3.58.40 \(\int \frac {4+(-3-\log (5)) \log (x)}{3+\log (5)} \, dx\)

Optimal. Leaf size=18 \[ x+\frac {4 (1+x)}{3+\log (5)}-x \log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {12, 2295} \begin {gather*} x+x (-\log (x))+\frac {4 x}{3+\log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (4*x)/(3 + Log[5]) - x*Log[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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int (4+(-3-\log (5)) \log (x)) \, dx}{3+\log (5)}\\ &=\frac {4 x}{3+\log (5)}-\int \log (x) \, dx\\ &=x+\frac {4 x}{3+\log (5)}-x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.89 \begin {gather*} x+\frac {4 x}{3+\log (5)}-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (4*x)/(3 + Log[5]) - x*Log[x]

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fricas [A]  time = 0.68, size = 27, normalized size = 1.50 \begin {gather*} \frac {x \log \relax (5) - {\left (x \log \relax (5) + 3 \, x\right )} \log \relax (x) + 7 \, x}{\log \relax (5) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5)-3)*log(x)+4)/(log(5)+3),x, algorithm="fricas")

[Out]

(x*log(5) - (x*log(5) + 3*x)*log(x) + 7*x)/(log(5) + 3)

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giac [A]  time = 0.15, size = 25, normalized size = 1.39 \begin {gather*} -\frac {{\left (x \log \relax (x) - x\right )} {\left (\log \relax (5) + 3\right )} - 4 \, x}{\log \relax (5) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((x*log(x) - x)*(log(5) + 3) - 4*x)/(log(5) + 3)

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maple [A]  time = 0.04, size = 19, normalized size = 1.06

method	result	size
norman	\(\frac {\left (7+\ln \relax (5)\right ) x}{\ln \relax (5)+3}-x \ln \relax (x )\)	\(19\)
risch	\(\frac {7 x}{\ln \relax (5)+3}+\frac {x \ln \relax (5)}{\ln \relax (5)+3}-x \ln \relax (x )\)	\(26\)
default	\(\frac {7 x -x \ln \relax (5) \ln \relax (x )-3 x \ln \relax (x )+x \ln \relax (5)}{\ln \relax (5)+3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7+ln(5))/(ln(5)+3)*x-x*ln(x)

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maxima [A]  time = 0.36, size = 25, normalized size = 1.39 \begin {gather*} -\frac {{\left (x \log \relax (x) - x\right )} {\left (\log \relax (5) + 3\right )} - 4 \, x}{\log \relax (5) + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-log(5)-3)*log(x)+4)/(log(5)+3),x, algorithm="maxima")

[Out]

-((x*log(x) - x)*(log(5) + 3) - 4*x)/(log(5) + 3)

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mupad [B]  time = 3.62, size = 18, normalized size = 1.00 \begin {gather*} \frac {x\,\left (\ln \relax (5)+7\right )}{\ln \relax (5)+3}-x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x*(log(5) + 7))/(log(5) + 3) - x*log(x)

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sympy [A]  time = 0.09, size = 15, normalized size = 0.83 \begin {gather*} - x \log {\relax (x )} + \frac {x \left (\log {\relax (5 )} + 7\right )}{\log {\relax (5 )} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*log(x) + x*(log(5) + 7)/(log(5) + 3)

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