3.1.82 \(\int \frac {a+b \log (e x)}{x} \, dx\) [82]

Optimal. Leaf size=17 \[ \frac {(a+b \log (e x))^2}{2 b} \]

[Out]

1/2*(a+b*ln(e*x))^2/b

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Rubi [A]
time = 0.01, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2338} \begin {gather*} \frac {(a+b \log (e x))^2}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[e*x])/x,x]

[Out]

(a + b*Log[e*x])^2/(2*b)

Rule 2338

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]

Rubi steps

\begin {align*} \int \frac {a+b \log (e x)}{x} \, dx &=\frac {(a+b \log (e x))^2}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.94 \begin {gather*} a \log (x)+\frac {1}{2} b \log ^2(e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[e*x])/x,x]

[Out]

a*Log[x] + (b*Log[e*x]^2)/2

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Maple [A]
time = 0.06, size = 17, normalized size = 1.00

method	result	size
risch	\(\frac {b \ln \left (e x \right )^{2}}{2}+\ln \left (x \right ) a\)	\(15\)
derivativedivides	\(\frac {b \ln \left (e x \right )^{2}}{2}+a \ln \left (e x \right )\)	\(17\)
default	\(\frac {b \ln \left (e x \right )^{2}}{2}+a \ln \left (e x \right )\)	\(17\)
norman	\(\frac {b \ln \left (e x \right )^{2}}{2}+a \ln \left (e x \right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(e*x))/x,x,method=_RETURNVERBOSE)

[Out]

1/2*b*ln(e*x)^2+a*ln(e*x)

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Maxima [A]
time = 0.29, size = 16, normalized size = 0.94 \begin {gather*} \frac {{\left (b \log \left (x e\right ) + a\right )}^{2}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x))/x,x, algorithm="maxima")

[Out]

1/2*(b*log(x*e) + a)^2/b

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Fricas [A]
time = 0.36, size = 18, normalized size = 1.06 \begin {gather*} \frac {1}{2} \, b \log \left (x e\right )^{2} + a \log \left (x e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x))/x,x, algorithm="fricas")

[Out]

1/2*b*log(x*e)^2 + a*log(x*e)

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Sympy [A]
time = 0.04, size = 14, normalized size = 0.82 \begin {gather*} a \log {\left (x \right )} + \frac {b \log {\left (e x \right )}^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(e*x))/x,x)

[Out]

a*log(x) + b*log(e*x)**2/2

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Giac [A]
time = 3.57, size = 14, normalized size = 0.82 \begin {gather*} \frac {1}{2} \, b \log \left (x\right )^{2} + {\left (a + b\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x))/x,x, algorithm="giac")

[Out]

1/2*b*log(x)^2 + (a + b)*log(x)

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Mupad [B]
time = 0.15, size = 14, normalized size = 0.82 \begin {gather*} \frac {b\,{\ln \left (e\,x\right )}^2}{2}+a\,\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(e*x))/x,x)

[Out]

a*log(x) + (b*log(e*x)^2)/2

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