3.28.94 \(\int \frac {-20+18 x+(5 e^3-20 x+9 x^2) \log (\frac {-5 e^3+20 x-9 x^2}{e^3})}{(5 e^3-20 x+9 x^2) \log (\frac {-5 e^3+20 x-9 x^2}{e^3})} \, dx\)

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

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 20, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 2, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6728, 6684} \begin {gather*} \log \left (\log \left (-\frac {9 x^2}{e^3}+\frac {20 x}{e^3}-5\right )\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-20 + 18*x + (5*E^3 - 20*x + 9*x^2)*Log[(-5*E^3 + 20*x - 9*x^2)/E^3])/((5*E^3 - 20*x + 9*x^2)*Log[(-5*E^3
+ 20*x - 9*x^2)/E^3]),x]

[Out]

x + Log[Log[-5 + (20*x)/E^3 - (9*x^2)/E^3]]

Rule 6684

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

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+\frac {2 (-10+9 x)}{\left (5 e^3-20 x+9 x^2\right ) \log \left (-5+\frac {20 x}{e^3}-\frac {9 x^2}{e^3}\right )}\right ) \, dx\\ &=x+2 \int \frac {-10+9 x}{\left (5 e^3-20 x+9 x^2\right ) \log \left (-5+\frac {20 x}{e^3}-\frac {9 x^2}{e^3}\right )} \, dx\\ &=x+\log \left (\log \left (-5+\frac {20 x}{e^3}-\frac {9 x^2}{e^3}\right )\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 16, normalized size = 0.64 \begin {gather*} x+\log \left (\log \left (-5+\frac {(20-9 x) x}{e^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-20 + 18*x + (5*E^3 - 20*x + 9*x^2)*Log[(-5*E^3 + 20*x - 9*x^2)/E^3])/((5*E^3 - 20*x + 9*x^2)*Log[(
-5*E^3 + 20*x - 9*x^2)/E^3]),x]

[Out]

x + Log[Log[-5 + ((20 - 9*x)*x)/E^3]]

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 21, normalized size = 0.84 \begin {gather*} x + \log \left (\log \left (-{\left (9 \, x^{2} - 20 \, x + 5 \, e^{3}\right )} e^{\left (-3\right )}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(3)+9*x^2-20*x)*log((-5*exp(3)-9*x^2+20*x)/exp(3))+18*x-20)/(5*exp(3)+9*x^2-20*x)/log((-5*exp
(3)-9*x^2+20*x)/exp(3)),x, algorithm="fricas")

[Out]

x + log(log(-(9*x^2 - 20*x + 5*e^3)*e^(-3)))

________________________________________________________________________________________

giac [A]  time = 0.62, size = 19, normalized size = 0.76 \begin {gather*} x + \log \left (\log \left (-9 \, x^{2} + 20 \, x - 5 \, e^{3}\right ) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(3)+9*x^2-20*x)*log((-5*exp(3)-9*x^2+20*x)/exp(3))+18*x-20)/(5*exp(3)+9*x^2-20*x)/log((-5*exp
(3)-9*x^2+20*x)/exp(3)),x, algorithm="giac")

[Out]

x + log(log(-9*x^2 + 20*x - 5*e^3) - 3)

________________________________________________________________________________________

maple [A]  time = 0.92, size = 21, normalized size = 0.84

method	result	size
risch	\(x +\ln \left (\ln \left (\left (-5 \,{\mathrm e}^{3}-9 x^{2}+20 x \right ) {\mathrm e}^{-3}\right )\right )\)	\(21\)
norman	\(x +\ln \left (\ln \left (\left (-5 \,{\mathrm e}^{3}-9 x^{2}+20 x \right ) {\mathrm e}^{-3}\right )\right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(3)+9*x^2-20*x)*ln((-5*exp(3)-9*x^2+20*x)/exp(3))+18*x-20)/(5*exp(3)+9*x^2-20*x)/ln((-5*exp(3)-9*x^
2+20*x)/exp(3)),x,method=_RETURNVERBOSE)

[Out]

x+ln(ln((-5*exp(3)-9*x^2+20*x)*exp(-3)))

________________________________________________________________________________________

maxima [A]  time = 0.71, size = 19, normalized size = 0.76 \begin {gather*} x + \log \left (\log \left (-9 \, x^{2} + 20 \, x - 5 \, e^{3}\right ) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(3)+9*x^2-20*x)*log((-5*exp(3)-9*x^2+20*x)/exp(3))+18*x-20)/(5*exp(3)+9*x^2-20*x)/log((-5*exp
(3)-9*x^2+20*x)/exp(3)),x, algorithm="maxima")

[Out]

x + log(log(-9*x^2 + 20*x - 5*e^3) - 3)

________________________________________________________________________________________

mupad [B]  time = 0.31, size = 18, normalized size = 0.72 \begin {gather*} x+\ln \left (\ln \left (-9\,{\mathrm {e}}^{-3}\,x^2+20\,{\mathrm {e}}^{-3}\,x-5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((18*x + log(-exp(-3)*(5*exp(3) - 20*x + 9*x^2))*(5*exp(3) - 20*x + 9*x^2) - 20)/(log(-exp(-3)*(5*exp(3) -
20*x + 9*x^2))*(5*exp(3) - 20*x + 9*x^2)),x)

[Out]

x + log(log(20*x*exp(-3) - 9*x^2*exp(-3) - 5))

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 20, normalized size = 0.80 \begin {gather*} x + \log {\left (\log {\left (\frac {- 9 x^{2} + 20 x - 5 e^{3}}{e^{3}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(3)+9*x**2-20*x)*ln((-5*exp(3)-9*x**2+20*x)/exp(3))+18*x-20)/(5*exp(3)+9*x**2-20*x)/ln((-5*ex
p(3)-9*x**2+20*x)/exp(3)),x)

[Out]

x + log(log((-9*x**2 + 20*x - 5*exp(3))*exp(-3)))

________________________________________________________________________________________