3.23.27 \(\int \frac {e^4 (-x+\log (x)) (10-2 x+\log ^2(x))}{2 (10 x-2 x^2+x \log ^2(x))} \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{2} e^4 \left (5-x+\frac {\log ^2(x)}{2}\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 6688, 2301} \begin {gather*} \frac {1}{4} e^4 \log ^2(x)-\frac {e^4 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(E^4*x) + (E^4*Log[x]^2)/4

Rule 12

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 20, normalized size = 1.00 \begin {gather*} -\frac {e^4 x}{2}+\frac {1}{4} e^4 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(E^4*x) + (E^4*Log[x]^2)/4

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fricas [A]  time = 1.00, size = 14, normalized size = 0.70 \begin {gather*} \frac {1}{4} \, e^{4} \log \relax (x)^{2} - \frac {1}{2} \, x e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^4*log(x)^2 - 1/2*x*e^4

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giac [A]  time = 0.28, size = 14, normalized size = 0.70 \begin {gather*} \frac {1}{4} \, e^{4} \log \relax (x)^{2} - \frac {1}{2} \, x e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e^4*log(x)^2 - 1/2*x*e^4

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maple [A]  time = 0.05, size = 15, normalized size = 0.75

method	result	size
default	\(\frac {{\mathrm e}^{4} \left (\frac {\ln \relax (x )^{2}}{2}-x \right )}{2}\)	\(15\)
norman	\(-\frac {x \,{\mathrm e}^{4}}{2}+\frac {{\mathrm e}^{4} \ln \relax (x )^{2}}{4}\)	\(15\)
risch	\(-\frac {x \,{\mathrm e}^{4}}{2}+\frac {{\mathrm e}^{4} \ln \relax (x )^{2}}{4}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 12, normalized size = 0.60 \begin {gather*} \frac {1}{4} \, {\left (\log \relax (x)^{2} - 2 \, x\right )} e^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(log(x)^2 - 2*x)*e^4

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mupad [B]  time = 1.28, size = 14, normalized size = 0.70 \begin {gather*} -\frac {{\mathrm {e}}^4\,\left (2\,x-{\ln \relax (x)}^2\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(exp(4)*(2*x - log(x)^2))/4

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sympy [A]  time = 0.12, size = 15, normalized size = 0.75 \begin {gather*} - \frac {x e^{4}}{2} + \frac {e^{4} \log {\relax (x )}^{2}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*exp(4)/2 + exp(4)*log(x)**2/4

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