relax (2)^{3} \ln \relax (3) {\mathrm e}^{20}}{9 \left (\ln \relax (2)+\frac {\ln \relax (3)}{3}-\frac {x}{3}\right )}+\frac {2 \ln \relax (2)^{2} \ln \relax (3)^{2} {\mathrm e}^{20}}{9 \left (\ln \relax (2)+\frac {\ln \relax (3)}{3}-\frac {x}{3}\right )}+\frac {4 \ln \relax (2) \ln \relax (3)^{3} {\mathrm e}^{20}}{81 \left (\ln \relax (2)+\frac {\ln \relax (3)}{3}-\frac {x}{3}\right )}+\frac {\ln \relax (3)^{4} {\mathrm e}^{20}}{243 \ln \relax (2)+81 \ln \relax (3)-81 x}\)	\(165\)
default	\(2 x -9 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} x \ln \relax (2)^{2}-6 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (2) \ln \relax (3) x -3 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} x^{2} \ln \relax (2)-{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (3)^{2} x -{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} x^{2} \ln \relax (3)-{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} x^{3}-\frac {81 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (2)^{4}}{-3 \ln \relax (2)-\ln \relax (3)+x}-\frac {108 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (2)^{3} \ln \relax (3)}{-3 \ln \relax (2)-\ln \relax (3)+x}-\frac {54 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (2)^{2} \ln \relax (3)^{2}}{-3 \ln \relax (2)-\ln \relax (3)+x}-\frac {12 \,{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (2) \ln \relax (3)^{3}}{-3 \ln \relax (2)-\ln \relax (3)+x}-\frac {{\mathrm e}^{20-4 \ln \left (\frac {3}{x}\right )-4 \ln \relax (x )} \ln \relax (3)^{4}}{-3 \ln \relax (2)-\ln \relax (3)+x}\)	\(308\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*ln(24)-3*x)*exp(-4*ln(3/x)+20)+2*x*ln(24)^2-4*x^2*ln(24)+2*x^3)/(x*ln(24)^2-2*x^2*ln(24)+x^3),x,method
=_RETURNVERBOSE)

[Out]

(-2*x^2+1/81*exp(20)*x^4+2*ln(24)^2)/(ln(24)-x)

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maxima [A]  time = 0.36, size = 45, normalized size = 1.61 \begin {gather*} -\frac {1}{81} \, x^{3} e^{20} - \frac {1}{81} \, x^{2} e^{20} \log \left (24\right ) - \frac {e^{20} \log \left (24\right )^{4}}{81 \, {\left (x - \log \left (24\right )\right )}} - \frac {1}{81} \, {\left (e^{20} \log \left (24\right )^{2} - 162\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*log(24)-3*x)*exp(-4*log(3/x)+20)+2*x*log(24)^2-4*x^2*log(24)+2*x^3)/(x*log(24)^2-2*x^2*log(24)+x
^3),x, algorithm="maxima")

[Out]

-1/81*x^3*e^20 - 1/81*x^2*e^20*log(24) - 1/81*e^20*log(24)^4/(x - log(24)) - 1/81*(e^20*log(24)^2 - 162)*x

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mupad [B]  time = 0.16, size = 29, normalized size = 1.04 \begin {gather*} -\frac {{\mathrm {e}}^{20}\,x^4-162\,x^2+162\,\ln \left (24\right )\,x}{81\,x-81\,\ln \left (24\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(20 - 4*log(3/x))*(3*x - 4*log(24)) - 2*x*log(24)^2 + 4*x^2*log(24) - 2*x^3)/(x*log(24)^2 - 2*x^2*log
(24) + x^3),x)

[Out]

-(162*x*log(24) + x^4*exp(20) - 162*x^2)/(81*x - 81*log(24))

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sympy [B]  time = 0.21, size = 51, normalized size = 1.82 \begin {gather*} - \frac {x^{3} e^{20}}{81} - \frac {x^{2} e^{20} \log {\left (24 \right )}}{81} - x \left (-2 + \frac {e^{20} \log {\left (24 \right )}^{2}}{81}\right ) - \frac {e^{20} \log {\left (24 \right )}^{4}}{81 x - 81 \log {\left (24 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*ln(24)-3*x)*exp(-4*ln(3/x)+20)+2*x*ln(24)**2-4*x**2*ln(24)+2*x**3)/(x*ln(24)**2-2*x**2*ln(24)+x*
*3),x)

[Out]

-x**3*exp(20)/81 - x**2*exp(20)*log(24)/81 - x*(-2 + exp(20)*log(24)**2/81) - exp(20)*log(24)**4/(81*x - 81*lo
g(24))

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