3.6.69 \(\int \frac {1+3 x+x \log (2)-x \log (5)}{x} \, dx\)

Optimal. Leaf size=13 \[ x (3+\log (2)-\log (5))+\log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6, 43} \begin {gather*} x (3+\log (2)-\log (5))+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 3*x + x*Log[2] - x*Log[5])/x,x]

[Out]

x*(3 + Log[2] - Log[5]) + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1+x (3+\log (2))-x \log (5)}{x} \, dx\\ &=\int \frac {1+x (3+\log (2)-\log (5))}{x} \, dx\\ &=\int \left (3+\frac {1}{x}+\log (2)-\log (5)\right ) \, dx\\ &=x (3+\log (2)-\log (5))+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.15 \begin {gather*} 3 x+x \log (2)-x \log (5)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 3*x + x*Log[2] - x*Log[5])/x,x]

[Out]

3*x + x*Log[2] - x*Log[5] + Log[x]

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fricas [A]  time = 0.67, size = 15, normalized size = 1.15 \begin {gather*} -x \log \relax (5) + x \log \relax (2) + 3 \, x + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(5)+x*log(2)+3*x+1)/x,x, algorithm="fricas")

[Out]

-x*log(5) + x*log(2) + 3*x + log(x)

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giac [A]  time = 0.38, size = 16, normalized size = 1.23 \begin {gather*} -x \log \relax (5) + x \log \relax (2) + 3 \, x + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(5)+x*log(2)+3*x+1)/x,x, algorithm="giac")

[Out]

-x*log(5) + x*log(2) + 3*x + log(abs(x))

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maple [A]  time = 0.03, size = 14, normalized size = 1.08

method	result	size
norman	\(\ln \relax (x )+x \left (3-\ln \relax (5)+\ln \relax (2)\right )\)	\(14\)
default	\(3 x -x \ln \relax (5)+x \ln \relax (2)+\ln \relax (x )\)	\(16\)
risch	\(3 x -x \ln \relax (5)+x \ln \relax (2)+\ln \relax (x )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*ln(5)+x*ln(2)+3*x+1)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)+x*(3-ln(5)+ln(2))

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maxima [A]  time = 0.37, size = 14, normalized size = 1.08 \begin {gather*} -x {\left (\log \relax (5) - \log \relax (2) - 3\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(5)+x*log(2)+3*x+1)/x,x, algorithm="maxima")

[Out]

-x*(log(5) - log(2) - 3) + log(x)

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mupad [B]  time = 0.44, size = 9, normalized size = 0.69 \begin {gather*} \ln \relax (x)+x\,\left (\ln \left (\frac {2}{5}\right )+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + x*log(2) - x*log(5) + 1)/x,x)

[Out]

log(x) + x*(log(2/5) + 3)

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sympy [A]  time = 0.08, size = 12, normalized size = 0.92 \begin {gather*} x \left (- \log {\relax (5 )} + \log {\relax (2 )} + 3\right ) + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(5)+x*ln(2)+3*x+1)/x,x)

[Out]

x*(-log(5) + log(2) + 3) + log(x)

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