3.143 \(\int \frac {\text {Li}_2(c (a+b x))}{(d+e x)^3} \, dx\)

Optimal. Leaf size=278 \[ \frac {b^2 \text {Li}_2(c (a+b x))}{2 e (b d-a e)^2}+\frac {b^2 \text {Li}_2\left (\frac {e (-a c-b x c+1)}{b c d-a c e+e}\right )}{2 e (b d-a e)^2}+\frac {b^2 c \log (-a c-b c x+1)}{2 e (b d-a e) (-a c e+b c d+e)}-\frac {b^2 c \log (d+e x)}{2 e (b d-a e) (-a c e+b c d+e)}+\frac {b^2 \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac {b \log (-a c-b c x+1)}{2 e (d+e x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.27, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {6598, 2418, 2393, 2391, 2395, 36, 31, 2394} \[ \frac {b^2 \text {PolyLog}(2,c (a+b x))}{2 e (b d-a e)^2}+\frac {b^2 \text {PolyLog}\left (2,\frac {e (-a c-b c x+1)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac {\text {PolyLog}(2,c (a+b x))}{2 e (d+e x)^2}+\frac {b^2 c \log (-a c-b c x+1)}{2 e (b d-a e) (-a c e+b c d+e)}-\frac {b^2 c \log (d+e x)}{2 e (b d-a e) (-a c e+b c d+e)}+\frac {b^2 \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )}{2 e (b d-a e)^2}-\frac {b \log (-a c-b c x+1)}{2 e (d+e x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[PolyLog[2, c*(a + b*x)]/(d + e*x)^3,x]

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 2395

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

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)^3} \, dx &=-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac {b \int \frac {\log (1-a c-b c x)}{(a+b x) (d+e x)^2} \, dx}{2 e}\\ &=-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac {b \int \left (\frac {b^2 \log (1-a c-b c x)}{(b d-a e)^2 (a+b x)}-\frac {e \log (1-a c-b c x)}{(b d-a e) (d+e x)^2}-\frac {b e \log (1-a c-b c x)}{(b d-a e)^2 (d+e x)}\right ) \, dx}{2 e}\\ &=-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}+\frac {b^2 \int \frac {\log (1-a c-b c x)}{d+e x} \, dx}{2 (b d-a e)^2}-\frac {b^3 \int \frac {\log (1-a c-b c x)}{a+b x} \, dx}{2 e (b d-a e)^2}+\frac {b \int \frac {\log (1-a c-b c x)}{(d+e x)^2} \, dx}{2 (b d-a e)}\\ &=-\frac {b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}+\frac {b^2 \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,a+b x\right )}{2 e (b d-a e)^2}+\frac {\left (b^3 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}{2 e (b d-a e)^2}-\frac {\left (b^2 c\right ) \int \frac {1}{(1-a c-b c x) (d+e x)} \, dx}{2 e (b d-a e)}\\ &=-\frac {b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}+\frac {b^2 \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}+\frac {b^2 \text {Li}_2(c (a+b x))}{2 e (b d-a e)^2}-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}-\frac {b^2 \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 )}{2 e (b d-a e)^2}-\frac {\left (b^2 c\right ) \int \frac {1}{d+e x} \, dx}{2 (b d-a e) (b c d+e-a c e)}-\frac {\left (b^3 c^2\right ) \int \frac {1}{1-a c-b c x} \, dx}{2 e (b d-a e) (b c d+e-a c e)}\\ &=\frac {b^2 c \log (1-a c-b c x)}{2 e (b d-a e) (b c d+e-a c e)}-\frac {b \log (1-a c-b c x)}{2 e (b d-a e) (d+e x)}-\frac {b^2 c \log (d+e x)}{2 e (b d-a e) (b c d+e-a c e)}+\frac {b^2 \log (1-a c-b c x) \log \left (\frac {b c (d+e x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}+\frac {b^2 \text {Li}_2(c (a+b x))}{2 e (b d-a e)^2}-\frac {\text {Li}_2(c (a+b x))}{2 e (d+e x)^2}+\frac {b^2 \text {Li}_2\left (\frac {e (1-a c-b c x)}{b c d+e-a c e}\right )}{2 e (b d-a e)^2}\\ \end {align*}

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Mathematica [A]  time = 0.31, size = 190, normalized size = 0.68 \[ \frac {\frac {b \left (b \text {Li}_2\left (\frac {e (a c+b x c-1)}{(a c-1) e-b c d}\right )+b \log (-a c-b c x+1) \log \left (\frac {b c (d+e x)}{-a c e+b c d+e}\right )-\frac {(b d-a e) \log (-a c-b c x+1)}{d+e x}+\frac {b c (b d-a e) (\log (-a c-b c x+1)-\log (d+e x))}{-a c e+b c d+e}+b \text {Li}_2(c (a+b x))\right )}{(b d-a e)^2}-\frac {\text {Li}_2(c (a+b x))}{(d+e x)^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

Integrate[PolyLog[2, c*(a + b*x)]/(d + e*x)^3,x]

[Out]

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

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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\rm Li}_2\left (b c x + a c\right )}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral(dilog(b*c*x + a*c)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\rm Li}_2\left ({\left (b x + a\right )} c\right )}{{\left (e x + d\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate(dilog((b*x + a)*c)/(e*x + d)^3, x)

