3.142 \(\int \frac {\text {Li}_2(c (a+b x))}{(d+e x)^2} \, dx\)

Optimal. Leaf size=138 \[ \frac {b \text {Li}_2(c (a+b x))}{e (b d-a e)}-\frac {\text {Li}_2(c (a+b x))}{e (d+e x)}+\frac {b \text {Li}_2\left (\frac {e (-a c-b x c+1)}{b c d-a c e+e}\right )}{e (b d-a e)}+\frac {b \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )}{e (b d-a e)} \]

[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 = 0.18, 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 = {6598, 2418, 2393, 2391, 2394} \[ \frac {b \text {PolyLog}(2,c (a+b x))}{e (b d-a e)}-\frac {\text {PolyLog}(2,c (a+b x))}{e (d+e x)}+\frac {b \text {PolyLog}\left (2,\frac {e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{e (b d-a e)}+\frac {b \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )}{e (b d-a e)} \]

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 2391

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 2393

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 2394

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 2418

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 6598

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 \operatorname {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 \operatorname {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 = 0.13, size = 108, normalized size = 0.78 \[ \frac {\frac {b \left (\text {Li}_2\left (\frac {e (a c+b x c-1)}{-b c d+a c e-e}\right )+\log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )+\text {Li}_2(c (a+b x))\right )}{b d-a e}-\frac {\text {Li}_2(c (a+b x))}{d+e x}}{e} \]

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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fricas [F]  time = 1.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm Li}_2\left (b c x + a c\right )}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

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)/(e^2*x^2 + 2*d*e*x + d^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{{\left (e x + d\right )}^{2}}\,{d x} \]

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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maple [A]  time = 0.03, size = 189, normalized size = 1.37 \[ -\frac {b c \polylog \left (2, b c x +a c \right )}{\left (b c e x +b c d \right ) e}-\frac {b \dilog \left (\frac {a c e -b c d +e \left (-b c x -a c +1\right )-e}{a c e -b c d -e}\right )}{e \left (a e -b d \right )}-\frac {b \ln \left (-b c x -a c +1\right ) \ln \left (\frac {a c e -b c d +e \left (-b c x -a c +1\right )-e}{a c e -b c d -e}\right )}{e \left (a e -b d \right )}-\frac {b \dilog \left (-b c x -a c +1\right )}{e \left (a e -b d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 166, normalized size = 1.20 \[ -\frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} b}{b d e - a e^{2}} + \frac {{\left (\log \left (-b c x - a c + 1\right ) \log \left (\frac {b c e x + {\left (a c - 1\right )} e}{b c d - {\left (a c - 1\right )} e} + 1\right ) + {\rm Li}_2\left (-\frac {b c e x + {\left (a c - 1\right )} e}{b c d - {\left (a c - 1\right )} e}\right )\right )} b}{b d e - a e^{2}} - \frac {{\rm Li}_2\left (b c x + a c\right )}{e^{2} x + d e} \]

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

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

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

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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