3.81.90 \(\int \frac {e^e (10-2 x)+2 e^e \log (x)+(5-x^2+x \log (x)) \log ^2(\frac {x^2}{9})}{(15 x-3 x^2+3 x \log (x)) \log ^2(\frac {x^2}{9})} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{3} \left (2+x-\frac {e^e}{\log \left (\frac {x^2}{9}\right )}+5 \log (5-x+\log (x))\right ) \]

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Rubi [A]  time = 1.34, antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 6, integrand size = 66, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6741, 12, 6742, 6684, 2302, 30} \begin {gather*} -\frac {e^e}{3 \log \left (\frac {x^2}{9}\right )}+\frac {x}{3}+\frac {5}{3} \log (-x+\log (x)+5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E*(10 - 2*x) + 2*E^E*Log[x] + (5 - x^2 + x*Log[x])*Log[x^2/9]^2)/((15*x - 3*x^2 + 3*x*Log[x])*Log[x^2/9
]^2),x]

[Out]

x/3 - E^E/(3*Log[x^2/9]) + (5*Log[5 - x + Log[x]])/3

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{3 x (5-x+\log (x)) \log ^2\left (\frac {x^2}{9}\right )} \, dx\\ &=\frac {1}{3} \int \frac {e^e (10-2 x)+2 e^e \log (x)+\left (5-x^2+x \log (x)\right ) \log ^2\left (\frac {x^2}{9}\right )}{x (5-x+\log (x)) \log ^2\left (\frac {x^2}{9}\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {-5+x^2-x \log (x)}{x (-5+x-\log (x))}+\frac {2 e^e}{x \log ^2\left (\frac {x^2}{9}\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {-5+x^2-x \log (x)}{x (-5+x-\log (x))} \, dx+\frac {1}{3} \left (2 e^e\right ) \int \frac {1}{x \log ^2\left (\frac {x^2}{9}\right )} \, dx\\ &=\frac {1}{3} \int \left (1+\frac {5 (-1+x)}{x (-5+x-\log (x))}\right ) \, dx+\frac {1}{3} e^e \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {x^2}{9}\right )\right )\\ &=\frac {x}{3}-\frac {e^e}{3 \log \left (\frac {x^2}{9}\right )}+\frac {5}{3} \int \frac {-1+x}{x (-5+x-\log (x))} \, dx\\ &=\frac {x}{3}-\frac {e^e}{3 \log \left (\frac {x^2}{9}\right )}+\frac {5}{3} \log (5-x+\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 31, normalized size = 0.97 \begin {gather*} \frac {1}{3} \left (x-\frac {e^e}{\log \left (\frac {x^2}{9}\right )}+5 \log (5-x+\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E*(10 - 2*x) + 2*E^E*Log[x] + (5 - x^2 + x*Log[x])*Log[x^2/9]^2)/((15*x - 3*x^2 + 3*x*Log[x])*Log
[x^2/9]^2),x]

[Out]

(x - E^E/Log[x^2/9] + 5*Log[5 - x + Log[x]])/3

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fricas [A]  time = 0.65, size = 42, normalized size = 1.31 \begin {gather*} \frac {2 \, x \log \relax (3) - 2 \, x \log \relax (x) + 10 \, {\left (\log \relax (3) - \log \relax (x)\right )} \log \left (-x + \log \relax (x) + 5\right ) + e^{e}}{6 \, {\left (\log \relax (3) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)*exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)
/log(1/9*x^2)^2,x, algorithm="fricas")

[Out]

1/6*(2*x*log(3) - 2*x*log(x) + 10*(log(3) - log(x))*log(-x + log(x) + 5) + e^e)/(log(3) - log(x))

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giac [A]  time = 0.18, size = 49, normalized size = 1.53 \begin {gather*} \frac {2 \, x \log \relax (3) - 2 \, x \log \relax (x) + 10 \, \log \relax (3) \log \left (-x + \log \relax (x) + 5\right ) - 10 \, \log \relax (x) \log \left (-x + \log \relax (x) + 5\right ) + e^{e}}{6 \, {\left (\log \relax (3) - \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)*exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)
/log(1/9*x^2)^2,x, algorithm="giac")

