3.1.77 \(\int \frac {(a+b \log (c x^n)) \log (d (e+f x)^m)}{x^4} \, dx\) [77]

Optimal. Leaf size=274 \[ -\frac {5 b f m n}{36 e x^2}+\frac {4 b f^2 m n}{9 e^2 x}+\frac {b f^3 m n \log (x)}{9 e^3}-\frac {b f^3 m n \log ^2(x)}{6 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {b f^3 m n \log (e+f x)}{9 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {b f^3 m n \text {Li}_2\left (1+\frac {f x}{e}\right )}{3 e^3} \]

[Out]

-5/36*b*f*m*n/e/x^2+4/9*b*f^2*m*n/e^2/x+1/9*b*f^3*m*n*ln(x)/e^3-1/6*b*f^3*m*n*ln(x)^2/e^3-1/6*f*m*(a+b*ln(c*x^
n))/e/x^2+1/3*f^2*m*(a+b*ln(c*x^n))/e^2/x+1/3*f^3*m*ln(x)*(a+b*ln(c*x^n))/e^3-1/9*b*f^3*m*n*ln(f*x+e)/e^3+1/3*
b*f^3*m*n*ln(-f*x/e)*ln(f*x+e)/e^3-1/3*f^3*m*(a+b*ln(c*x^n))*ln(f*x+e)/e^3-1/9*b*n*ln(d*(f*x+e)^m)/x^3-1/3*(a+
b*ln(c*x^n))*ln(d*(f*x+e)^m)/x^3+1/3*b*f^3*m*n*polylog(2,1+f*x/e)/e^3

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Rubi [A]
time = 0.13, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2442, 46, 2423, 2338, 2441, 2352} \begin {gather*} \frac {b f^3 m n \text {PolyLog}\left (2,\frac {f x}{e}+1\right )}{3 e^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {f^3 m \log (e+f x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {b f^3 m n \log ^2(x)}{6 e^3}+\frac {b f^3 m n \log (x)}{9 e^3}-\frac {b f^3 m n \log (e+f x)}{9 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}+\frac {4 b f^2 m n}{9 e^2 x}-\frac {5 b f m n}{36 e x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*f*m*n)/(36*e*x^2) + (4*b*f^2*m*n)/(9*e^2*x) + (b*f^3*m*n*Log[x])/(9*e^3) - (b*f^3*m*n*Log[x]^2)/(6*e^3)
- (f*m*(a + b*Log[c*x^n]))/(6*e*x^2) + (f^2*m*(a + b*Log[c*x^n]))/(3*e^2*x) + (f^3*m*Log[x]*(a + b*Log[c*x^n])
)/(3*e^3) - (b*f^3*m*n*Log[e + f*x])/(9*e^3) + (b*f^3*m*n*Log[-((f*x)/e)]*Log[e + f*x])/(3*e^3) - (f^3*m*(a +
b*Log[c*x^n])*Log[e + f*x])/(3*e^3) - (b*n*Log[d*(e + f*x)^m])/(9*x^3) - ((a + b*Log[c*x^n])*Log[d*(e + f*x)^m
])/(3*x^3) + (b*f^3*m*n*PolyLog[2, 1 + (f*x)/e])/(3*e^3)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2338

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 2352

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

Rule 2423

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 2441

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 2442

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]

