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

Optimal. Leaf size=234 \[ -\frac{b f^2 m n \text{PolyLog}\left (2,\frac{f x}{e}+1\right )}{2 e^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{2 x^2}-\frac{f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{f^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac{f m \left (a+b \log \left (c x^n\right )\right )}{2 e x}-\frac{b n \log \left (d (e+f x)^m\right )}{4 x^2}+\frac{b f^2 m n \log ^2(x)}{4 e^2}-\frac{b f^2 m n \log (x)}{4 e^2}+\frac{b f^2 m n \log (e+f x)}{4 e^2}-\frac{b f^2 m n \log \left (-\frac{f x}{e}\right ) \log (e+f x)}{2 e^2}-\frac{3 b f m n}{4 e x} \]

[Out]

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

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Rubi [A]  time = 0.162672, antiderivative size = 234, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2395, 44, 2376, 2301, 2394, 2315} \[ -\frac{b f^2 m n \text{PolyLog}\left (2,\frac{f x}{e}+1\right )}{2 e^2}-\frac{\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{2 x^2}-\frac{f^2 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{f^2 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac{f m \left (a+b \log \left (c x^n\right )\right )}{2 e x}-\frac{b n \log \left (d (e+f x)^m\right )}{4 x^2}+\frac{b f^2 m n \log ^2(x)}{4 e^2}-\frac{b f^2 m n \log (x)}{4 e^2}+\frac{b f^2 m n \log (e+f x)}{4 e^2}-\frac{b f^2 m n \log \left (-\frac{f x}{e}\right ) \log (e+f x)}{2 e^2}-\frac{3 b f m n}{4 e x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -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 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.148011, size = 232, normalized size = 0.99 \[ -\frac{-2 b f^2 m n x^2 \text{PolyLog}\left (2,-\frac{f x}{e}\right )+f^2 m x^2 \log (x) \left (2 a+2 b \log \left (c x^n\right )+2 b n \log (e+f x)-2 b n \log \left (\frac{f x}{e}+1\right )+b n\right )+2 a e^2 \log \left (d (e+f x)^m\right )-2 a f^2 m x^2 \log (e+f x)+2 a e f m x+2 b e^2 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )-2 b f^2 m x^2 \log \left (c x^n\right ) \log (e+f x)+2 b e f m x \log \left (c x^n\right )+b e^2 n \log \left (d (e+f x)^m\right )-b f^2 m n x^2 \log (e+f x)+3 b e f m n x-b f^2 m n x^2 \log ^2(x)}{4 e^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.348, size = 2100, normalized size = 9. \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))*ln(d*(f*x+e)^m)/x^3,x)

[Out]

