3.120 \(\int \frac{x \log ^2(x)}{(d+e x)^4} \, dx\)

Optimal. Leaf size=107 \[ -\frac{\text{PolyLog}\left (2,-\frac{e x}{d}\right )}{3 d^2 e^2}-\frac{\log (x) \log \left (\frac{e x}{d}+1\right )}{3 d^2 e^2}+\frac{x^2 \log ^2(x) (3 d+e x)}{6 d^2 (d+e x)^3}-\frac{x}{3 d^2 e (d+e x)}+\frac{x \log (x)}{3 d e (d+e x)^2} \]

[Out]

-x/(3*d^2*e*(d + e*x)) + (x*Log[x])/(3*d*e*(d + e*x)^2) + (x^2*(3*d + e*x)*Log[x]^2)/(6*d^2*(d + e*x)^3) - (Lo
g[x]*Log[1 + (e*x)/d])/(3*d^2*e^2) - PolyLog[2, -((e*x)/d)]/(3*d^2*e^2)

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Rubi [A]  time = 0.406657, antiderivative size = 157, normalized size of antiderivative = 1.47, number of steps used = 22, number of rules used = 10, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.769, Rules used = {2353, 2319, 2347, 2344, 2301, 2317, 2391, 2314, 31, 44} \[ -\frac{\text{PolyLog}\left (2,-\frac{e x}{d}\right )}{3 d^2 e^2}+\frac{\log ^2(x)}{6 d^2 e^2}-\frac{\log (x) \log \left (\frac{e x}{d}+1\right )}{3 d^2 e^2}+\frac{\log (x)}{3 d^2 e^2}-\frac{x \log (x)}{3 d^2 e (d+e x)}+\frac{1}{3 d e^2 (d+e x)}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}+\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log (x)}{3 e^2 (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(x*Log[x]^2)/(d + e*x)^4,x]

[Out]

1/(3*d*e^2*(d + e*x)) + Log[x]/(3*d^2*e^2) - Log[x]/(3*e^2*(d + e*x)^2) - (x*Log[x])/(3*d^2*e*(d + e*x)) + Log
[x]^2/(6*d^2*e^2) + (d*Log[x]^2)/(3*e^2*(d + e*x)^3) - Log[x]^2/(2*e^2*(d + e*x)^2) - (Log[x]*Log[1 + (e*x)/d]
)/(3*d^2*e^2) - PolyLog[2, -((e*x)/d)]/(3*d^2*e^2)

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 0]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 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 2314

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

Rule 31

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x \log ^2(x)}{(d+e x)^4} \, dx &=\int \left (-\frac{d \log ^2(x)}{e (d+e x)^4}+\frac{\log ^2(x)}{e (d+e x)^3}\right ) \, dx\\ &=\frac{\int \frac{\log ^2(x)}{(d+e x)^3} \, dx}{e}-\frac{d \int \frac{\log ^2(x)}{(d+e x)^4} \, dx}{e}\\ &=\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}+\frac{\int \frac{\log (x)}{x (d+e x)^2} \, dx}{e^2}-\frac{(2 d) \int \frac{\log (x)}{x (d+e x)^3} \, dx}{3 e^2}\\ &=\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}-\frac{2 \int \frac{\log (x)}{x (d+e x)^2} \, dx}{3 e^2}+\frac{\int \frac{\log (x)}{x (d+e x)} \, dx}{d e^2}+\frac{2 \int \frac{\log (x)}{(d+e x)^3} \, dx}{3 e}-\frac{\int \frac{\log (x)}{(d+e x)^2} \, dx}{d e}\\ &=-\frac{\log (x)}{3 e^2 (d+e x)^2}-\frac{x \log (x)}{d^2 e (d+e x)}+\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}+\frac{\int \frac{1}{x (d+e x)^2} \, dx}{3 e^2}+\frac{\int \frac{\log (x)}{x} \, dx}{d^2 e^2}-\frac{2 \int \frac{\log (x)}{x (d+e x)} \, dx}{3 d e^2}+\frac{\int \frac{1}{d+e x} \, dx}{d^2 e}-\frac{\int \frac{\log (x)}{d+e x} \, dx}{d^2 e}+\frac{2 \int \frac{\log (x)}{(d+e x)^2} \, dx}{3 d e}\\ &=-\frac{\log (x)}{3 e^2 (d+e x)^2}-\frac{x \log (x)}{3 d^2 e (d+e x)}+\frac{\log ^2(x)}{2 d^2 e^2}+\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}+\frac{\log (d+e x)}{d^2 e^2}-\frac{\log (x) \log \left (1+\frac{e x}{d}\right )}{d^2 e^2}+\frac{\int \left (\frac{1}{d^2 x}-\frac{e}{d (d+e x)^2}-\frac{e}{d^2 (d+e x)}\right ) \, dx}{3 e^2}-\frac{2 \int \frac{\log (x)}{x} \, dx}{3 d^2 e^2}+\frac{\int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{d^2 e^2}-\frac{2 \int \frac{1}{d+e x} \, dx}{3 d^2 e}+\frac{2 \int \frac{\log (x)}{d+e x} \, dx}{3 d^2 e}\\ &=\frac{1}{3 d e^2 (d+e x)}+\frac{\log (x)}{3 d^2 e^2}-\frac{\log (x)}{3 e^2 (d+e x)^2}-\frac{x \log (x)}{3 d^2 e (d+e x)}+\frac{\log ^2(x)}{6 d^2 e^2}+\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}-\frac{\log (x) \log \left (1+\frac{e x}{d}\right )}{3 d^2 e^2}-\frac{\text{Li}_2\left (-\frac{e x}{d}\right )}{d^2 e^2}-\frac{2 \int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{3 d^2 e^2}\\ &=\frac{1}{3 d e^2 (d+e x)}+\frac{\log (x)}{3 d^2 e^2}-\frac{\log (x)}{3 e^2 (d+e x)^2}-\frac{x \log (x)}{3 d^2 e (d+e x)}+\frac{\log ^2(x)}{6 d^2 e^2}+\frac{d \log ^2(x)}{3 e^2 (d+e x)^3}-\frac{\log ^2(x)}{2 e^2 (d+e x)^2}-\frac{\log (x) \log \left (1+\frac{e x}{d}\right )}{3 d^2 e^2}-\frac{\text{Li}_2\left (-\frac{e x}{d}\right )}{3 d^2 e^2}\\ \end{align*}