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^4} \, dx &=-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}-(b n) \int \left (-\frac {f m}{6 e x^3}+\frac {f^2 m}{3 e^2 x^2}+\frac {f^3 m \log (x)}{3 e^3 x}-\frac {f^3 m \log (e+f x)}{3 e^3 x}-\frac {\log \left (d (e+f x)^m\right )}{3 x^4}\right ) \, dx\\ &=-\frac {b f m n}{12 e x^2}+\frac {b f^2 m n}{3 e^2 x}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {1}{3} (b n) \int \frac {\log \left (d (e+f x)^m\right )}{x^4} \, dx-\frac {\left (b f^3 m n\right ) \int \frac {\log (x)}{x} \, dx}{3 e^3}+\frac {\left (b f^3 m n\right ) \int \frac {\log (e+f x)}{x} \, dx}{3 e^3}\\ &=-\frac {b f m n}{12 e x^2}+\frac {b f^2 m n}{3 e^2 x}-\frac {b f^3 m n \log ^2(x)}{6 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {1}{9} (b f m n) \int \frac {1}{x^3 (e+f x)} \, dx-\frac {\left (b f^4 m n\right ) \int \frac {\log \left (-\frac {f x}{e}\right )}{e+f x} \, dx}{3 e^3}\\ &=-\frac {b f m n}{12 e x^2}+\frac {b f^2 m n}{3 e^2 x}-\frac {b f^3 m n \log ^2(x)}{6 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {b f^3 m n \text {Li}_2\left (1+\frac {f x}{e}\right )}{3 e^3}+\frac {1}{9} (b f m n) \int \left (\frac {1}{e x^3}-\frac {f}{e^2 x^2}+\frac {f^2}{e^3 x}-\frac {f^3}{e^3 (e+f x)}\right ) \, dx\\ &=-\frac {5 b f m n}{36 e x^2}+\frac {4 b f^2 m n}{9 e^2 x}+\frac {b f^3 m n \log (x)}{9 e^3}-\frac {b f^3 m n \log ^2(x)}{6 e^3}-\frac {f m \left (a+b \log \left (c x^n\right )\right )}{6 e x^2}+\frac {f^2 m \left (a+b \log \left (c x^n\right )\right )}{3 e^2 x}+\frac {f^3 m \log (x) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {b f^3 m n \log (e+f x)}{9 e^3}+\frac {b f^3 m n \log \left (-\frac {f x}{e}\right ) \log (e+f x)}{3 e^3}-\frac {f^3 m \left (a+b \log \left (c x^n\right )\right ) \log (e+f x)}{3 e^3}-\frac {b n \log \left (d (e+f x)^m\right )}{9 x^3}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d (e+f x)^m\right )}{3 x^3}+\frac {b f^3 m n \text {Li}_2\left (1+\frac {f x}{e}\right )}{3 e^3}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 280, normalized size = 1.02 \begin {gather*} -\frac {6 a e^2 f m x+5 b e^2 f m n x-12 a e f^2 m x^2-16 b e f^2 m n x^2+6 b f^3 m n x^3 \log ^2(x)+6 b e^2 f m x \log \left (c x^n\right )-12 b e f^2 m x^2 \log \left (c x^n\right )+12 a f^3 m x^3 \log (e+f x)+4 b f^3 m n x^3 \log (e+f x)+12 b f^3 m x^3 \log \left (c x^n\right ) \log (e+f x)+12 a e^3 \log \left (d (e+f x)^m\right )+4 b e^3 n \log \left (d (e+f x)^m\right )+12 b e^3 \log \left (c x^n\right ) \log \left (d (e+f x)^m\right )-4 f^3 m x^3 \log (x) \left (3 a+b n+3 b \log \left (c x^n\right )+3 b n \log (e+f x)-3 b n \log \left (1+\frac {f x}{e}\right )\right )+12 b f^3 m n x^3 \text {Li}_2\left (-\frac {f x}{e}\right )}{36 e^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/36*(6*a*e^2*f*m*x + 5*b*e^2*f*m*n*x - 12*a*e*f^2*m*x^2 - 16*b*e*f^2*m*n*x^2 + 6*b*f^3*m*n*x^3*Log[x]^2 + 6*
b*e^2*f*m*x*Log[c*x^n] - 12*b*e*f^2*m*x^2*Log[c*x^n] + 12*a*f^3*m*x^3*Log[e + f*x] + 4*b*f^3*m*n*x^3*Log[e + f
*x] + 12*b*f^3*m*x^3*Log[c*x^n]*Log[e + f*x] + 12*a*e^3*Log[d*(e + f*x)^m] + 4*b*e^3*n*Log[d*(e + f*x)^m] + 12
*b*e^3*Log[c*x^n]*Log[d*(e + f*x)^m] - 4*f^3*m*x^3*Log[x]*(3*a + b*n + 3*b*Log[c*x^n] + 3*b*n*Log[e + f*x] - 3
*b*n*Log[1 + (f*x)/e]) + 12*b*f^3*m*n*x^3*PolyLog[2, -((f*x)/e)])/(e^3*x^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.36, size = 2282, normalized size = 8.33

method	result	size
risch	\(\text {Expression too large to display}\)	\(2282\)

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^4,x,method=_RETURNVERBOSE)

[Out]

