3.154 \(\int (g x)^{-1-2 m} (a+b \log (c x^n)) \log (d (e+f x^m)^k) \, dx\)

Optimal. Leaf size=414 \[ -\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \text{PolyLog}\left (2,\frac{f x^m}{e}+1\right )}{2 e^2 g m^2}-\frac{(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac{f^2 k x^{2 m} \log (x) (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac{f^2 k x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g m}-\frac{f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-\frac{b n (g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}-\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (-\frac{f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}+\frac{b f^2 k n x^{2 m} \log ^2(x) (g x)^{-2 m}}{4 e^2 g}-\frac{b f^2 k n x^{2 m} \log (x) (g x)^{-2 m}}{4 e^2 g m}-\frac{3 b f k n x^m (g x)^{-2 m}}{4 e g m^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.521206, antiderivative size = 414, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 12, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2455, 20, 266, 44, 2376, 30, 19, 2301, 2454, 2394, 2315, 16} \[ -\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \text{PolyLog}\left (2,\frac{f x^m}{e}+1\right )}{2 e^2 g m^2}-\frac{(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac{f^2 k x^{2 m} \log (x) (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac{f^2 k x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g m}-\frac{f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-\frac{b n (g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}-\frac{b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (-\frac{f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}+\frac{b f^2 k n x^{2 m} \log ^2(x) (g x)^{-2 m}}{4 e^2 g}-\frac{b f^2 k n x^{2 m} \log (x) (g x)^{-2 m}}{4 e^2 g m}-\frac{3 b f k n x^m (g x)^{-2 m}}{4 e g m^2} \]

Antiderivative was successfully verified.

[In]

Int[(g*x)^(-1 - 2*m)*(a + b*Log[c*x^n])*Log[d*(e + f*x^m)^k],x]

[Out]

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

Rule 2455

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

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 19

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m + n)*(b*v)^n)/(a*v)^n, Int[u*v^(m + n),
x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] && IntegerQ[m + n]

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 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

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]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rubi steps

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

Mathematica [A]  time = 0.355511, size = 302, normalized size = 0.73 \[ \frac{(g x)^{-2 m} \left (2 b f^2 k n x^{2 m} \text{PolyLog}\left (2,-\frac{f x^m}{e}\right )-f^2 k m x^{2 m} \log (x) \left (2 a m+2 b m \log \left (c x^n\right )-2 b n \log \left (\frac{f x^m}{e}+1\right )+2 b n \log \left (f-f x^{-m}\right )+b n\right )-2 a e^2 m \log \left (d \left (e+f x^m\right )^k\right )-2 a e f k m x^m+2 a f^2 k m x^{2 m} \log \left (f-f x^{-m}\right )-2 b e^2 m \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )-2 b e f k m x^m \log \left (c x^n\right )+2 b f^2 k m x^{2 m} \log \left (c x^n\right ) \log \left (f-f x^{-m}\right )-b e^2 n \log \left (d \left (e+f x^m\right )^k\right )-3 b e f k n x^m+b f^2 k m^2 n x^{2 m} \log ^2(x)+b f^2 k n x^{2 m} \log \left (f-f x^{-m}\right )\right )}{4 e^2 g m^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(g*x)^(-1 - 2*m)*(a + b*Log[c*x^n])*Log[d*(e + f*x^m)^k],x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.237, size = 0, normalized size = 0. \begin{align*} \int \left ( gx \right ) ^{-1-2\,m} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \ln \left ( d \left ( e+f{x}^{m} \right ) ^{k} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x)^(-1-2*m)*(a+b*ln(c*x^n))*ln(d*(e+f*x^m)^k),x)

[Out]

int((g*x)^(-1-2*m)*(a+b*ln(c*x^n))*ln(d*(e+f*x^m)^k),x)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^(-1-2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 0.900203, size = 828, normalized size = 2. \begin{align*} \frac{2 \, b f^{2} g^{-2 \, m - 1} k m n x^{2 \, m} \log \left (x\right ) \log \left (\frac{f x^{m} + e}{e}\right ) + 2 \, b f^{2} g^{-2 \, m - 1} k n x^{2 \, m}{\rm Li}_2\left (-\frac{f x^{m} + e}{e} + 1\right ) -{\left (b f^{2} k m^{2} n \log \left (x\right )^{2} +{\left (2 \, b f^{2} k m^{2} \log \left (c\right ) + 2 \, a f^{2} k m^{2} + b f^{2} k m n\right )} \log \left (x\right )\right )} g^{-2 \, m - 1} x^{2 \, m} -{\left (2 \, b e f k m n \log \left (x\right ) + 2 \, b e f k m \log \left (c\right ) + 2 \, a e f k m + 3 \, b e f k n\right )} g^{-2 \, m - 1} x^{m} -{\left (2 \, b e^{2} m n \log \left (d\right ) \log \left (x\right ) +{\left (2 \, b e^{2} m \log \left (c\right ) + 2 \, a e^{2} m + b e^{2} n\right )} \log \left (d\right )\right )} g^{-2 \, m - 1} +{\left ({\left (2 \, b f^{2} k m \log \left (c\right ) + 2 \, a f^{2} k m + b f^{2} k n\right )} g^{-2 \, m - 1} x^{2 \, m} -{\left (2 \, b e^{2} k m n \log \left (x\right ) + 2 \, b e^{2} k m \log \left (c\right ) + 2 \, a e^{2} k m + b e^{2} k n\right )} g^{-2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, e^{2} m^{2} x^{2 \, m}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^(-1-2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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((g*x)**(-1-2*m)*(a+b*ln(c*x**n))*ln(d*(e+f*x**m)**k),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{-2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x)^(-1-2*m)*(a+b*log(c*x^n))*log(d*(e+f*x^m)^k),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*(g*x)^(-2*m - 1)*log((f*x^m + e)^k*d), x)