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maple [A]  time = 0.02, size = 437, normalized size = 1.57 \[ -\frac {b^{2} c^{2} \polylog \left (2, b c x +a c \right )}{2 \left (b c e x +b c d \right )^{2} e}+\frac {b^{2} \dilog \left (\frac {a c e -b c d +e \left (-b c x -a c +1\right )-e}{a c e -b c d -e}\right )}{2 e \left (a e -b d \right )^{2}}+\frac {b^{2} \ln \left (-b c x -a c +1\right ) \ln \left (\frac {a c e -b c d +e \left (-b c x -a c +1\right )-e}{a c e -b c d -e}\right )}{2 e \left (a e -b d \right )^{2}}+\frac {b^{2} \dilog \left (-b c x -a c +1\right )}{2 e \left (a e -b d \right )^{2}}-\frac {b^{2} c \ln \left (a c e -b c d +e \left (-b c x -a c +1\right )-e \right )}{2 e \left (a e -b d \right ) \left (a c e -b c d -e \right )}-\frac {b^{3} c^{2} \ln \left (-b c x -a c +1\right ) x}{2 \left (a e -b d \right ) \left (a c e -b c d -e \right ) \left (-b c e x -b c d \right )}-\frac {b^{2} c^{2} \ln \left (-b c x -a c +1\right ) a}{2 \left (a e -b d \right ) \left (a c e -b c d -e \right ) \left (-b c e x -b c d \right )}+\frac {b^{2} c \ln \left (-b c x -a c +1\right )}{2 \left (a e -b d \right ) \left (a c e -b c d -e \right ) \left (-b c e x -b c d \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2,c*(b*x+a))/(e*x+d)^3,x)

[Out]

-1/2*b^2*c^2/(b*c*e*x+b*c*d)^2/e*polylog(2,b*c*x+a*c)+1/2*b^2/e/(a*e-b*d)^2*dilog((a*c*e-b*c*d+e*(-b*c*x-a*c+1
)-e)/(a*c*e-b*c*d-e))+1/2*b^2/e/(a*e-b*d)^2*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))+1/2*b^2/e*dilog(-b*c*x-a*c+1)/(a*e-b*d)^2-1/2*b^2*c/e/(a*e-b*d)/(a*c*e-b*c*d-e)*ln(a*c*e-b*c*d+e*(-b*c*x-a
*c+1)-e)-1/2*b^3*c^2/(a*e-b*d)*ln(-b*c*x-a*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-b*c*d)*x-1/2*b^2*c^2/(a*e-b*d)*ln(-b
*c*x-a*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-b*c*d)*a+1/2*b^2*c/(a*e-b*d)*ln(-b*c*x-a*c+1)/(a*c*e-b*c*d-e)/(-b*c*e*x-
b*c*d)

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maxima [A]  time = 0.32, size = 379, normalized size = 1.36 \[ \frac {b^{2} c \log \left (b c x + a c - 1\right )}{2 \, {\left (b^{2} c d^{2} e - {\left (2 \, a b c - b\right )} d e^{2} + {\left (a^{2} c - a\right )} e^{3}\right )}} - \frac {b^{2} c \log \left (e x + d\right )}{2 \, {\left (b^{2} c d^{2} e - {\left (2 \, a b c - b\right )} d e^{2} + {\left (a^{2} c - a\right )} e^{3}\right )}} - \frac {{\left (\log \left (b c x + a c\right ) \log \left (-b c x - a c + 1\right ) + {\rm Li}_2\left (-b c x - a c + 1\right )\right )} b^{2}}{2 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )}} + \frac {{\left (\log \left (-b c x - a c + 1\right ) \log \left (\frac {b c e x + {\left (a c - 1\right )} e}{b c d - {\left (a c - 1\right )} e} + 1\right ) + {\rm Li}_2\left (-\frac {b c e x + {\left (a c - 1\right )} e}{b c d - {\left (a c - 1\right )} e}\right )\right )} b^{2}}{2 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )}} - \frac {{\left (b d - a e\right )} {\rm Li}_2\left (b c x + a c\right ) + {\left (b e x + b d\right )} \log \left (-b c x - a c + 1\right )}{2 \, {\left (b d^{3} e - a d^{2} e^{2} + {\left (b d e^{3} - a e^{4}\right )} x^{2} + 2 \, {\left (b d^{2} e^{2} - a d e^{3}\right )} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)^3,x, algorithm="maxima")

[Out]

1/2*b^2*c*log(b*c*x + a*c - 1)/(b^2*c*d^2*e - (2*a*b*c - b)*d*e^2 + (a^2*c - a)*e^3) - 1/2*b^2*c*log(e*x + d)/
(b^2*c*d^2*e - (2*a*b*c - b)*d*e^2 + (a^2*c - a)*e^3) - 1/2*(log(b*c*x + a*c)*log(-b*c*x - a*c + 1) + dilog(-b
*c*x - a*c + 1))*b^2/(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3) + 1/2*(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^2/(b^2*d^2*e - 2*a*b
*d*e^2 + a^2*e^3) - 1/2*((b*d - a*e)*dilog(b*c*x + a*c) + (b*e*x + b*d)*log(-b*c*x - a*c + 1))/(b*d^3*e - a*d^
2*e^2 + (b*d*e^3 - a*e^4)*x^2 + 2*(b*d^2*e^2 - a*d*e^3)*x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {polylog}\left (2,c\,\left (a+b\,x\right )\right )}{{\left (d+e\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(polylog(2, c*(a + b*x))/(d + e*x)^3,x)

[Out]

int(polylog(2, c*(a + b*x))/(d + e*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {Li}_{2}\left (a c + b c x\right )}{\left (d + e x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(polylog(2,c*(b*x+a))/(e*x+d)**3,x)

[Out]

Integral(polylog(2, a*c + b*c*x)/(d + e*x)**3, x)

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