3.2.26 \(\int \text {PolyLog}(2,c (a+b x)) \, dx\) [126]

Optimal. Leaf size=60 \[ -x-\frac {(1-a c-b c x) \log (1-a c-b c x)}{b c}+\frac {a \text {PolyLog}(2,c (a+b x))}{b}+x \text {PolyLog}(2,c (a+b x)) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {6730, 2494, 2436, 2332, 2468, 2440, 2438} \begin {gather*} x \text {Li}_2(c (a+b x))+\frac {a \text {Li}_2(c (a+b x))}{b}-\frac {(-a c-b c x+1) \log (-a c-b c x+1)}{b c}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)],x]

[Out]

-x - ((1 - a*c - b*c*x)*Log[1 - a*c - b*c*x])/(b*c) + (a*PolyLog[2, c*(a + b*x)])/b + x*PolyLog[2, c*(a + b*x)
]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, 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 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2468

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rule 2494

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rule 6730

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)], x_Symbol] :> Simp[x*PolyLog[n, c*(a + b*x)^p], x] + (-Dist[
p, Int[PolyLog[n - 1, c*(a + b*x)^p], x], x] + Dist[a*p, Int[PolyLog[n - 1, c*(a + b*x)^p]/(a + b*x), x], x])
/; FreeQ[{a, b, c, p}, x] && GtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 53, normalized size = 0.88 \begin {gather*} \frac {-c (a+b x)+(-1+c (a+b x)) \log (1-c (a+b x))+c (a+b x) \text {PolyLog}(2,c (a+b x))}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, c*(a + b*x)],x]

[Out]

(-(c*(a + b*x)) + (-1 + c*(a + b*x))*Log[1 - c*(a + b*x)] + c*(a + b*x)*PolyLog[2, c*(a + b*x)])/(b*c)

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Maple [A]
time = 0.42, size = 63, normalized size = 1.05

method	result	size
derivativedivides	\(\frac {\left (x b c +a c \right ) \polylog \left (2, x b c +a c \right )-\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )+1-x b c -a c}{b c}\)	\(63\)
default	\(\frac {\left (x b c +a c \right ) \polylog \left (2, x b c +a c \right )-\left (-x b c -a c +1\right ) \ln \left (-x b c -a c +1\right )+1-x b c -a c}{b c}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,c*(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 90, normalized size = 1.50 \begin {gather*} -\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} a}{b} + \frac {b c x {\rm Li}_2\left (b c x + a c\right ) - b c x + {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a)),x, algorithm="maxima")

[Out]

-(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*a/b + (b*c*x*dilog(b*c*x + a*c) - b*c*x +
(b*c*x + a*c - 1)*log(-b*c*x - a*c + 1))/(b*c)

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Fricas [A]
time = 0.35, size = 55, normalized size = 0.92 \begin {gather*} -\frac {b c x - {\left (b c x + a c\right )} {\rm Li}_2\left (b c x + a c\right ) - {\left (b c x + a c - 1\right )} \log \left (-b c x - a c + 1\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a)),x, algorithm="fricas")

[Out]

-(b*c*x - (b*c*x + a*c)*dilog(b*c*x + a*c) - (b*c*x + a*c - 1)*log(-b*c*x - a*c + 1))/(b*c)

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Sympy [A]
time = 0.93, size = 75, normalized size = 1.25 \begin {gather*} \begin {cases} 0 & \text {for}\: c = 0 \wedge \left (b = 0 \vee c = 0\right ) \\x \operatorname {Li}_{2}\left (a c\right ) & \text {for}\: b = 0 \\- \frac {a \operatorname {Li}_{1}\left (a c + b c x\right )}{b} + \frac {a \operatorname {Li}_{2}\left (a c + b c x\right )}{b} - x \operatorname {Li}_{1}\left (a c + b c x\right ) + x \operatorname {Li}_{2}\left (a c + b c x\right ) - x + \frac {\operatorname {Li}_{1}\left (a c + b c x\right )}{b c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a)),x)

[Out]

Piecewise((0, Eq(c, 0) & (Eq(b, 0) | Eq(c, 0))), (x*polylog(2, a*c), Eq(b, 0)), (-a*polylog(1, a*c + b*c*x)/b
+ a*polylog(2, a*c + b*c*x)/b - x*polylog(1, a*c + b*c*x) + x*polylog(2, a*c + b*c*x) - x + polylog(1, a*c + b
*c*x)/(b*c), True))

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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(polylog(2,c*(b*x+a)),x, algorithm="giac")

[Out]

integrate(dilog((b*x + a)*c), x)

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Mupad [B]
time = 0.59, size = 61, normalized size = 1.02 \begin {gather*} \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b}-x-\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )}{b\,c}+\frac {\ln \left (1-c\,\left (a+b\,x\right )\right )\,\left (a+b\,x\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x)),x)

[Out]

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

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