3.364 \(\int \frac{\log (f x^m) (a+b \log (c (d+e x)^n))}{x^3} \, dx\)
Optimal. Leaf size=156 \[ -\frac{b e^2 m n \text{PolyLog}\left (2,-\frac{d}{e x}\right )}{2 d^2}-\frac{1}{4} \left (\frac{2 \log \left (f x^m\right )}{x^2}+\frac{m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b e^2 n \log \left (\frac{d}{e x}+1\right ) \log \left (f x^m\right )}{2 d^2}-\frac{b e^2 m n \log (x)}{4 d^2}+\frac{b e^2 m n \log (d+e x)}{4 d^2}-\frac{b e n \log \left (f x^m\right )}{2 d x}-\frac{3 b e m n}{4 d x} \]
[Out]
(-3*b*e*m*n)/(4*d*x) - (b*e^2*m*n*Log[x])/(4*d^2) - (b*e*n*Log[f*x^m])/(2*d*x) + (b*e^2*n*Log[1 + d/(e*x)]*Log
[f*x^m])/(2*d^2) + (b*e^2*m*n*Log[d + e*x])/(4*d^2) - ((m/x^2 + (2*Log[f*x^m])/x^2)*(a + b*Log[c*(d + e*x)^n])
)/4 - (b*e^2*m*n*PolyLog[2, -(d/(e*x))])/(2*d^2)
________________________________________________________________________________________
Rubi [A] time = 0.151057, antiderivative size = 175, normalized size of antiderivative =
1.12, number of steps used = 9, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules
used = {2426, 44, 2351, 2304, 2301, 2317, 2391} \[ \frac{b e^2 m n \text{PolyLog}\left (2,-\frac{e x}{d}\right )}{2 d^2}-\frac{1}{4} \left (\frac{2 \log \left (f x^m\right )}{x^2}+\frac{m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac{b e^2 n \log ^2\left (f x^m\right )}{4 d^2 m}+\frac{b e^2 n \log \left (\frac{e x}{d}+1\right ) \log \left (f x^m\right )}{2 d^2}-\frac{b e^2 m n \log (x)}{4 d^2}+\frac{b e^2 m n \log (d+e x)}{4 d^2}-\frac{b e n \log \left (f x^m\right )}{2 d x}-\frac{3 b e m n}{4 d x} \]
Antiderivative was successfully verified.
[In]
Int[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]))/x^3,x]
[Out]
(-3*b*e*m*n)/(4*d*x) - (b*e^2*m*n*Log[x])/(4*d^2) - (b*e*n*Log[f*x^m])/(2*d*x) - (b*e^2*n*Log[f*x^m]^2)/(4*d^2
*m) + (b*e^2*m*n*Log[d + e*x])/(4*d^2) - ((m/x^2 + (2*Log[f*x^m])/x^2)*(a + b*Log[c*(d + e*x)^n]))/4 + (b*e^2*
n*Log[f*x^m]*Log[1 + (e*x)/d])/(2*d^2) + (b*e^2*m*n*PolyLog[2, -((e*x)/d)])/(2*d^2)
Rule 2426
Int[Log[(f_.)*(x_)^(m_.)]*((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :
> -Simp[(((m*(g*x)^(q + 1))/(q + 1) - (g*x)^(q + 1)*Log[f*x^m])*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] +
(-Dist[(b*e*n)/(g*(q + 1)), Int[((g*x)^(q + 1)*Log[f*x^m])/(d + e*x), x], x] + Dist[(b*e*m*n)/(g*(q + 1)^2), I
nt[(g*x)^(q + 1)/(d + e*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, m, n, q}, x] && 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])