[Out]

1/6*(2*x*log(3) - 2*x*log(x) + 10*log(3)*log(-x + log(x) + 5) - 10*log(x)*log(-x + log(x) + 5) + e^e)/(log(3)
- log(x))

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maple [C]  time = 0.13, size = 80, normalized size = 2.50

method	result	size
risch	\(\frac {x}{3}+\frac {2 i {\mathrm e}^{{\mathrm e}}}{3 \left (-\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+4 i \ln \relax (3)-4 i \ln \relax (x )\right )}+\frac {5 \ln \left (\ln \relax (x )-x +5\right )}{3}\)	\(80\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*ln(x)-x^2+5)*ln(1/9*x^2)^2+2*exp(exp(1))*ln(x)+(-2*x+10)*exp(exp(1)))/(3*x*ln(x)-3*x^2+15*x)/ln(1/9*x^
2)^2,x,method=_RETURNVERBOSE)

[Out]

1/3*x+2/3*I*exp(exp(1))/(-Pi*csgn(I*x^2)^3-Pi*csgn(I*x)^2*csgn(I*x^2)+2*Pi*csgn(I*x)*csgn(I*x^2)^2+4*I*ln(3)-4
*I*ln(x))+5/3*ln(ln(x)-x+5)

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maxima [A]  time = 0.61, size = 36, normalized size = 1.12 \begin {gather*} \frac {2 \, x \log \relax (3) - 2 \, x \log \relax (x) + e^{e}}{6 \, {\left (\log \relax (3) - \log \relax (x)\right )}} + \frac {5}{3} \, \log \left (-x + \log \relax (x) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*log(x)-x^2+5)*log(1/9*x^2)^2+2*exp(exp(1))*log(x)+(-2*x+10)*exp(exp(1)))/(3*x*log(x)-3*x^2+15*x)
/log(1/9*x^2)^2,x, algorithm="maxima")

[Out]

1/6*(2*x*log(3) - 2*x*log(x) + e^e)/(log(3) - log(x)) + 5/3*log(-x + log(x) + 5)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (x\,\ln \relax (x)-x^2+5\right )\,{\ln \left (\frac {x^2}{9}\right )}^2+2\,{\mathrm {e}}^{\mathrm {e}}\,\ln \relax (x)-{\mathrm {e}}^{\mathrm {e}}\,\left (2\,x-10\right )}{{\ln \left (\frac {x^2}{9}\right )}^2\,\left (15\,x+3\,x\,\ln \relax (x)-3\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*exp(exp(1))*log(x) - exp(exp(1))*(2*x - 10) + log(x^2/9)^2*(x*log(x) - x^2 + 5))/(log(x^2/9)^2*(15*x +
3*x*log(x) - 3*x^2)),x)

[Out]

int((2*exp(exp(1))*log(x) - exp(exp(1))*(2*x - 10) + log(x^2/9)^2*(x*log(x) - x^2 + 5))/(log(x^2/9)^2*(15*x +
3*x*log(x) - 3*x^2)), x)

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sympy [A]  time = 0.40, size = 29, normalized size = 0.91 \begin {gather*} \frac {x}{3} + \frac {5 \log {\left (- x + \log {\relax (x )} + 5 \right )}}{3} - \frac {e^{e}}{6 \log {\relax (x )} - 6 \log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*ln(x)-x**2+5)*ln(1/9*x**2)**2+2*exp(exp(1))*ln(x)+(-2*x+10)*exp(exp(1)))/(3*x*ln(x)-3*x**2+15*x)
/ln(1/9*x**2)**2,x)

[Out]

x/3 + 5*log(-x + log(x) + 5)/3 - exp(E)/(6*log(x) - 6*log(3))

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