-1/4*I/e^2*f^2*m*ln(f*x+e)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/2/e^2*f^2*m*ln(f*x+e)*b*ln(c)+1/4*I/x^2*
Pi*a*csgn(I*d*(f*x+e)^m)^3+1/8*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^2*b*csgn(I*c*x^n)^3-1/2*b*f^2*m*n*ln(-f*x/e)*ln(f*
x+e)/e^2-1/4*b*f^2*m*n*ln(x)/e^2+1/4*b*f^2*m*n*ln(x)^2/e^2+1/4*b*f^2*m*n*ln(f*x+e)/e^2-3/4*b*f*m*n/e/x-1/2/e^2
*f^2*m*ln(x)*a+1/2/e^2*f^2*m*ln(f*x+e)*a-1/2*ln(d)*b/x^2*ln(x^n)-1/2/x^2*ln(c)*ln(d)*b-1/4/x^2*ln(d)*b*n+(-1/2
*b/x^2*ln(x^n)-1/4*(-I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+I*b*Pi*csgn(I
*x^n)*csgn(I*c*x^n)^2-I*b*Pi*csgn(I*c*x^n)^3+2*b*ln(c)+b*n+2*a)/x^2)*ln((f*x+e)^m)-1/4*I/e*f*m/x*b*Pi*csgn(I*x
^n)*csgn(I*c*x^n)^2-1/4*I/e^2*f^2*m*ln(x)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/8*Pi^2*csgn(I*d)*csgn(I*(f*x+e)^m)*
csgn(I*d*(f*x+e)^m)/x^2*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/4*I/e^2*f^2*m*ln(x)*b*Pi*csgn(I*x^n)*csgn(I*c*
x^n)^2+1/4*I/e^2*f^2*m*ln(f*x+e)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/4*I/e^2*f^2*m*ln(f*x+e)*b*Pi*csgn(I*x^n)*csg
n(I*c*x^n)^2-1/4*I/e*f*m/x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/4*I/e*f*m/x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^
n)+1/4*I/e^2*f^2*m*ln(x)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/4*I*Pi*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*
d*(f*x+e)^m)*b/x^2*ln(x^n)-1/8*Pi^2*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^2*b*csgn(I*c)*csgn(I*c*x
^n)^2-1/2/e*f*m/x*a-1/2*n*f^2*b*m/e^2*dilog(-f*x/e)+1/4*I/e*f*m/x*b*Pi*csgn(I*c*x^n)^3+1/4*I/e^2*f^2*m*ln(x)*b
*Pi*csgn(I*c*x^n)^3-1/4*I/e^2*f^2*m*ln(f*x+e)*b*Pi*csgn(I*c*x^n)^3-1/2/e*f*m/x*b*ln(c)-1/2/e^2*f^2*m*ln(x)*b*l
n(c)-1/2*m*f^2*b*ln(x^n)/e^2*ln(x)+1/2*m*f^2*b*ln(x^n)/e^2*ln(f*x+e)-1/2*m*f*b*ln(x^n)/e/x+1/4*I*Pi*csgn(I*d*(
f*x+e)^m)^3*b/x^2*ln(x^n)-1/4*I/x^2*Pi*a*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2-1/8*I/x^2*Pi*b*n*csgn(I*d)*csgn(I*d*(
f*x+e)^m)^2-1/8*I/x^2*Pi*b*n*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2+1/4*I/x^2*Pi*a*csgn(I*d)*csgn(I*(f*x+e)^m
)*csgn(I*d*(f*x+e)^m)-1/4*I*Pi*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2*b/x^2*ln(x^n)-1/4*I/x^2*ln(c)*Pi*b*csgn
(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2-1/4*I/x^2*Pi*ln(d)*b*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I/x^2*Pi*ln(d)*b*csgn(I
*x^n)*csgn(I*c*x^n)^2+1/8*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^2*b*csgn(I*c)*csgn(I*c*x^n)^2+1/8*Pi^2*csgn(I
*d)*csgn(I*d*(f*x+e)^m)^2/x^2*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2*b/x^2*ln(
x^n)-1/4*I/x^2*ln(c)*Pi*b*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2+1/8*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2/x^2
*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/8*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^2*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/8*Pi
^2*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^2*b*csgn(I*c*x^n)^3+1/8*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(
f*x+e)^m)^2/x^2*b*csgn(I*c)*csgn(I*c*x^n)^2-1/4*I/x^2*Pi*a*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2+1/4*I/x^2*l
n(c)*Pi*b*csgn(I*d*(f*x+e)^m)^3+1/4*I/x^2*Pi*ln(d)*b*csgn(I*c*x^n)^3+1/8*I/x^2*Pi*b*n*csgn(I*d*(f*x+e)^m)^3-1/
8*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^2*b*csgn(I*c)*csgn(I*c*x^n)^2-1/8*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^2*b*csgn(I*x^n)*
csgn(I*c*x^n)^2-1/8*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^2*b*csgn(I*c*x^n)^3-1/8*Pi^2*csgn(I*(f*x+e)^m)*csgn
(I*d*(f*x+e)^m)^2/x^2*b*csgn(I*c*x^n)^3-1/2/x^2*ln(d)*a-1/8*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2/x^2*b
*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/4*I/x^2*ln(c)*Pi*b*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)+1/4*
I/x^2*Pi*ln(d)*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/8*I/x^2*Pi*b*n*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*
x+e)^m)-1/8*Pi^2*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^2*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/8*Pi^2*cs
gn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^2*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)

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Maxima [A]  time = 1.69233, size = 385, normalized size = 1.65 \begin{align*} \frac{{\left (\log \left (\frac{f x}{e} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{f x}{e}\right )\right )} b f^{2} m n}{2 \, e^{2}} + \frac{{\left (2 \, a f^{2} m +{\left (f^{2} m n + 2 \, f^{2} m \log \left (c\right )\right )} b\right )} \log \left (f x + e\right )}{4 \, e^{2}} - \frac{2 \, b f^{2} m n x^{2} \log \left (f x + e\right ) \log \left (x\right ) - b f^{2} m n x^{2} \log \left (x\right )^{2} + 2 \, a e^{2} \log \left (d\right ) +{\left (2 \, a f^{2} m +{\left (f^{2} m n + 2 \, f^{2} m \log \left (c\right )\right )} b\right )} x^{2} \log \left (x\right ) +{\left (e^{2} n \log \left (d\right ) + 2 \, e^{2} \log \left (c\right ) \log \left (d\right )\right )} b +{\left (2 \, a e f m +{\left (3 \, e f m n + 2 \, e f m \log \left (c\right )\right )} b\right )} x +{\left (2 \, b e^{2} \log \left (x^{n}\right ) + 2 \, a e^{2} +{\left (e^{2} n + 2 \, e^{2} \log \left (c\right )\right )} b\right )} \log \left ({\left (f x + e\right )}^{m}\right ) - 2 \,{\left (b f^{2} m x^{2} \log \left (f x + e\right ) - b f^{2} m x^{2} \log \left (x\right ) - b e f m x - b e^{2} \log \left (d\right )\right )} \log \left (x^{n}\right )}{4 \, e^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)*log((f*x + e)^m*d)/x^3, x)

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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((a+b*ln(c*x**n))*ln(d*(f*x+e)**m)/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 (c x^{n}\right ) + a\right )} \log \left ({\left (f x + e\right )}^{m} d\right )}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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