3.2.42 \(\int \frac {\text {PolyLog}(2,c (a+b x))}{(d+e x)^2} \, dx\) [142]

Optimal. Leaf size=138 \[ \frac {b \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}+\frac {b \text {PolyLog}(2,c (a+b x))}{e (b d-a e)}-\frac {\text {PolyLog}(2,c (a+b x))}{e (d+e x)}+\frac {b \text {PolyLog}\left (2,\frac {e (1-a c-b c x)}{b c d+e-a c e}\right )}{e (b d-a e)} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {6733, 2465, 2440, 2438, 2441} \begin {gather*} \frac {b \text {Li}_2(c (a+b x))}{e (b d-a e)}-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}+\frac {b \text {Li}_2\left (\frac {e (-a c-b x c+1)}{b c d-a c e+e}\right )}{e (b d-a e)}+\frac {b \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )}{e (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/(d + e*x)^2,x]

[Out]

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

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 2441

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

Rule 2465

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 6733

Int[((d_.) + (e_.)*(x_))^(m_.)*PolyLog[2, (c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> Simp[(d + e*x)^(m + 1)*(Po
lyLog[2, c*(a + b*x)]/(e*(m + 1))), x] + Dist[b/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(Log[1 - a*c - b*c*x]/(a +
b*x)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\text {Li}_2(c (a+b x))}{(d+e x)^2} \, dx &=-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}-\frac {b \int \frac {\log (1-a c-b c x)}{(a+b x) (d+e x)} \, dx}{e}\\ &=-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}-\frac {b \int \left (\frac {b \log (1-a c-b c x)}{(b d-a e) (a+b x)}-\frac {e \log (1-a c-b c x)}{(b d-a e) (d+e x)}\right ) \, dx}{e}\\ &=-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}+\frac {b \int \frac {\log (1-a c-b c x)}{d+e x} \, dx}{b d-a e}-\frac {b^2 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{e (b d-a e)}\\ &=\frac {b \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}-\frac {b \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{e (b d-a e)}+\frac {\left (b^2 c\right ) \int \frac {\log \left (-\frac {b c (d+e x)}{-b c d-(1-a c) e}\right )}{1-a c-b c x} \, dx}{e (b d-a e)}\\ &=\frac {b \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}+\frac {b \text {Li}_2(c (a+b x))}{e (b d-a e)}-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}-\frac {b \text {Subst}\left (\int \frac {\log \left (1+\frac {e x}{-b c d-(1-a c) e}\right )}{x} \, dx,x,1-a c-b c x\right )}{e (b d-a e)}\\ &=\frac {b \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{e (b d-a e)}+\frac {b \text {Li}_2(c (a+b x))}{e (b d-a e)}-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}+\frac {b \text {Li}_2\left (\frac {e (1-a c-b c x)}{b c d+e-a c e}\right )}{e (b d-a e)}\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 108, normalized size = 0.78 \begin {gather*} \frac {-\frac {\text {PolyLog}(2,c (a+b x))}{d+e x}+\frac {b \left (\log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )+\text {PolyLog}(2,c (a+b x))+\text {PolyLog}\left (2,\frac {e (-1+a c+b c x)}{-b c d-e+a c e}\right )\right )}{b d-a e}}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, c*(a + b*x)]/(d + e*x)^2,x]

[Out]

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

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Maple [A]
time = 2.11, size = 213, normalized size = 1.54

method	result	size
derivativedivides	\(\frac {\frac {c^{2} b^{2} \polylog \left (2, x b c +a c \right )}{\left (a e c -b c d -e \left (x b c +a c \right )\right ) e}+\frac {c^{2} b^{2} \left (-\frac {\dilog \left (-x b c -a c +1\right )}{c \left (a e -b d \right )}-\frac {\left (\frac {\dilog \left (\frac {a e c -b c d +e \left (-x b c -a c +1\right )-e}{a e c -b c d -e}\right )}{e}+\frac {\ln \left (-x b c -a c +1\right ) \ln \left (\frac {a e c -b c d +e \left (-x b c -a c +1\right )-e}{a e c -b c d -e}\right )}{e}\right ) e}{c \left (a e -b d \right )}\right )}{e}}{b c}\)	\(213\)
default	\(\frac {\frac {c^{2} b^{2} \polylog \left (2, x b c +a c \right )}{\left (a e c -b c d -e \left (x b c +a c \right )\right ) e}+\frac {c^{2} b^{2} \left (-\frac {\dilog \left (-x b c -a c +1\right )}{c \left (a e -b d \right )}-\frac {\left (\frac {\dilog \left (\frac {a e c -b c d +e \left (-x b c -a c +1\right )-e}{a e c -b c d -e}\right )}{e}+\frac {\ln \left (-x b c -a c +1\right ) \ln \left (\frac {a e c -b c d +e \left (-x b c -a c +1\right )-e}{a e c -b c d -e}\right )}{e}\right ) e}{c \left (a e -b d \right )}\right )}{e}}{b c}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,c*(b*x+a))/(e*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/b/c*(c^2*b^2/(a*e*c-b*c*d-e*(b*c*x+a*c))/e*polylog(2,b*c*x+a*c)+c^2*b^2/e*(-dilog(-b*c*x-a*c+1)/c/(a*e-b*d)-
(dilog((a*e*c-b*c*d+e*(-b*c*x-a*c+1)-e)/(a*c*e-b*c*d-e))/e+ln(-b*c*x-a*c+1)*ln((a*e*c-b*c*d+e*(-b*c*x-a*c+1)-e
)/(a*c*e-b*c*d-e))/e)*e/c/(a*e-b*d)))

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Maxima [A]
time = 0.27, size = 172, normalized size = 1.25 \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 )} b}{b d e - a e^{2}} + \frac {{\left (\log \left (-b c x - a c + 1\right ) \log \left (\frac {b c x e + a c e - e}{b c d - a c e + e} + 1\right ) + {\rm Li}_2\left (-\frac {b c x e + a c e - e}{b c d - a c e + e}\right )\right )} b}{b d e - a e^{2}} - \frac {{\rm Li}_2\left (b c x + a c\right )}{x e^{2} + d e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b*c*x - a*c + 1))*b/(b*d*e - a*e^2) + (log(-b*c*x - a*c + 1)
*log((b*c*x*e + a*c*e - e)/(b*c*d - a*c*e + e) + 1) + dilog(-(b*c*x*e + a*c*e - e)/(b*c*d - a*c*e + e)))*b/(b*
d*e - a*e^2) - dilog(b*c*x + a*c)/(x*e^2 + d*e)

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Fricas [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))/(e*x+d)^2,x, algorithm="fricas")

[Out]

integral(dilog(b*c*x + a*c)/(x^2*e^2 + 2*d*x*e + d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(polylog(2, a*c + b*c*x)/(d + e*x)**2, 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(polylog(2,c*(b*x+a))/(e*x+d)^2,x, algorithm="giac")

[Out]

integrate(dilog((b*x + a)*c)/(e*x + d)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x))/(d + e*x)^2,x)

[Out]

int(polylog(2, c*(a + b*x))/(d + e*x)^2, x)

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