3.103.14 \(\int \frac {e^{e^3} (-3200+2400 x-600 x^2+50 x^3)+10 x \log (x)+(-3200+2400 x-600 x^2+50 x^3) \log (\log (x))}{(-64 x+48 x^2-12 x^3+x^4) \log (x)} \, dx\)

Optimal. Leaf size=24 \[ -5-\frac {5}{(4-x)^2}+25 \left (e^{e^3}+\log (\log (x))\right )^2 \]

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Rubi [A]  time = 0.34, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 12, 2301} \begin {gather*} 25 \left (\log (\log (x))+e^{e^3}\right )^2-\frac {5}{(4-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^3*(-3200 + 2400*x - 600*x^2 + 50*x^3) + 10*x*Log[x] + (-3200 + 2400*x - 600*x^2 + 50*x^3)*Log[Log[x]]
)/((-64*x + 48*x^2 - 12*x^3 + x^4)*Log[x]),x]

[Out]

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

Rule 12

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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} &=\int 10 \left (\frac {1}{(-4+x)^3}+\frac {5 \left (e^{e^3}+\log (\log (x))\right )}{x \log (x)}\right ) \, dx\\ &=10 \int \left (\frac {1}{(-4+x)^3}+\frac {5 \left (e^{e^3}+\log (\log (x))\right )}{x \log (x)}\right ) \, dx\\ &=-\frac {5}{(4-x)^2}+50 \int \frac {e^{e^3}+\log (\log (x))}{x \log (x)} \, dx\\ &=-\frac {5}{(4-x)^2}+50 \operatorname {Subst}\left (\int \frac {e^{e^3}+\log (x)}{x} \, dx,x,\log (x)\right )\\ &=-\frac {5}{(4-x)^2}+25 \left (e^{e^3}+\log (\log (x))\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.12 \begin {gather*} 10 \left (-\frac {1}{2 (-4+x)^2}+\frac {5}{2} \left (e^{e^3}+\log (\log (x))\right )^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^3*(-3200 + 2400*x - 600*x^2 + 50*x^3) + 10*x*Log[x] + (-3200 + 2400*x - 600*x^2 + 50*x^3)*Log[L
og[x]])/((-64*x + 48*x^2 - 12*x^3 + x^4)*Log[x]),x]

[Out]

10*(-1/2*1/(-4 + x)^2 + (5*(E^E^3 + Log[Log[x]])^2)/2)

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fricas [B]  time = 0.82, size = 45, normalized size = 1.88 \begin {gather*} \frac {5 \, {\left (10 \, {\left (x^{2} - 8 \, x + 16\right )} e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 5 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (\log \relax (x)\right )^{2} - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-600*x^2+2400*x-3200)*log(log(x))+10*x*log(x)+(50*x^3-600*x^2+2400*x-3200)*exp(exp(3)))/(x^4
-12*x^3+48*x^2-64*x)/log(x),x, algorithm="fricas")

[Out]

5*(10*(x^2 - 8*x + 16)*e^(e^3)*log(log(x)) + 5*(x^2 - 8*x + 16)*log(log(x))^2 - 1)/(x^2 - 8*x + 16)

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giac [B]  time = 0.13, size = 67, normalized size = 2.79 \begin {gather*} \frac {5 \, {\left (10 \, x^{2} e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 5 \, x^{2} \log \left (\log \relax (x)\right )^{2} - 80 \, x e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) - 40 \, x \log \left (\log \relax (x)\right )^{2} + 160 \, e^{\left (e^{3}\right )} \log \left (\log \relax (x)\right ) + 80 \, \log \left (\log \relax (x)\right )^{2} - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-600*x^2+2400*x-3200)*log(log(x))+10*x*log(x)+(50*x^3-600*x^2+2400*x-3200)*exp(exp(3)))/(x^4
-12*x^3+48*x^2-64*x)/log(x),x, algorithm="giac")

[Out]

5*(10*x^2*e^(e^3)*log(log(x)) + 5*x^2*log(log(x))^2 - 80*x*e^(e^3)*log(log(x)) - 40*x*log(log(x))^2 + 160*e^(e
^3)*log(log(x)) + 80*log(log(x))^2 - 1)/(x^2 - 8*x + 16)

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maple [A]  time = 0.04, size = 24, normalized size = 1.00

method	result	size
default	\(25 \ln \left (\ln \relax (x )\right )^{2}+50 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \left (\ln \relax (x )\right )-\frac {5}{\left (x -4\right )^{2}}\)	\(24\)
risch	\(25 \ln \left (\ln \relax (x )\right )^{2}+\frac {50 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{3}} x^{2}-400 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{{\mathrm e}^{3}} x +800 \,{\mathrm e}^{{\mathrm e}^{3}} \ln \left (\ln \relax (x )\right )-5}{x^{2}-8 x +16}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((50*x^3-600*x^2+2400*x-3200)*ln(ln(x))+10*x*ln(x)+(50*x^3-600*x^2+2400*x-3200)*exp(exp(3)))/(x^4-12*x^3+4
8*x^2-64*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

25*ln(ln(x))^2+50*exp(exp(3))*ln(ln(x))-5/(x-4)^2

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maxima [B]  time = 0.40, size = 53, normalized size = 2.21 \begin {gather*} \frac {5 \, {\left (5 \, {\left (x^{2} - 8 \, x + 16\right )} \log \left (\log \relax (x)\right )^{2} + 10 \, {\left (x^{2} e^{\left (e^{3}\right )} - 8 \, x e^{\left (e^{3}\right )} + 16 \, e^{\left (e^{3}\right )}\right )} \log \left (\log \relax (x)\right ) - 1\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x^3-600*x^2+2400*x-3200)*log(log(x))+10*x*log(x)+(50*x^3-600*x^2+2400*x-3200)*exp(exp(3)))/(x^4
-12*x^3+48*x^2-64*x)/log(x),x, algorithm="maxima")

[Out]

5*(5*(x^2 - 8*x + 16)*log(log(x))^2 + 10*(x^2*e^(e^3) - 8*x*e^(e^3) + 16*e^(e^3))*log(log(x)) - 1)/(x^2 - 8*x
+ 16)

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mupad [B]  time = 7.20, size = 28, normalized size = 1.17 \begin {gather*} 50\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{{\mathrm {e}}^3}-\frac {5}{x^2-8\,x+16}+25\,{\ln \left (\ln \relax (x)\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x))*(2400*x - 600*x^2 + 50*x^3 - 3200) + 10*x*log(x) + exp(exp(3))*(2400*x - 600*x^2 + 50*x^3 -
3200))/(log(x)*(64*x - 48*x^2 + 12*x^3 - x^4)),x)

[Out]

50*log(log(x))*exp(exp(3)) - 5/(x^2 - 8*x + 16) + 25*log(log(x))^2

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sympy [A]  time = 0.38, size = 31, normalized size = 1.29 \begin {gather*} 25 \log {\left (\log {\relax (x )} \right )}^{2} + 50 e^{e^{3}} \log {\left (\log {\relax (x )} \right )} - \frac {10}{2 x^{2} - 16 x + 32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((50*x**3-600*x**2+2400*x-3200)*ln(ln(x))+10*x*ln(x)+(50*x**3-600*x**2+2400*x-3200)*exp(exp(3)))/(x*
*4-12*x**3+48*x**2-64*x)/ln(x),x)

[Out]

25*log(log(x))**2 + 50*exp(exp(3))*log(log(x)) - 10/(2*x**2 - 16*x + 32)

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