3.51 \(\int \frac{a+b \log (c x^n)}{x^2 (d+e x)^3} \, dx\)

Optimal. Leaf size=171 \[ -\frac{3 b e n \text{PolyLog}\left (2,-\frac{d}{e x}\right )}{d^4}+\frac{2 e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}+\frac{3 e \log \left (\frac{d}{e x}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 (d+e x)^2}-\frac{a+b \log \left (c x^n\right )}{d^3 x}+\frac{b e n}{2 d^3 (d+e x)}+\frac{b e n \log (x)}{2 d^4}-\frac{5 b e n \log (d+e x)}{2 d^4}-\frac{b n}{d^3 x} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.247753, antiderivative size = 193, normalized size of antiderivative = 1.13, number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {44, 2351, 2304, 2301, 2319, 2314, 31, 2317, 2391} \[ \frac{3 b e n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{d^4}+\frac{2 e^2 x \left (a+b \log \left (c x^n\right )\right )}{d^4 (d+e x)}-\frac{3 e \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^4 n}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 (d+e x)^2}+\frac{3 e \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{d^4}-\frac{a+b \log \left (c x^n\right )}{d^3 x}+\frac{b e n}{2 d^3 (d+e x)}+\frac{b e n \log (x)}{2 d^4}-\frac{5 b e n \log (d+e x)}{2 d^4}-\frac{b n}{d^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 2351

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

Rule 2304

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

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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.169734, size = 173, normalized size = 1.01 \[ \frac{6 b e n \text{PolyLog}\left (2,-\frac{e x}{d}\right )-\frac{d^2 e \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^2}-\frac{4 d e \left (a+b \log \left (c x^n\right )\right )}{d+e x}+6 e \log \left (\frac{e x}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{x}-\frac{3 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}+4 b e n (\log (x)-\log (d+e x))+b e n \left (\frac{d}{d+e x}-\log (d+e x)+\log (x)\right )-\frac{2 b d n}{x}}{2 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.168, size = 894, normalized size = 5.2 \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^2/(e*x+d)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, a{\left (\frac{6 \, e^{2} x^{2} + 9 \, d e x + 2 \, d^{2}}{d^{3} e^{2} x^{3} + 2 \, d^{4} e x^{2} + d^{5} x} - \frac{6 \, e \log \left (e x + d\right )}{d^{4}} + \frac{6 \, e \log \left (x\right )}{d^{4}}\right )} + b \int \frac{\log \left (c\right ) + \log \left (x^{n}\right )}{e^{3} x^{5} + 3 \, d e^{2} x^{4} + 3 \, d^{2} e x^{3} + d^{3} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*((6*e^2*x^2 + 9*d*e*x + 2*d^2)/(d^3*e^2*x^3 + 2*d^4*e*x^2 + d^5*x) - 6*e*log(e*x + d)/d^4 + 6*e*log(x)/
d^4) + b*integrate((log(c) + log(x^n))/(e^3*x^5 + 3*d*e^2*x^4 + 3*d^2*e*x^3 + d^3*x^2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 97.3874, size = 425, normalized size = 2.49 \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**2/(e*x+d)**3,x)

[Out]

a*e**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), True))/d**2 + 2*a*e**2*Piecewise((x/d**2, Eq(e, 0
)), (-1/(d*e + e**2*x), True))/d**3 - a/(d**3*x) + 3*a*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))
/d**4 - 3*a*e*log(x)/d**4 - b*e**2*n*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*d**2*e + 2*d*e**2*x) - log(x)/(2*d**
2*e) + log(d/e + x)/(2*d**2*e), True))/d**2 + b*e**2*Piecewise((x/d**3, Eq(e, 0)), (-1/(2*e*(d + e*x)**2), Tru
e))*log(c*x**n)/d**2 - 2*b*e**2*n*Piecewise((x/d**2, Eq(e, 0)), (-log(x)/(d*e) + log(d/e + x)/(d*e), True))/d*
*3 + 2*b*e**2*Piecewise((x/d**2, Eq(e, 0)), (-1/(d*e + e**2*x), True))*log(c*x**n)/d**3 - b*n/(d**3*x) - b*log
(c*x**n)/(d**3*x) - 3*b*e**2*n*Piecewise((x/d, Eq(e, 0)), (Piecewise((log(d)*log(x) - polylog(2, e*x*exp_polar
(I*pi)/d), Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, e*x*exp_polar(I*pi)/d), 1/Abs(x) < 1), (-meijerg(((), (
1, 1)), ((0, 0), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x*exp_polar(I*p
i)/d), True))/e, True))/d**4 + 3*b*e**2*Piecewise((x/d, Eq(e, 0)), (log(d + e*x)/e, True))*log(c*x**n)/d**4 +
3*b*e*n*log(x)**2/(2*d**4) - 3*b*e*log(x)*log(c*x**n)/d**4

________________________________________________________________________________________

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 )}^{3} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log(c*x^n) + a)/((e*x + d)^3*x^2), x)