3.1.72 \(\int \frac {a+b \log (c (\frac {1}{c}+e x))}{x} \, dx\) [72]

Optimal. Leaf size=15 \[ a \log (x)-b \text {Li}_2(-c e x) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2439, 2438} \begin {gather*} a \log (x)-b \text {PolyLog}(2,-c e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*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 \left (c \left (\frac {1}{c}+e x\right )\right )}{x} \, dx &=a \log (x)+b \int \frac {\log (1+c e x)}{x} \, dx\\ &=a \log (x)-b \text {Li}_2(-c e x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} a \log (x)-b \text {Li}_2(-c e x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(c^(-1) + e*x)])/x,x]

[Out]

a*Log[x] - b*PolyLog[2, -(c*e*x)]

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Maple [A]
time = 0.16, size = 19, normalized size = 1.27

method	result	size
risch	\(\ln \left (x \right ) a -b \dilog \left (c e x +1\right )\)	\(16\)
derivativedivides	\(a \ln \left (c e x \right )-b \dilog \left (c e x +1\right )\)	\(19\)
default	\(a \ln \left (c e x \right )-b \dilog \left (c e x +1\right )\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 15, normalized size = 1.00 \begin {gather*} -b {\rm Li}_2\left (-c 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(c*(1/c+e*x)))/x,x, algorithm="fricas")

[Out]

-b*dilog(-c*x*e) + a*log(x)

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Sympy [C] Result contains complex when optimal does not.
time = 3.24, size = 17, normalized size = 1.13 \begin {gather*} a \log {\left (x \right )} - b \operatorname {Li}_{2}\left (c e x e^{i \pi }\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*log(x) - b*polylog(2, c*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(c*(1/c+e*x)))/x,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.08, size = 15, normalized size = 1.00 \begin {gather*} a\,\ln \left (x\right )-b\,\mathrm {polylog}\left (2,-c\,e\,x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*log(c*(e*x + 1/c)))/x,x)

[Out]

a*log(x) - b*polylog(2, -c*e*x)

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