\(\int \frac {\log (f x^m) (a+b \log (c (d+e x)^n))}{x^3} \, dx\) [364]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {3 b e m n}{4 d x}-\frac {b e^2 m n \log (x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}+\frac {b e^2 n \log \left (1+\frac {d}{e x}\right ) \log \left (f x^m\right )}{2 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {1}{4} \left (\frac {m}{x^2}+\frac {2 \log \left (f x^m\right )}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )-\frac {b e^2 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{2 d^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2473, 2380, 2341, 2379, 2438, 46} \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {1}{4} \left (\frac {2 \log \left (f x^m\right )}{x^2}+\frac {m}{x^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {b e^2 n \log \left (\frac {d}{e x}+1\right ) \log \left (f x^m\right )}{2 d^2}-\frac {b e^2 m n \operatorname {PolyLog}\left (2,-\frac {d}{e x}\right )}{2 d^2}-\frac {b e^2 m n \log (x)}{4 d^2}+\frac {b e^2 m n \log (d+e x)}{4 d^2}-\frac {b e n \log \left (f x^m\right )}{2 d x}-\frac {3 b e m n}{4 d x} \]

[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[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)

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 2341

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 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2380

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.))/((d_) + (e_.)*(x_)^(r_.)), x_Symbol] :> Dist[1/d,
Int[x^m*(a + b*Log[c*x^n])^p, x], x] - Dist[e/d, Int[(x^(m + r)*(a + b*Log[c*x^n])^p)/(d + e*x^r), x], x] /;
FreeQ[{a, b, c, d, e, m, n, r}, x] && IGtQ[p, 0] && IGtQ[r, 0] && ILtQ[m, -1]

Rule 2438

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 2473

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

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.31 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=-\frac {a d^2 m+3 b d e m n x-b e^2 m n x^2 \log ^2(x)+2 a d^2 \log \left (f x^m\right )+2 b d e n x \log \left (f x^m\right )-b e^2 m n x^2 \log (d+e x)-2 b e^2 n x^2 \log \left (f x^m\right ) \log (d+e x)+b d^2 m \log \left (c (d+e x)^n\right )+2 b d^2 \log \left (f x^m\right ) \log \left (c (d+e x)^n\right )+b e^2 n x^2 \log (x) \left (m+2 \log \left (f x^m\right )+2 m \log (d+e x)-2 m \log \left (1+\frac {e x}{d}\right )\right )-2 b e^2 m n x^2 \operatorname {PolyLog}\left (2,-\frac {e x}{d}\right )}{4 d^2 x^2} \]

[In]

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

[Out]

-1/4*(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)])/(d^2*x^2)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.39 (sec) , antiderivative size = 901, normalized size of antiderivative = 5.78

method	result	size
risch	\(\text {Expression too large to display}\)	\(901\)

[In]

int(ln(f*x^m)*(a+b*ln(c*(e*x+d)^n))/x^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \]

[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)]

Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.27 \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\frac {1}{4} \, {\left (\frac {2 \, {\left (\log \left (\frac {e x}{d} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {e x}{d}\right )\right )} b e^{2} n}{d^{2}} + \frac {b e^{2} n \log \left (e x + d\right )}{d^{2}} - \frac {2 \, b e^{2} n x^{2} \log \left (e x + d\right ) \log \left (x\right ) - b e^{2} n x^{2} \log \left (x\right )^{2} + b e^{2} n x^{2} \log \left (x\right ) + 3 \, b d e n x + b d^{2} \log \left ({\left (e x + d\right )}^{n}\right ) + b d^{2} \log \left (c\right ) + a d^{2}}{d^{2} x^{2}}\right )} m + \frac {1}{2} \, {\left (b e n {\left (\frac {e \log \left (e x + d\right )}{d^{2}} - \frac {e \log \left (x\right )}{d^{2}} - \frac {1}{d x}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{x^{2}} - \frac {a}{x^{2}}\right )} \log \left (f x^{m}\right ) \]

[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)

Giac [F]

\[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} \log \left (f x^{m}\right )}{x^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (f x^m\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{x^3} \, dx=\int \frac {\ln \left (f\,x^m\right )\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{x^3} \,d x \]

[In]

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

[Out]

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