\(\int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} (4-5 x+x^2)+e^{10} (-8+10 x-2 x^2) \log (-1+x)+e^{10} (4-5 x+x^2) \log ^2(-1+x)} \, dx\) [5895]

Optimal result

Integrand size = 88, antiderivative size = 22 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=-3+\log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6820, 12, 6860, 2437, 2339, 30} \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\log (4-x)+\frac {1}{e^{10} (1-\log (x-1))} \]

[In]

Int[(-4 + E^10*(-1 + x) + x + E^10*(2 - 2*x)*Log[-1 + x] + E^10*(-1 + x)*Log[-1 + x]^2)/(E^10*(4 - 5*x + x^2)
+ E^10*(-8 + 10*x - 2*x^2)*Log[-1 + x] + E^10*(4 - 5*x + x^2)*Log[-1 + x]^2),x]

[Out]

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

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 2339

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 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6820

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

Rule 6860

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{align*} \text {integral}& = \int \frac {-4+e^{10} (-1+x)+x-2 e^{10} (-1+x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right ) (1-\log (-1+x))^2} \, dx \\ & = \frac {\int \frac {-4+e^{10} (-1+x)+x-2 e^{10} (-1+x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{\left (4-5 x+x^2\right ) (1-\log (-1+x))^2} \, dx}{e^{10}} \\ & = \frac {\int \left (\frac {e^{10}}{-4+x}+\frac {1}{(-1+x) (-1+\log (-1+x))^2}\right ) \, dx}{e^{10}} \\ & = \log (4-x)+\frac {\int \frac {1}{(-1+x) (-1+\log (-1+x))^2} \, dx}{e^{10}} \\ & = \log (4-x)+\frac {\text {Subst}\left (\int \frac {1}{x (-1+\log (x))^2} \, dx,x,-1+x\right )}{e^{10}} \\ & = \log (4-x)+\frac {\text {Subst}\left (\int \frac {1}{x^2} \, dx,x,-1+\log (-1+x)\right )}{e^{10}} \\ & = \log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log (4-x)-\frac {1}{-1+\log (-1+x)}}{e^{10}} \]

[In]

Integrate[(-4 + E^10*(-1 + x) + x + E^10*(2 - 2*x)*Log[-1 + x] + E^10*(-1 + x)*Log[-1 + x]^2)/(E^10*(4 - 5*x +
x^2) + E^10*(-8 + 10*x - 2*x^2)*Log[-1 + x] + E^10*(4 - 5*x + x^2)*Log[-1 + x]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {{\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right )\)	\(18\)
norman	\(-\frac {{\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}+\ln \left (x -4\right )\)	\(20\)
derivativedivides	\({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (-1+x \right )-1}+{\mathrm e}^{10} \ln \left (x -4\right )\right )\)	\(26\)
default	\({\mathrm e}^{-10} \left (-\frac {1}{\ln \left (-1+x \right )-1}+{\mathrm e}^{10} \ln \left (x -4\right )\right )\)	\(26\)
parallelrisch	\(\frac {\left (-1+{\mathrm e}^{10} \ln \left (x -4\right ) \ln \left (-1+x \right )-{\mathrm e}^{10} \ln \left (x -4\right )\right ) {\mathrm e}^{-10}}{\ln \left (-1+x \right )-1}\)	\(39\)

[In]

int(((-1+x)*exp(5)^2*ln(-1+x)^2+(2-2*x)*exp(5)^2*ln(-1+x)+(-1+x)*exp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*ln(-1+x)^
2+(-2*x^2+10*x-8)*exp(5)^2*ln(-1+x)+(x^2-5*x+4)*exp(5)^2),x,method=_RETURNVERBOSE)

[Out]

-exp(-10)/(ln(-1+x)-1)+ln(x-4)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \]

[In]

integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*exp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*l
og(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*log(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="fricas")

[Out]

(e^10*log(x - 1)*log(x - 4) - e^10*log(x - 4) - 1)/(e^10*log(x - 1) - e^10)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\log {\left (x - 4 \right )} - \frac {1}{e^{10} \log {\left (x - 1 \right )} - e^{10}} \]

[In]

integrate(((-1+x)*exp(5)**2*ln(-1+x)**2+(2-2*x)*exp(5)**2*ln(-1+x)+(-1+x)*exp(5)**2+x-4)/((x**2-5*x+4)*exp(5)*
*2*ln(-1+x)**2+(-2*x**2+10*x-8)*exp(5)**2*ln(-1+x)+(x**2-5*x+4)*exp(5)**2),x)

[Out]

log(x - 4) - 1/(exp(10)*log(x - 1) - exp(10))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=-\frac {1}{e^{10} \log \left (x - 1\right ) - e^{10}} + \log \left (x - 4\right ) \]

[In]

integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*exp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*l
og(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*log(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="maxima")

[Out]

-1/(e^10*log(x - 1) - e^10) + log(x - 4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\frac {e^{10} \log \left (x - 1\right ) \log \left (x - 4\right ) - e^{10} \log \left (x - 4\right ) - 1}{e^{10} \log \left (x - 1\right ) - e^{10}} \]

[In]

integrate(((-1+x)*exp(5)^2*log(-1+x)^2+(2-2*x)*exp(5)^2*log(-1+x)+(-1+x)*exp(5)^2+x-4)/((x^2-5*x+4)*exp(5)^2*l
og(-1+x)^2+(-2*x^2+10*x-8)*exp(5)^2*log(-1+x)+(x^2-5*x+4)*exp(5)^2),x, algorithm="giac")

[Out]

(e^10*log(x - 1)*log(x - 4) - e^10*log(x - 4) - 1)/(e^10*log(x - 1) - e^10)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-4+e^{10} (-1+x)+x+e^{10} (2-2 x) \log (-1+x)+e^{10} (-1+x) \log ^2(-1+x)}{e^{10} \left (4-5 x+x^2\right )+e^{10} \left (-8+10 x-2 x^2\right ) \log (-1+x)+e^{10} \left (4-5 x+x^2\right ) \log ^2(-1+x)} \, dx=\ln \left (x-4\right )-\frac {{\mathrm {e}}^{-10}}{\ln \left (x-1\right )-1} \]

[In]

int((x + exp(10)*(x - 1) - log(x - 1)*exp(10)*(2*x - 2) + log(x - 1)^2*exp(10)*(x - 1) - 4)/(exp(10)*(x^2 - 5*
x + 4) - log(x - 1)*exp(10)*(2*x^2 - 10*x + 8) + log(x - 1)^2*exp(10)*(x^2 - 5*x + 4)),x)

[Out]

log(x - 4) - exp(-10)/(log(x - 1) - 1)