3.1.81 \(\int \frac {a+b \log (1+e x)}{x} \, dx\) [81]

Optimal. Leaf size=14 \[ a \log (x)-b \text {Li}_2(-e x) \]

[Out]

a*ln(x)-b*polylog(2,-e*x)

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Rubi [A]
time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2439, 2438} \begin {gather*} a \log (x)-b \text {PolyLog}(2,-e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[1 + e*x])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(e*x)]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + e*(x/d)]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rubi steps

\begin {align*} \int \frac {a+b \log (1+e x)}{x} \, dx &=a \log (x)+b \int \frac {\log (1+e x)}{x} \, dx\\ &=a \log (x)-b \text {Li}_2(-e x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 14, normalized size = 1.00 \begin {gather*} a \log (x)-b \text {Li}_2(-e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[1 + e*x])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(e*x)]

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Maple [A]
time = 0.18, size = 17, normalized size = 1.21

method	result	size
risch	\(\ln \left (x \right ) a -b \dilog \left (e x +1\right )\)	\(15\)
derivativedivides	\(a \ln \left (e x \right )-b \dilog \left (e x +1\right )\)	\(17\)
default	\(a \ln \left (e x \right )-b \dilog \left (e x +1\right )\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(e*x+1))/x,x,method=_RETURNVERBOSE)

[Out]

a*ln(e*x)-b*dilog(e*x+1)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (14) = 28\).
time = 0.30, size = 29, normalized size = 2.07 \begin {gather*} {\left (\log \left (x e + 1\right ) \log \left (-x e\right ) + {\rm Li}_2\left (x e + 1\right )\right )} b + a \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="maxima")

[Out]

(log(x*e + 1)*log(-x*e) + dilog(x*e + 1))*b + a*log(x)

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Fricas [A]
time = 0.36, size = 14, normalized size = 1.00 \begin {gather*} -b {\rm Li}_2\left (-x e\right ) + a \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="fricas")

[Out]

-b*dilog(-x*e) + a*log(x)

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Sympy [C] Result contains complex when optimal does not.
time = 1.95, size = 15, normalized size = 1.07 \begin {gather*} a \log {\left (x \right )} - b \operatorname {Li}_{2}\left (e x e^{i \pi }\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(e*x+1))/x,x)

[Out]

a*log(x) - b*polylog(2, e*x*exp_polar(I*pi))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(e*x+1))/x,x, algorithm="giac")

[Out]

integrate((b*log(x*e + 1) + a)/x, x)

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Mupad [B]
time = 0.06, size = 14, normalized size = 1.00 \begin {gather*} a\,\ln \left (x\right )-b\,\mathrm {polylog}\left (2,-e\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(e*x + 1))/x,x)

[Out]

a*log(x) - b*polylog(2, -e*x)

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