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 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{\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx &=-\frac{1}{4} \left (\frac{m}{x^2}+\frac{2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{1}{2} (b e n) \int \frac{\log \left (f x^m\right )}{x^2 (d+e x)} \, dx+\frac{1}{4} (b e m n) \int \frac{1}{x^2 (d+e x)} \, dx\\ &=-\frac{1}{4} \left (\frac{m}{x^2}+\frac{2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{1}{2} (b e n) \int \left (\frac{\log \left (f x^m\right )}{d x^2}-\frac{e \log \left (f x^m\right )}{d^2 x}+\frac{e^2 \log \left (f x^m\right )}{d^2 (d+e x)}\right ) \, dx+\frac{1}{4} (b e m n) \int \left (\frac{1}{d x^2}-\frac{e}{d^2 x}+\frac{e^2}{d^2 (d+e x)}\right ) \, dx\\ &=-\frac{b e m n}{4 d x}-\frac{b e^2 m n \log (x)}{4 d^2}+\frac{b e^2 m n \log (d+e x)}{4 d^2}-\frac{1}{4} \left (\frac{m}{x^2}+\frac{2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{(b e n) \int \frac{\log \left (f x^m\right )}{x^2} \, dx}{2 d}-\frac{\left (b e^2 n\right ) \int \frac{\log \left (f x^m\right )}{x} \, dx}{2 d^2}+\frac{\left (b e^3 n\right ) \int \frac{\log \left (f x^m\right )}{d+e x} \, dx}{2 d^2}\\ &=-\frac{3 b e m n}{4 d x}-\frac{b e^2 m n \log (x)}{4 d^2}-\frac{b e n \log \left (f x^m\right )}{2 d x}-\frac{b e^2 n \log ^2\left (f x^m\right )}{4 d^2 m}+\frac{b e^2 m n \log (d+e x)}{4 d^2}-\frac{1}{4} \left (\frac{m}{x^2}+\frac{2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b e^2 n \log \left (f x^m\right ) \log \left (1+\frac{e x}{d}\right )}{2 d^2}-\frac{\left (b e^2 m n\right ) \int \frac{\log \left (1+\frac{e x}{d}\right )}{x} \, dx}{2 d^2}\\ &=-\frac{3 b e m n}{4 d x}-\frac{b e^2 m n \log (x)}{4 d^2}-\frac{b e n \log \left (f x^m\right )}{2 d x}-\frac{b e^2 n \log ^2\left (f x^m\right )}{4 d^2 m}+\frac{b e^2 m n \log (d+e x)}{4 d^2}-\frac{1}{4} \left (\frac{m}{x^2}+\frac{2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac{b e^2 n \log \left (f x^m\right ) \log \left (1+\frac{e x}{d}\right )}{2 d^2}+\frac{b e^2 m n \text{Li}_2\left (-\frac{e x}{d}\right )}{2 d^2}\\ \end{align*}
Mathematica [A] time = 0.125022, size = 204, normalized size = 1.31 \[ -\frac{-2 b e^2 m n x^2 \text{PolyLog}\left (2,-\frac{e x}{d}\right )+2 a d^2 \log \left (f x^m\right )+a d^2 m+2 b d^2 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b d^2 m \log \left (c (d+e x)^n\right )-2 b e^2 n x^2 \log (d+e x) \log \left (f x^m\right )+b e^2 n x^2 \log (x) \left (2 m \log (d+e x)-2 m \log \left (\frac{e x}{d}+1\right )+2 \log \left (f x^m\right )+m\right )-b e^2 m n x^2 \log (d+e x)+2 b d e n x \log \left (f x^m\right )+3 b d e m n x-b e^2 m n x^2 \log ^2(x)}{4 d^2 x^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(Log[f*x^m]*(a + b*Log[c*(d + e*x)^n]))/x^3,x]
[Out]
-(a*d^2*m + 3*b*d*e*m*n*x - b*e^2*m*n*x^2*Log[x]^2 + 2*a*d^2*Log[f*x^m] + 2*b*d*e*n*x*Log[f*x^m] - b*e^2*m*n*x
^2*Log[d + e*x] - 2*b*e^2*n*x^2*Log[f*x^m]*Log[d + e*x] + b*d^2*m*Log[c*(d + e*x)^n] + 2*b*d^2*Log[f*x^m]*Log[
c*(d + e*x)^n] + b*e^2*n*x^2*Log[x]*(m + 2*Log[f*x^m] + 2*m*Log[d + e*x] - 2*m*Log[1 + (e*x)/d]) - 2*b*e^2*m*n
*x^2*PolyLog[2, -((e*x)/d)])/(4*d^2*x^2)
