3.59 \(\int \frac{a+b \log (c x^n)}{x (d+e x)^4} \, dx\)

Optimal. Leaf size=174 \[ \frac{b n \text{PolyLog}\left (2,-\frac{d}{e x}\right )}{d^4}-\frac{\log \left (\frac{d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac{e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}+\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}-\frac{5 b n}{6 d^3 (d+e x)}-\frac{b n}{6 d^2 (d+e x)^2}+\frac{11 b n \log (d+e x)}{6 d^4}-\frac{5 b n \log (x)}{6 d^4} \]

[Out]

-(b*n)/(6*d^2*(d + e*x)^2) - (5*b*n)/(6*d^3*(d + e*x)) - (5*b*n*Log[x])/(6*d^4) + (a + b*Log[c*x^n])/(3*d*(d +
e*x)^3) + (a + b*Log[c*x^n])/(2*d^2*(d + e*x)^2) - (e*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) - (Log[1 + d/(e*x
)]*(a + b*Log[c*x^n]))/d^4 + (11*b*n*Log[d + e*x])/(6*d^4) + (b*n*PolyLog[2, -(d/(e*x))])/d^4

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Rubi [A]  time = 0.357489, antiderivative size = 196, normalized size of antiderivative = 1.13, number of steps used = 15, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2347, 2344, 2301, 2317, 2391, 2314, 31, 2319, 44} \[ -\frac{b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{d^4}-\frac{\log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac{e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}+\frac{\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}-\frac{5 b n}{6 d^3 (d+e x)}-\frac{b n}{6 d^2 (d+e x)^2}+\frac{11 b n \log (d+e x)}{6 d^4}-\frac{5 b n \log (x)}{6 d^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x*(d + e*x)^4),x]

[Out]

-(b*n)/(6*d^2*(d + e*x)^2) - (5*b*n)/(6*d^3*(d + e*x)) - (5*b*n*Log[x])/(6*d^4) + (a + b*Log[c*x^n])/(3*d*(d +
e*x)^3) + (a + b*Log[c*x^n])/(2*d^2*(d + e*x)^2) - (e*x*(a + b*Log[c*x^n]))/(d^4*(d + e*x)) + (a + b*Log[c*x^
n])^2/(2*b*d^4*n) + (11*b*n*Log[d + e*x])/(6*d^4) - ((a + b*Log[c*x^n])*Log[1 + (e*x)/d])/d^4 - (b*n*PolyLog[2
, -((e*x)/d)])/d^4

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 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 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{a+b \log \left (c x^n\right )}{x (d+e x)^4} \, dx &=\frac{\int \frac{a+b \log \left (c x^n\right )}{x (d+e x)^3} \, dx}{d}-\frac{e \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^4} \, dx}{d}\\ &=\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac{\int \frac{a+b \log \left (c x^n\right )}{x (d+e x)^2} \, dx}{d^2}-\frac{e \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^3} \, dx}{d^2}-\frac{(b n) \int \frac{1}{x (d+e x)^3} \, dx}{3 d}\\ &=\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}+\frac{\int \frac{a+b \log \left (c x^n\right )}{x (d+e x)} \, dx}{d^3}-\frac{e \int \frac{a+b \log \left (c x^n\right )}{(d+e x)^2} \, dx}{d^3}-\frac{(b n) \int \frac{1}{x (d+e x)^2} \, dx}{2 d^2}-\frac{(b n) \int \left (\frac{1}{d^3 x}-\frac{e}{d (d+e x)^3}-\frac{e}{d^2 (d+e x)^2}-\frac{e}{d^3 (d+e x)}\right ) \, dx}{3 d}\\ &=-\frac{b n}{6 d^2 (d+e x)^2}-\frac{b n}{3 d^3 (d+e x)}-\frac{b n \log (x)}{3 d^4}+\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac{e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{b n \log (d+e x)}{3 d^4}+\frac{\int \frac{a+b \log \left (c x^n\right )}{x} \, dx}{d^4}-\frac{e \int \frac{a+b \log \left (c x^n\right )}{d+e x} \, dx}{d^4}-\frac{(b n) \int \left (\frac{1}{d^2 x}-\frac{e}{d (d+e x)^2}-\frac{e}{d^2 (d+e x)}\right ) \, dx}{2 d^2}+\frac{(b e n) \int \frac{1}{d+e x} \, dx}{d^4}\\ &=-\frac{b n}{6 d^2 (d+e x)^2}-\frac{5 b n}{6 d^3 (d+e x)}-\frac{5 b n \log (x)}{6 d^4}+\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac{e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac{11 b n \log (d+e x)}{6 d^4}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{d^4}+\frac{(b n) \int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{d^4}\\ &=-\frac{b n}{6 d^2 (d+e x)^2}-\frac{5 b n}{6 d^3 (d+e x)}-\frac{5 b n \log (x)}{6 d^4}+\frac{a+b \log \left (c x^n\right )}{3 d (d+e x)^3}+\frac{a+b \log \left (c x^n\right )}{2 d^2 (d+e x)^2}-\frac{e x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{\left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}+\frac{11 b n \log (d+e x)}{6 d^4}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x}{d}\right )}{d^4}-\frac{b n \text{Li}_2\left (-\frac{e x}{d}\right )}{d^4}\\ \end{align*}