1/3*b*f^3*m*n*ln(-f*x/e)*ln(f*x+e)/e^3+1/12*Pi^2*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^3*b*csgn(I*
c)*csgn(I*x^n)*csgn(I*c*x^n)+1/9*b*f^3*m*n*ln(x)/e^3-1/6*b*f^3*m*n*ln(x)^2/e^3-1/9*b*f^3*m*n*ln(f*x+e)/e^3-1/6
*I/e^3*f^3*m*ln(f*x+e)*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/12*I/e*f*m/x^2*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/12*I
/e*f*m/x^2*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/6*I/e^3*f^3*m*ln(f*x+e)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2-1/6/e*f*m
/x^2*b*ln(c)+1/3/e^2*f^2*m/x*b*ln(c)+1/3/e^3*f^3*m*ln(x)*b*ln(c)-1/3/e^3*f^3*m*ln(f*x+e)*b*ln(c)-1/3/x^3*ln(d)
*a-1/6*I/x^3*Pi*a*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2-1/6*I/x^3*Pi*a*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2-1/6*I
/x^3*Pi*ln(c)*b*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2-1/18*I/x^3*Pi*b*n*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2-1/18
*I/x^3*Pi*b*n*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2+1/3*n*f^3*b*m/e^3*dilog(-f*x/e)+(-1/3*b/x^3*ln(x^n)-1/18
*(-3*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+3*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+3*I*b*Pi*csgn(I*x^n)*csgn(I
*c*x^n)^2-3*I*b*Pi*csgn(I*c*x^n)^3+6*b*ln(c)+2*b*n+6*a)/x^3)*ln((f*x+e)^m)-1/6*I/x^3*Pi*ln(d)*b*csgn(I*x^n)*cs
gn(I*c*x^n)^2-1/6*I/x^3*Pi*ln(c)*b*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2-1/6/e*f*m/x^2*a+1/3/e^3*f^3*m*ln(x)*a-1/3/e
^3*f^3*m*ln(f*x+e)*a-1/6*I/e^3*f^3*m*ln(x)*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/6*I/e^3*f^3*m*ln(f*x+e)*
b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/6*I/e^2*f^2*m/x*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/12*Pi^2*
csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^3*b*csgn(I*c*x^n)^3+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^
2/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2+1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+
1/12*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/3*ln(d)*b/x^3*ln(x^n)+1/3/
e^2*f^2*m/x*a-1/12*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2/x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/6*
I/x^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/6*I*Pi*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)*
b/x^3*ln(x^n)-5/36*b*f*m*n/e/x^2+4/9*b*f^2*m*n/e^2/x+1/12*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2/x^3*b*c
sgn(I*x^n)*csgn(I*c*x^n)^2+1/3*m*f^3*b*ln(x^n)/e^3*ln(x)+1/3*m*f^2*b*ln(x^n)/e^2/x-1/3*m*f^3*b*ln(x^n)/e^3*ln(
f*x+e)-1/6*m*f*b*ln(x^n)/e/x^2+1/6*I/x^3*Pi*a*csgn(I*d*(f*x+e)^m)^3+1/12*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^3*b*csgn
(I*c*x^n)^3+1/12*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^3*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/12*Pi^2*csgn(I*d)*csgn
(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I
*d*(f*x+e)^m)/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^3*b*csgn(I*c)*csgn
(I*x^n)*csgn(I*c*x^n)-1/9/x^3*ln(d)*b*n-1/3/x^3*ln(d)*ln(c)*b-1/6*I*Pi*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2*b/x^3*l
n(x^n)-1/6*I*Pi*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)^2*b/x^3*ln(x^n)-1/12*Pi^2*csgn(I*(f*x+e)^m)*csgn(I*d*(f*
x+e)^m)^2/x^3*b*csgn(I*c*x^n)^3-1/12*Pi^2*csgn(I*d*(f*x+e)^m)^3/x^3*b*csgn(I*c)*csgn(I*c*x^n)^2-1/12*Pi^2*csgn
(I*d*(f*x+e)^m)^3/x^3*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6*I*Pi*csgn(I*d*(f*x+e)^m)^3*b/x^3*ln(x^n)+1/6*I/x^3*Pi*
ln(d)*b*csgn(I*c*x^n)^3+1/6*I/x^3*Pi*ln(c)*b*csgn(I*d*(f*x+e)^m)^3+1/18*I/x^3*Pi*b*n*csgn(I*d*(f*x+e)^m)^3+1/6
*I/e^3*f^3*m*ln(x)*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/6*I/e^2*f^2*m/x*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/6*I/x^3*P
i*ln(c)*b*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)+1/18*I/x^3*Pi*b*n*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I
*d*(f*x+e)^m)+1/6*I/e^2*f^2*m/x*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6*I/e^3*f^3*m*ln(x)*b*Pi*csgn(I*x^n)*csgn(I
*c*x^n)^2-1/6*I/e^3*f^3*m*ln(x)*b*Pi*csgn(I*c*x^n)^3+1/6*I/e^3*f^3*m*ln(f*x+e)*b*Pi*csgn(I*c*x^n)^3+1/6*I/x^3*
Pi*a*csgn(I*d)*csgn(I*(f*x+e)^m)*csgn(I*d*(f*x+e)^m)-1/6*I/x^3*Pi*ln(d)*b*csgn(I*c)*csgn(I*c*x^n)^2+1/12*I/e*f
*m/x^2*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-1/6*I/e^2*f^2*m/x*b*Pi*csgn(I*c*x^n)^3+1/12*I/e*f*m/x^2*b*Pi*c
sgn(I*c*x^n)^3-1/12*Pi^2*csgn(I*d)*csgn(I*d*(f*x+e)^m)^2/x^3*b*csgn(I*c*x^n)^3