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Maple [C] time = 0.841, size = 2051, normalized size = 13.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(ln(f*x^m)*(a+b*ln(c*(e*x+d)^n))/x^3,x)
[Out]
-1/4*I/d*e*b*n/x*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2
*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2+1/4/d^2*b*e^2*
m*n*ln(x)^2+1/4*I/d^2*b*e^2*n*ln(x)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+(-1/2*b/x^2*ln(x^m)-1/4*(-I*Pi*b*cs
gn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+I*Pi*b*csgn(I*f)*csgn(I*f*x^m)^2+I*Pi*b*csgn(I*x^m)*csgn(I*f*x^m)^2-I*Pi*b*c
sgn(I*f*x^m)^3+2*b*ln(f)+b*m)/x^2)*ln((e*x+d)^n)+1/4*I/d*e*b*n/x*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*I/
x^2*Pi*b*m*csgn(I*c*(e*x+d)^n)^3-1/4*I/x^2*Pi*a*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/x^2*Pi*a*csgn(I*x^m)*csgn(I*f*
x^m)^2+1/4*I*b*Pi*csgn(I*c*(e*x+d)^n)^3/x^2*ln(x^m)-1/4/x^2*a*m-1/4*I/x^2*ln(f)*Pi*b*csgn(I*(e*x+d)^n)*csgn(I*
c*(e*x+d)^n)^2-1/4*I/x^2*Pi*ln(c)*b*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/x^2*Pi*ln(c)*b*csgn(I*x^m)*csgn(I*f*x^m)^2
-1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/2/x^2*ln(f)*a+1/4*I/x^2*Pi*a*csgn(I*f)*c
sgn(I*x^m)*csgn(I*f*x^m)-1/4*I/x^2*ln(f)*Pi*b*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x
+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2-1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn
(I*f)*csgn(I*x^m)*csgn(I*f*x^m)-1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*x^m)*c
sgn(I*f*x^m)-1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*f*x^m)^3-1/2*b*ln(c)/x^2*ln(x^m)-1/2/x^2*ln(f)*ln(c)*b-1/4/
x^2*ln(c)*b*m+1/4*I/x^2*ln(f)*Pi*b*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/2*a/x^2*ln(x^m)-1/2*m*b*e
^2*n/d^2*dilog(-e*x/d)+1/4*I/d^2*b*e^2*n*ln(e*x+d)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/4*I/d*e*b*n/x*Pi*csgn(I*f)*c
sgn(I*f*x^m)^2-1/4*I/d^2*b*e^2*n*ln(x)*Pi*csgn(I*x^m)*csgn(I*f*x^m)^2-1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csg
n(I*f)*csgn(I*f*x^m)^2-1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2+1/2*e^2*n*b*ln(x^m)/d^
2*ln(e*x+d)-1/2*e^2*n*b*ln(x^m)/d^2*ln(x)-1/2*e*n*b*ln(x^m)/d/x-1/2*m*b*e^2*n/d^2*ln(e*x+d)*ln(-e*x/d)-1/8*I/x
^2*Pi*b*m*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f*x^m)
^3-1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f*x^m)^3+1/4*I/x^2*ln(f)*Pi*b*csgn(I*c*(e*x+d
)^n)^3+1/4*I/x^2*Pi*ln(c)*b*csgn(I*f*x^m)^3+1/4*I/d^2*b*e^2*n*ln(x)*Pi*csgn(I*f*x^m)^3-1/8*b*Pi^2*csgn(I*c)*cs
gn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*f)*csgn(I*f*x^m)^2-1/8*I/x^2*Pi*b*m*csgn(I*c)*csgn(I*c*(e*x+d)^
n)^2-1/4*I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*ln(x^m)+1/4*I/d*e*b*n/x*Pi*csgn(I*f*x^m)^3+1/4*I/x^2*Pi*a*
csgn(I*f*x^m)^3+1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*f*x^m)^3+1/4*I/x^2*Pi*ln(c)*b*csgn(I*f)*csgn(I*x^m
