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

Optimal. Leaf size=13 \[ \log \left (4 e^{-1+\log ^2(x)}+x\right ) \]

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Rubi [A]  time = 0.30, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {6741, 12, 6684} \begin {gather*} \log \left (e x+4 e^{\log ^2(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4*E^Log[x]^2 + E*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 6684

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

Rule 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e \left (x+8 e^{-1+\log ^2(x)} \log (x)\right )}{x \left (4 e^{\log ^2(x)}+e x\right )} \, dx\\ &=e \int \frac {x+8 e^{-1+\log ^2(x)} \log (x)}{x \left (4 e^{\log ^2(x)}+e x\right )} \, dx\\ &=\log \left (4 e^{\log ^2(x)}+e x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 13, normalized size = 1.00 \begin {gather*} \log \left (4 e^{\log ^2(x)}+e x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4*E^Log[x]^2 + E*x]

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fricas [A]  time = 0.57, size = 12, normalized size = 0.92 \begin {gather*} \log \left (x + 4 \, e^{\left (\log \relax (x)^{2} - 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 4*e^(log(x)^2 - 1))

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giac [A]  time = 0.22, size = 13, normalized size = 1.00 \begin {gather*} \log \left (x e + 4 \, e^{\left (\log \relax (x)^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x*e + 4*e^(log(x)^2))

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maple [A]  time = 0.03, size = 13, normalized size = 1.00

method	result	size
norman	\(\ln \left (x +4 \,{\mathrm e}^{\ln \relax (x )^{2}-1}\right )\)	\(13\)
risch	\(1+\ln \left (\frac {x}{4}+{\mathrm e}^{\left (\ln \relax (x )-1\right ) \left (\ln \relax (x )+1\right )}\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 12, normalized size = 0.92 \begin {gather*} \log \left (\frac {1}{4} \, x e + e^{\left (\log \relax (x)^{2}\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/4*x*e + e^(log(x)^2))

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mupad [B]  time = 5.45, size = 12, normalized size = 0.92 \begin {gather*} \ln \left (x+4\,{\mathrm {e}}^{{\ln \relax (x)}^2-1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 4*exp(log(x)^2 - 1))

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sympy [A]  time = 0.27, size = 12, normalized size = 0.92 \begin {gather*} \log {\left (\frac {x}{4} + e^{\log {\relax (x )}^{2} - 1} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x/4 + exp(log(x)**2 - 1))

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