3.2.65 \(\int \frac {\log (1-c x) \text {PolyLog}(2,c x)}{x} \, dx\) [165]

Optimal. Leaf size=11 \[ -\frac {1}{2} \text {PolyLog}(2,c x)^2 \]

[Out]

-1/2*polylog(2,c*x)^2

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Rubi [A]
time = 0.02, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6724, 6736} \begin {gather*} -\frac {1}{2} \text {Li}_2(c x){}^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[1 - c*x]*PolyLog[2, c*x])/x,x]

[Out]

-1/2*PolyLog[2, c*x]^2

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 6736

Int[(Log[1 + (e_.)*(x_)]*PolyLog[2, (c_.)*(x_)])/(x_), x_Symbol] :> Simp[-PolyLog[2, c*x]^2/2, x] /; FreeQ[{c,
e}, x] && EqQ[c + e, 0]

Rubi steps

\begin {align*} \int \frac {\log (1-c x) \text {Li}_2(c x)}{x} \, dx &=-\frac {1}{2} \text {Li}_2(c x){}^2\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \text {PolyLog}(2,c x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[1 - c*x]*PolyLog[2, c*x])/x,x]

[Out]

-1/2*PolyLog[2, c*x]^2

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Maple [A]
time = 0.06, size = 10, normalized size = 0.91

method	result	size
derivativedivides	\(-\frac {\polylog \left (2, c x \right )^{2}}{2}\)	\(10\)
default	\(-\frac {\polylog \left (2, c x \right )^{2}}{2}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(-c*x+1)*polylog(2,c*x)/x,x,method=_RETURNVERBOSE)

[Out]

-1/2*polylog(2,c*x)^2

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Maxima [A]
time = 0.25, size = 8, normalized size = 0.73 \begin {gather*} -\frac {1}{2} \, {\rm Li}_2\left (c x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*dilog(c*x)^2

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Fricas [A]
time = 0.36, size = 8, normalized size = 0.73 \begin {gather*} -\frac {1}{2} \, {\rm Li}_2\left (c x\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*dilog(c*x)^2

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\log {\left (- c x + 1 \right )} \operatorname {Li}_{2}\left (c x\right )}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(-c*x+1)*polylog(2,c*x)/x,x)

[Out]

Integral(log(-c*x + 1)*polylog(2, c*x)/x, x)

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

[Out]

integrate(dilog(c*x)*log(-c*x + 1)/x, x)

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Mupad [B]
time = 0.24, size = 9, normalized size = 0.82 \begin {gather*} -\frac {{\mathrm {polylog}\left (2,c\,x\right )}^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(1 - c*x)*polylog(2, c*x))/x,x)

[Out]

-polylog(2, c*x)^2/2

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