)*csgn(I*f*x^m)+1/8*I/x^2*Pi*b*m*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I*b*Pi*csgn(I*c)*csgn(I*(
e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*ln(x^m)-1/4*I/d^2*b*e^2*n*ln(x)*Pi*csgn(I*f)*csgn(I*f*x^m)^2-1/2/d*e*b*n/x*l
n(f)+1/2/d^2*b*e^2*n*ln(e*x+d)*ln(f)-1/2/d^2*b*e^2*n*ln(x)*ln(f)+1/8*b*Pi^2*csgn(I*c*(e*x+d)^n)^3/x^2*csgn(I*f
)*csgn(I*x^m)*csgn(I*f*x^m)+1/8*b*Pi^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)/x^2*csgn(I*f*x^m)^3+1/8
*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)
^2/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2+1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*csgn(I*f)*csgn(I*f*x
^m)^2-1/4*I*b*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2/x^2*ln(x^m)+1/8*b*Pi^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*
x+d)^n)^2/x^2*csgn(I*x^m)*csgn(I*f*x^m)^2-1/4*b*e^2*m*n*ln(x)/d^2+1/4*b*e^2*m*n*ln(e*x+d)/d^2-3/4*b*e*m*n/d/x
________________________________________________________________________________________
Maxima [A] time = 1.22992, size = 267, normalized size = 1.71 \begin{align*} \frac{1}{4} \,{\left (\frac{2 \,{\left (\log \left (\frac{e x}{d} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{e x}{d}\right )\right )} b e^{2} n}{d^{2}} + \frac{b e^{2} n \log \left (e x + d\right )}{d^{2}} - \frac{2 \, b e^{2} n x^{2} \log \left (e x + d\right ) \log \left (x\right ) - b e^{2} n x^{2} \log \left (x\right )^{2} + b e^{2} n x^{2} \log \left (x\right ) + 3 \, b d e n x + b d^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b d^{2} \log \left (c\right ) + a d^{2}}{d^{2} x^{2}}\right )} m + \frac{1}{2} \,{\left (b e n{\left (\frac{e \log \left (e x + d\right )}{d^{2}} - \frac{e \log \left (x\right )}{d^{2}} - \frac{1}{d x}\right )} - \frac{b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{2}} - \frac{a}{x^{2}}\right )} \log \left (f x^{m}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x^3,x, algorithm="maxima")
[Out]
1/4*(2*(log(e*x/d + 1)*log(x) + dilog(-e*x/d))*b*e^2*n/d^2 + b*e^2*n*log(e*x + d)/d^2 - (2*b*e^2*n*x^2*log(e*x
+ d)*log(x) - b*e^2*n*x^2*log(x)^2 + b*e^2*n*x^2*log(x) + 3*b*d*e*n*x + b*d^2*log((e*x + d)^n) + b*d^2*log(c)
+ a*d^2)/(d^2*x^2))*m + 1/2*(b*e*n*(e*log(e*x + d)/d^2 - e*log(x)/d^2 - 1/(d*x)) - b*log((e*x + d)^n*c)/x^2 -
a/x^2)*log(f*x^m)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x + d\right )}^{n} c\right ) \log \left (f x^{m}\right ) + a \log \left (f x^{m}\right )}{x^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x^3,x, algorithm="fricas")
[Out]
integral((b*log((e*x + d)^n*c)*log(f*x^m) + a*log(f*x^m))/x^3, x)
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(ln(f*x**m)*(a+b*ln(c*(e*x+d)**n))/x**3,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(log(f*x^m)*(a+b*log(c*(e*x+d)^n))/x^3,x, algorithm="giac")
[Out]
integrate((b*log((e*x + d)^n*c) + a)*log(f*x^m)/x^3, x)