Mathematica [A]  time = 0.170912, size = 222, normalized size = 1.28 \[ \frac{-6 b n \text{PolyLog}\left (2,-\frac{e x}{d}\right )+\frac{3 a^2}{b n}+\frac{6 a \log \left (c x^n\right )}{n}+\frac{2 a d^3}{(d+e x)^3}+\frac{3 a d^2}{(d+e x)^2}+\frac{6 a d}{d+e x}-6 a \log \left (\frac{e x}{d}+1\right )+\frac{2 b d^3 \log \left (c x^n\right )}{(d+e x)^3}+\frac{3 b d^2 \log \left (c x^n\right )}{(d+e x)^2}+\frac{6 b d \log \left (c x^n\right )}{d+e x}-6 b \log \left (c x^n\right ) \log \left (\frac{e x}{d}+1\right )+\frac{3 b \log ^2\left (c x^n\right )}{n}-\frac{b d^2 n}{(d+e x)^2}-\frac{5 b d n}{d+e x}+11 b n \log (d+e x)-11 b n \log (x)}{6 d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*x^n])/(x*(d + e*x)^4),x]

[Out]

((3*a^2)/(b*n) + (2*a*d^3)/(d + e*x)^3 + (3*a*d^2)/(d + e*x)^2 - (b*d^2*n)/(d + e*x)^2 + (6*a*d)/(d + e*x) - (
5*b*d*n)/(d + e*x) - 11*b*n*Log[x] + (6*a*Log[c*x^n])/n + (2*b*d^3*Log[c*x^n])/(d + e*x)^3 + (3*b*d^2*Log[c*x^
n])/(d + e*x)^2 + (6*b*d*Log[c*x^n])/(d + e*x) + (3*b*Log[c*x^n]^2)/n + 11*b*n*Log[d + e*x] - 6*a*Log[1 + (e*x
)/d] - 6*b*Log[c*x^n]*Log[1 + (e*x)/d] - 6*b*n*PolyLog[2, -((e*x)/d)])/(6*d^4)

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Maple [C]  time = 0.164, size = 884, normalized size = 5.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))/x/(e*x+d)^4,x)

[Out]