Mathematica [A]  time = 0.126762, size = 96, normalized size = 0.9 \[ \frac{-2 (d+e x)^3 \text{PolyLog}\left (2,-\frac{e x}{d}\right )+e^2 x^2 \log ^2(x) (3 d+e x)+2 d (d+e x)^2-2 \log (x) (d+e x) \left ((d+e x)^2 \log \left (\frac{e x}{d}+1\right )-d e x\right )}{6 d^2 e^2 (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*Log[x]^2)/(d + e*x)^4,x]

[Out]

(2*d*(d + e*x)^2 + e^2*x^2*(3*d + e*x)*Log[x]^2 - 2*(d + e*x)*Log[x]*(-(d*e*x) + (d + e*x)^2*Log[1 + (e*x)/d])
- 2*(d + e*x)^3*PolyLog[2, -((e*x)/d)])/(6*d^2*e^2*(d + e*x)^3)

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Maple [F]  time = 0.345, size = 0, normalized size = 0. \begin{align*} \int{\frac{x \left ( \ln \left ( x \right ) \right ) ^{2}}{ \left ( ex+d \right ) ^{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*ln(x)^2/(e*x+d)^4,x)

[Out]

int(x*ln(x)^2/(e*x+d)^4,x)

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Maxima [A]  time = 1.18798, size = 178, normalized size = 1.66 \begin{align*} -\frac{d^{2} \log \left (x\right )^{2} - 2 \,{\left (e^{2} \log \left (x\right ) + e^{2}\right )} x^{2} - 2 \, d^{2} +{\left (3 \, d e \log \left (x\right )^{2} - 2 \, d e \log \left (x\right ) - 4 \, d e\right )} x}{6 \,{\left (d e^{5} x^{3} + 3 \, d^{2} e^{4} x^{2} + 3 \, d^{3} e^{3} x + d^{4} e^{2}\right )}} + \frac{\log \left (x\right )^{2}}{6 \, d^{2} e^{2}} - \frac{\log \left (\frac{e x}{d} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{e x}{d}\right )}{3 \, d^{2} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x)^2/(e*x+d)^4,x, algorithm="maxima")

[Out]

-1/6*(d^2*log(x)^2 - 2*(e^2*log(x) + e^2)*x^2 - 2*d^2 + (3*d*e*log(x)^2 - 2*d*e*log(x) - 4*d*e)*x)/(d*e^5*x^3
+ 3*d^2*e^4*x^2 + 3*d^3*e^3*x + d^4*e^2) + 1/6*log(x)^2/(d^2*e^2) - 1/3*(log(e*x/d + 1)*log(x) + dilog(-e*x/d)
)/(d^2*e^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x \log \left (x\right )^{2}}{e^{4} x^{4} + 4 \, d e^{3} x^{3} + 6 \, d^{2} e^{2} x^{2} + 4 \, d^{3} e x + d^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x)^2/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral(x*log(x)^2/(e^4*x^4 + 4*d*e^3*x^3 + 6*d^2*e^2*x^2 + 4*d^3*e*x + d^4), x)

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Sympy [A]  time = 34.4248, size = 357, normalized size = 3.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*ln(x)**2/(e*x+d)**4,x)

[Out]

(-d - 3*e*x)*log(x)**2/(6*d**3*e**2 + 18*d**2*e**3*x + 18*d*e**4*x**2 + 6*e**5*x**3) + Piecewise((x/d**3, Eq(e
, 0)), (-1/(2*e*(d + e*x)**2), True))*log(x)/e - Piecewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - l
og(x)/(2*d**2*e) + log(d/e + x)/(2*d**2*e), True))/e + Piecewise((-1/(e**3*x), Eq(d, 0)), (-1/(2*d*e**2 + 2*e*
*3*x) - log(d + e*x)/(2*d*e**2), True))/(3*d) - Piecewise((1/(e**3*x), Eq(d, 0)), (-1/(2*d*(d/x + e)**2), True
))*log(x)/(3*d) - 2*Piecewise((-1/(e**2*x), Eq(d, 0)), (-log(d**2 + d*e*x)/(d*e), True))/(3*d*e) + 2*Piecewise
((1/(e**2*x), Eq(d, 0)), (-1/(d**2/x + d*e), True))*log(x)/(3*d*e) + Piecewise((-1/(e*x), Eq(d, 0)), (Piecewis
e((log(e)*log(x) + polylog(2, d*exp_polar(I*pi)/(e*x)), Abs(x) < 1), (-log(e)*log(1/x) + polylog(2, d*exp_pola
r(I*pi)/(e*x)), 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(e) + meijerg(((1, 1), ()), ((), (0
, 0)), x)*log(e) + polylog(2, d*exp_polar(I*pi)/(e*x)), True))/d, True))/(3*d*e**2) - Piecewise((1/(e*x), Eq(d
, 0)), (log(d/x + e)/d, True))*log(x)/(3*d*e**2)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log \left (x\right )^{2}}{{\left (e x + d\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*log(x)^2/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate(x*log(x)^2/(e*x + d)^4, x)