3.25.99 \(\int \frac {2 e^5 (120-60 x-4 x^3)}{(-3 x+x^2) (-3 x^2+x^3+(-3 x^2+x^3) \log (4))} \, dx\)

Optimal. Leaf size=24 \[ \frac {8 e^5 \left (5+x^2\right )}{(-3+x) x^2 (1+\log (4))} \]

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Rubi [B]  time = 0.27, antiderivative size = 53, normalized size of antiderivative = 2.21, number of steps used = 6, number of rules used = 4, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.085, Rules used = {12, 1593, 6688, 1620} \begin {gather*} -\frac {40 e^5}{3 x^2 (1+\log (4))}-\frac {40 e^5}{9 x (1+\log (4))}-\frac {112 e^5}{9 (3-x) (1+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^5*(120 - 60*x - 4*x^3))/((-3*x + x^2)*(-3*x^2 + x^3 + (-3*x^2 + x^3)*Log[4])),x]

[Out]

(-112*E^5)/(9*(3 - x)*(1 + Log[4])) - (40*E^5)/(3*x^2*(1 + Log[4])) - (40*E^5)/(9*x*(1 + Log[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 1593

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (2 e^5\right ) \int \frac {120-60 x-4 x^3}{\left (-3 x+x^2\right ) \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx\\ &=\left (2 e^5\right ) \int \frac {120-60 x-4 x^3}{(-3+x) x \left (-3 x^2+x^3+\left (-3 x^2+x^3\right ) \log (4)\right )} \, dx\\ &=\left (2 e^5\right ) \int \frac {4 \left (30-15 x-x^3\right )}{(3-x)^2 x^3 (1+\log (4))} \, dx\\ &=\frac {\left (8 e^5\right ) \int \frac {30-15 x-x^3}{(3-x)^2 x^3} \, dx}{1+\log (4)}\\ &=\frac {\left (8 e^5\right ) \int \left (-\frac {14}{9 (-3+x)^2}+\frac {10}{3 x^3}+\frac {5}{9 x^2}\right ) \, dx}{1+\log (4)}\\ &=-\frac {112 e^5}{9 (3-x) (1+\log (4))}-\frac {40 e^5}{3 x^2 (1+\log (4))}-\frac {40 e^5}{9 x (1+\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \begin {gather*} \frac {8 e^5 \left (5+x^2\right )}{(-3+x) x^2 (1+\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^5*(120 - 60*x - 4*x^3))/((-3*x + x^2)*(-3*x^2 + x^3 + (-3*x^2 + x^3)*Log[4])),x]

[Out]

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

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fricas [A]  time = 0.83, size = 33, normalized size = 1.38 \begin {gather*} \frac {8 \, {\left (x^{2} + 5\right )} e^{5}}{x^{3} - 3 \, x^{2} + 2 \, {\left (x^{3} - 3 \, x^{2}\right )} \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5)/(x^2-3*x),x, algorithm="fricas")

[Out]

8*(x^2 + 5)*e^5/(x^3 - 3*x^2 + 2*(x^3 - 3*x^2)*log(2))

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giac [A]  time = 0.33, size = 40, normalized size = 1.67 \begin {gather*} \frac {112 \, e^{5}}{9 \, {\left (x - 3\right )} {\left (2 \, \log \relax (2) + 1\right )}} - \frac {40 \, {\left (x e^{5} + 3 \, e^{5}\right )}}{9 \, x^{2} {\left (2 \, \log \relax (2) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5)/(x^2-3*x),x, algorithm="giac")

[Out]

112/9*e^5/((x - 3)*(2*log(2) + 1)) - 40/9*(x*e^5 + 3*e^5)/(x^2*(2*log(2) + 1))

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maple [A]  time = 0.36, size = 28, normalized size = 1.17

method	result	size
risch	\(\frac {4 \,{\mathrm e}^{5} \left (2 x^{2}+10\right )}{x^{2} \left (x -3\right ) \left (1+2 \ln \relax (2)\right )}\)	\(28\)
gosper	\(4 \left (x^{2}+5\right ) {\mathrm e}^{\ln \left (\frac {2}{x^{2} \left (2 x \ln \relax (2)-6 \ln \relax (2)+x -3\right )}\right )+5}\)	\(31\)
norman	\(\frac {\frac {8 \,{\mathrm e}^{5} x^{2}}{1+2 \ln \relax (2)}+\frac {40 \,{\mathrm e}^{5}}{1+2 \ln \relax (2)}}{x^{2} \left (x -3\right )}\)	\(38\)
default	\(-\frac {20 \,{\mathrm e}^{5+\ln \left (\frac {2}{2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}}\right )+\ln \left (2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}\right )}}{3 \left (1+2 \ln \relax (2)\right ) x^{2}}-\frac {20 \,{\mathrm e}^{5+\ln \left (\frac {2}{2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}}\right )+\ln \left (2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}\right )}}{9 \left (1+2 \ln \relax (2)\right ) x}+\frac {56 \,{\mathrm e}^{5+\ln \left (\frac {2}{2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}}\right )+\ln \left (2 \left (x^{3}-3 x^{2}\right ) \ln \relax (2)+x^{3}-3 x^{2}\right )}}{9 \left (1+2 \ln \relax (2)\right ) \left (x -3\right )}\)	\(202\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^3-60*x+120)*exp(ln(2/(2*(x^3-3*x^2)*ln(2)+x^3-3*x^2))+5)/(x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

4*exp(5)*(2*x^2+10)/x^2/(x-3)/(1+2*ln(2))

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maxima [A]  time = 0.41, size = 33, normalized size = 1.38 \begin {gather*} \frac {8 \, {\left (x^{2} + 5\right )} e^{5}}{x^{3} {\left (2 \, \log \relax (2) + 1\right )} - 3 \, x^{2} {\left (2 \, \log \relax (2) + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^3-60*x+120)*exp(log(2/(2*(x^3-3*x^2)*log(2)+x^3-3*x^2))+5)/(x^2-3*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.17, size = 41, normalized size = 1.71 \begin {gather*} -\frac {\frac {8\,{\mathrm {e}}^5\,x^3}{3}+40\,{\mathrm {e}}^5}{6\,x^2\,\ln \relax (2)-2\,x^3\,\ln \relax (2)+3\,x^2-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(-2/(2*log(2)*(3*x^2 - x^3) + 3*x^2 - x^3)) + 5)*(60*x + 4*x^3 - 120))/(3*x - x^2),x)

[Out]

-(40*exp(5) + (8*x^3*exp(5))/3)/(6*x^2*log(2) - 2*x^3*log(2) + 3*x^2 - x^3)

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sympy [A]  time = 0.59, size = 37, normalized size = 1.54 \begin {gather*} - \frac {- 8 x^{2} e^{5} - 40 e^{5}}{x^{3} \left (1 + 2 \log {\relax (2 )}\right ) + x^{2} \left (- 6 \log {\relax (2 )} - 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**3-60*x+120)*exp(ln(2/(2*(x**3-3*x**2)*ln(2)+x**3-3*x**2))+5)/(x**2-3*x),x)

[Out]

-(-8*x**2*exp(5) - 40*exp(5))/(x**3*(1 + 2*log(2)) + x**2*(-6*log(2) - 3))

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