-1/6*I*b*Pi*csgn(I*c*x^n)^3/d/(e*x+d)^3-1/2*I*b*Pi*csgn(I*c*x^n)^3/d^4*ln(x)-1/2*b*n/d^4*ln(x)^2+b*n/d^4*dilog
(-e*x/d)-1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^4*ln(e*x+d)+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^2/(e*x+d)
^2+a/d^3/(e*x+d)+1/2*a/d^2/(e*x+d)^2+1/3*a/d/(e*x+d)^3-a/d^4*ln(e*x+d)+a/d^4*ln(x)+1/2*I*b*Pi*csgn(I*c*x^n)^3/
d^4*ln(e*x+d)+b*n/d^4*ln(e*x+d)*ln(-e*x/d)+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*ln(x)+1/2*I*b*Pi*csgn(I*
c*x^n)^2*csgn(I*c)/d^4*ln(x)+1/2*b*ln(x^n)/d^2/(e*x+d)^2+1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/(e*x+d)^3+b*
ln(x^n)/d^3/(e*x+d)-1/2*I*b*Pi*csgn(I*c*x^n)^3/d^3/(e*x+d)-1/4*I*b*Pi*csgn(I*c*x^n)^3/d^2/(e*x+d)^2-1/4*I*b*Pi
*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^2/(e*x+d)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^4*ln(e*x+d
)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^3/(e*x+d)-1/6*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d/
(e*x+d)^3-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d^4*ln(x)+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^3/(e
*x+d)+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/d^2/(e*x+d)^2-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^4*ln(e*x+d)+
b*ln(c)/d^4*ln(x)-b*ln(c)/d^4*ln(e*x+d)+1/3*b*ln(x^n)/d/(e*x+d)^3+b*ln(x^n)/d^4*ln(x)+1/6*I*b*Pi*csgn(I*c*x^n)
^2*csgn(I*c)/d/(e*x+d)^3+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d^3/(e*x+d)+b*ln(c)/d^3/(e*x+d)+1/2*b*ln(c)/d^
2/(e*x+d)^2+1/3*b*ln(c)/d/(e*x+d)^3-b*ln(x^n)/d^4*ln(e*x+d)-11/6*b*n*ln(x)/d^4+11/6*b*n*ln(e*x+d)/d^4-1/6*b*n/
d^2/(e*x+d)^2-5/6*b*n/d^3/(e*x+d)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{6} \, a{\left (\frac{6 \, e^{2} x^{2} + 15 \, d e x + 11 \, d^{2}}{d^{3} e^{3} x^{3} + 3 \, d^{4} e^{2} x^{2} + 3 \, d^{5} e x + d^{6}} - \frac{6 \, \log \left (e x + d\right )}{d^{4}} + \frac{6 \, \log \left (x\right )}{d^{4}}\right )} + b \int \frac{\log \left (c\right ) + \log \left (x^{n}\right )}{e^{4} x^{5} + 4 \, d e^{3} x^{4} + 6 \, d^{2} e^{2} x^{3} + 4 \, d^{3} e x^{2} + d^{4} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*a*((6*e^2*x^2 + 15*d*e*x + 11*d^2)/(d^3*e^3*x^3 + 3*d^4*e^2*x^2 + 3*d^5*e*x + d^6) - 6*log(e*x + d)/d^4 +
6*log(x)/d^4) + b*integrate((log(c) + log(x^n))/(e^4*x^5 + 4*d*e^3*x^4 + 6*d^2*e^2*x^3 + 4*d^3*e*x^2 + d^4*x),
x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left (c x^{n}\right ) + a}{e^{4} x^{5} + 4 \, d e^{3} x^{4} + 6 \, d^{2} e^{2} x^{3} + 4 \, d^{3} e x^{2} + d^{4} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n) + a)/(e^4*x^5 + 4*d*e^3*x^4 + 6*d^2*e^2*x^3 + 4*d^3*e*x^2 + d^4*x), x)

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Sympy [A]  time = 130.542, size = 493, normalized size = 2.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x/(e*x+d)**4,x)

[Out]

-a*e*Piecewise((x/d**4, Eq(e, 0)), (-1/(3*e*(d + e*x)**3), True))/d - a*e*Piecewise((x/d**3, Eq(e, 0)), (-1/(2
*e*(d + e*x)**2), True))/d**2 - a*e*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))/d**3 - a*e*Piecew
ise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))/d**4 + a*log(x)/d**4 - b*e**3*n*Piecewise((-1/(e**4*x), Eq(d, 0))
, (-3*d/(6*d**2*e**3 + 12*d*e**4*x + 6*e**5*x**2) - 4*e*x/(6*d**2*e**3 + 12*d*e**4*x + 6*e**5*x**2) - log(d +
e*x)/(3*d*e**3), True))/d**3 + b*e**3*Piecewise((1/(e**4*x), Eq(d, 0)), (-1/(3*d*(d/x + e)**3), True))*log(c*x
**n)/d**3 + 3*b*e**2*n*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))/d**3 - 3*b*e**2*Piecewise((1/(e**3*x), Eq(d, 0)), (-1/(2*d*(d/x + e)**2), True))*log(c*x**n)/d**3 - 3*
b*e*n*Piecewise((-1/(e**2*x), Eq(d, 0)), (-log(d**2 + d*e*x)/(d*e), True))/d**3 + 3*b*e*Piecewise((1/(e**2*x),
Eq(d, 0)), (-1/(d**2/x + d*e), True))*log(c*x**n)/d**3 + b*n*Piecewise((-1/(e*x), Eq(d, 0)), (Piecewise((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_polar(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))/d**3 - b*Piecewise((1/(e*x), Eq(d, 0)), (log
(d/x + e)/d, True))*log(c*x**n)/d**3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))/x/(e*x+d)^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x + d)^4*x), x)