3.10.14 \(\int \frac {x-x^2+(e^4 (5-5 x)+x^3) \log (x)}{e^4 (2 x^2-2 x^3+e^2 (-x^2+x^3)) \log (x)} \, dx\)

Optimal. Leaf size=35 \[ \frac {-3+\frac {5}{x}-\frac {-x+\log \left (\frac {\log (x)}{1-x}\right )}{e^4}}{-2+e^2} \]

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Rubi [A]  time = 0.57, antiderivative size = 66, normalized size of antiderivative = 1.89, number of steps used = 8, number of rules used = 6, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {12, 6741, 6742, 1620, 2302, 29} \begin {gather*} -\frac {x}{e^4 \left (2-e^2\right )}-\frac {5}{\left (2-e^2\right ) x}-\frac {\log (1-x)}{e^4 \left (2-e^2\right )}+\frac {\log (\log (x))}{e^4 \left (2-e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5/((2 - E^2)*x) - x/(E^4*(2 - E^2)) - Log[1 - x]/(E^4*(2 - E^2)) + Log[Log[x]]/(E^4*(2 - E^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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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

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Mathematica [A]  time = 0.09, size = 32, normalized size = 0.91 \begin {gather*} \frac {\frac {5 e^4}{x}+x+\log (1-x)-\log (\log (x))}{e^4 \left (-2+e^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 33, normalized size = 0.94 \begin {gather*} \frac {x^{2} + x \log \left (x - 1\right ) - x \log \left (\log \relax (x)\right ) + 5 \, e^{4}}{x e^{6} - 2 \, x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+5)*exp(4)+x^3)*log(x)-x^2+x)/((x^3-x^2)*exp(2)-2*x^3+2*x^2)/exp(4)/log(x),x, algorithm="fric
as")

[Out]

(x^2 + x*log(x - 1) - x*log(log(x)) + 5*e^4)/(x*e^6 - 2*x*e^4)

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giac [A]  time = 0.28, size = 33, normalized size = 0.94 \begin {gather*} \frac {{\left (x^{2} + x \log \left (x - 1\right ) - x \log \left (\log \relax (x)\right ) + 5 \, e^{4}\right )} e^{\left (-4\right )}}{x e^{2} - 2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+5)*exp(4)+x^3)*log(x)-x^2+x)/((x^3-x^2)*exp(2)-2*x^3+2*x^2)/exp(4)/log(x),x, algorithm="giac
")

[Out]

(x^2 + x*log(x - 1) - x*log(log(x)) + 5*e^4)*e^(-4)/(x*e^2 - 2*x)

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maple [A]  time = 0.16, size = 41, normalized size = 1.17

method	result	size
risch	\(\frac {{\mathrm e}^{-4} \left (\ln \left (x -1\right ) x +5 \,{\mathrm e}^{4}+x^{2}\right )}{\left ({\mathrm e}^{2}-2\right ) x}-\frac {{\mathrm e}^{-4} \ln \left (\ln \relax (x )\right )}{{\mathrm e}^{2}-2}\)	\(41\)
default	\({\mathrm e}^{-4} \left (-\frac {\ln \left (\ln \relax (x )\right )}{{\mathrm e}^{2}-2}+\frac {x}{{\mathrm e}^{2}-2}+\frac {\ln \left (x -1\right )}{{\mathrm e}^{2}-2}+\frac {5 \,{\mathrm e}^{4}}{\left ({\mathrm e}^{2}-2\right ) x}\right )\)	\(50\)
norman	\(\frac {\frac {{\mathrm e}^{-4} x^{2}}{{\mathrm e}^{2}-2}+\frac {5}{{\mathrm e}^{2}-2}}{x}+\frac {{\mathrm e}^{-4} \ln \left (x -1\right )}{{\mathrm e}^{2}-2}-\frac {{\mathrm e}^{-4} \ln \left (\ln \relax (x )\right )}{{\mathrm e}^{2}-2}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x+5)*exp(4)+x^3)*ln(x)-x^2+x)/((x^3-x^2)*exp(2)-2*x^3+2*x^2)/exp(4)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.59, size = 44, normalized size = 1.26 \begin {gather*} {\left (\frac {\log \left (x - 1\right )}{e^{2} - 2} - \frac {\log \left (\log \relax (x)\right )}{e^{2} - 2} + \frac {x^{2} + 5 \, e^{4}}{x {\left (e^{2} - 2\right )}}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+5)*exp(4)+x^3)*log(x)-x^2+x)/((x^3-x^2)*exp(2)-2*x^3+2*x^2)/exp(4)/log(x),x, algorithm="maxi
ma")

[Out]

(log(x - 1)/(e^2 - 2) - log(log(x))/(e^2 - 2) + (x^2 + 5*e^4)/(x*(e^2 - 2)))*e^(-4)

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mupad [B]  time = 0.88, size = 41, normalized size = 1.17 \begin {gather*} \frac {5}{x\,\left ({\mathrm {e}}^2-2\right )}+\frac {{\mathrm {e}}^{-4}\,\left (\ln \left (x-1\right )-\ln \left (\ln \relax (x)\right )\right )}{{\mathrm {e}}^2-2}+\frac {x\,{\mathrm {e}}^{-4}}{{\mathrm {e}}^2-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-4)*(x + log(x)*(x^3 - exp(4)*(5*x - 5)) - x^2))/(log(x)*(exp(2)*(x^2 - x^3) - 2*x^2 + 2*x^3)),x)

[Out]

5/(x*(exp(2) - 2)) + (exp(-4)*(log(x - 1) - log(log(x))))/(exp(2) - 2) + (x*exp(-4))/(exp(2) - 2)

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sympy [A]  time = 0.63, size = 44, normalized size = 1.26 \begin {gather*} \frac {x}{- 2 e^{4} + e^{6}} + \frac {\log {\left (x - 1 \right )}}{\left (-2 + e^{2}\right ) e^{4}} - \frac {\log {\left (\log {\relax (x )} \right )}}{\left (-2 + e^{2}\right ) e^{4}} + \frac {5}{x \left (-2 + e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+5)*exp(4)+x**3)*ln(x)-x**2+x)/((x**3-x**2)*exp(2)-2*x**3+2*x**2)/exp(4)/ln(x),x)

[Out]

x/(-2*exp(4) + exp(6)) + exp(-4)*log(x - 1)/(-2 + exp(2)) - exp(-4)*log(log(x))/(-2 + exp(2)) + 5/(x*(-2 + exp
(2)))

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