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Maxima [A]
time = 0.40, size = 320, normalized size = 1.17 \begin {gather*} -\frac {1}{3} \, {\left (\log \left (f x e^{\left (-1\right )} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-f x e^{\left (-1\right )}\right )\right )} b f^{3} m n e^{\left (-3\right )} - \frac {1}{9} \, {\left (3 \, a f^{3} m + {\left (f^{3} m n + 3 \, f^{3} m \log \left (c\right )\right )} b\right )} e^{\left (-3\right )} \log \left (f x + e\right ) + \frac {{\left (12 \, b f^{3} m n x^{3} \log \left (f x + e\right ) \log \left (x\right ) - 6 \, b f^{3} m n x^{3} \log \left (x\right )^{2} + 4 \, {\left (3 \, a f^{3} m + {\left (f^{3} m n + 3 \, f^{3} m \log \left (c\right )\right )} b\right )} x^{3} \log \left (x\right ) + 4 \, {\left (3 \, a f^{2} m + {\left (4 \, f^{2} m n + 3 \, f^{2} m \log \left (c\right )\right )} b\right )} x^{2} e - {\left (6 \, a f m + {\left (5 \, f m n + 6 \, f m \log \left (c\right )\right )} b\right )} x e^{2} - 4 \, {\left ({\left (n \log \left (d\right ) + 3 \, \log \left (c\right ) \log \left (d\right )\right )} b + 3 \, a \log \left (d\right )\right )} e^{3} - 4 \, {\left (3 \, b e^{3} \log \left (x^{n}\right ) + {\left (b {\left (n + 3 \, \log \left (c\right )\right )} + 3 \, a\right )} e^{3}\right )} \log \left ({\left (f x + e\right )}^{m}\right ) - 6 \, {\left (2 \, b f^{3} m x^{3} \log \left (f x + e\right ) - 2 \, b f^{3} m x^{3} \log \left (x\right ) - 2 \, b f^{2} m x^{2} e + b f m x e^{2} + 2 \, b e^{3} \log \left (d\right )\right )} \log \left (x^{n}\right )\right )} e^{\left (-3\right )}}{36 \, x^{3}} \end {gather*}

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^4,x, algorithm="maxima")

[Out]

-1/3*(log(f*x*e^(-1) + 1)*log(x) + dilog(-f*x*e^(-1)))*b*f^3*m*n*e^(-3) - 1/9*(3*a*f^3*m + (f^3*m*n + 3*f^3*m*
log(c))*b)*e^(-3)*log(f*x + e) + 1/36*(12*b*f^3*m*n*x^3*log(f*x + e)*log(x) - 6*b*f^3*m*n*x^3*log(x)^2 + 4*(3*
a*f^3*m + (f^3*m*n + 3*f^3*m*log(c))*b)*x^3*log(x) + 4*(3*a*f^2*m + (4*f^2*m*n + 3*f^2*m*log(c))*b)*x^2*e - (6
*a*f*m + (5*f*m*n + 6*f*m*log(c))*b)*x*e^2 - 4*((n*log(d) + 3*log(c)*log(d))*b + 3*a*log(d))*e^3 - 4*(3*b*e^3*
log(x^n) + (b*(n + 3*log(c)) + 3*a)*e^3)*log((f*x + e)^m) - 6*(2*b*f^3*m*x^3*log(f*x + e) - 2*b*f^3*m*x^3*log(
x) - 2*b*f^2*m*x^2*e + b*f*m*x*e^2 + 2*b*e^3*log(d))*log(x^n))*e^(-3)/x^3

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^4,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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**4,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (d\,{\left (e+f\,x\right )}^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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