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

Optimal. Leaf size=22 \[ -3+\log (4-x)+\frac {1}{e^{10} (1-\log (-1+x))} \]

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Rubi [A]  time = 0.34, antiderivative size = 21, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, integrand size = 88, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {6688, 12, 6728, 2390, 2302, 30} \begin {gather*} \log (4-x)+\frac {1}{e^{10} (1-\log (x-1))} \end {gather*}

Antiderivative was successfully verified.

[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 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 2390

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 6688

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

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

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Mathematica [A]  time = 0.06, size = 25, normalized size = 1.14 \begin {gather*} \frac {e^{10} \log (4-x)-\frac {1}{-1+\log (-1+x)}}{e^{10}} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.64, size = 36, normalized size = 1.64 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [A]  time = 3.00, size = 36, normalized size = 1.64 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 0.17, size = 18, normalized size = 0.82

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 21, normalized size = 0.95 \begin {gather*} -\frac {1}{e^{10} \log \left (x - 1\right ) - e^{10}} + \log \left (x - 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 17, normalized size = 0.77 \begin {gather*} \ln \left (x-4\right )-\frac {{\mathrm {e}}^{-10}}{\ln \left (x-1\right )-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.13, size = 17, normalized size = 0.77 \begin {gather*} \log {\left (x - 4 \right )} - \frac {1}{e^{10} \log {\left (x - 1 \right )} - e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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