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 \[ -\frac {(g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac {f^2 k x^{2 m} \log (x) (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g}+\frac {f^2 k x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2 g m}-\frac {f k x^m (g x)^{-2 m} \left (a+b \log \left (c x^n\right )\right )}{2 e g m}-\frac {b n (g x)^{-2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \text {Li}_2\left (\frac {f x^m}{e}+1\right )}{2 e^2 g m^2}+\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (e+f x^m\right )}{4 e^2 g m^2}-\frac {b f^2 k n x^{2 m} (g x)^{-2 m} \log \left (-\frac {f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 e^2 g m^2}+\frac {b f^2 k n x^{2 m} \log ^2(x) (g x)^{-2 m}}{4 e^2 g}-\frac {b f^2 k n x^{2 m} \log (x) (g x)^{-2 m}}{4 e^2 g m}-\frac {3 b f k n x^m (g x)^{-2 m}}{4 e g m^2} \]

[Out]

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

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Rubi [A]  time = 0.52, antiderivative size = 414, normalized size of antiderivative = 1.00, 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 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]

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 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 30

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

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*}

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Mathematica [A]  time = 0.37, size = 302, normalized size = 0.73 \[ \frac {(g x)^{-2 m} \left (-f^2 k m x^{2 m} \log (x) \left (2 a m+2 b m \log \left (c x^n\right )-2 b n \log \left (\frac {f x^m}{e}+1\right )+2 b n \log \left (f-f x^{-m}\right )+b n\right )-2 a e^2 m \log \left (d \left (e+f x^m\right )^k\right )-2 a e f k m x^m+2 a f^2 k m x^{2 m} \log \left (f-f x^{-m}\right )-2 b e^2 m \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )-2 b e f k m x^m \log \left (c x^n\right )+2 b f^2 k m x^{2 m} \log \left (c x^n\right ) \log \left (f-f x^{-m}\right )-b e^2 n \log \left (d \left (e+f x^m\right )^k\right )+2 b f^2 k n x^{2 m} \text {Li}_2\left (-\frac {f x^m}{e}\right )-3 b e f k n x^m+b f^2 k m^2 n x^{2 m} \log ^2(x)+b f^2 k n x^{2 m} \log \left (f-f x^{-m}\right )\right )}{4 e^2 g m^2} \]

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))

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fricas [A]  time = 0.67, size = 338, normalized size = 0.82 \[ \frac {2 \, b f^{2} g^{-2 \, m - 1} k m n x^{2 \, m} \log \relax (x) \log \left (\frac {f x^{m} + e}{e}\right ) + 2 \, b f^{2} g^{-2 \, m - 1} k n x^{2 \, m} {\rm Li}_2\left (-\frac {f x^{m} + e}{e} + 1\right ) - {\left (b f^{2} k m^{2} n \log \relax (x)^{2} + {\left (2 \, b f^{2} k m^{2} \log \relax (c) + 2 \, a f^{2} k m^{2} + b f^{2} k m n\right )} \log \relax (x)\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e f k m n \log \relax (x) + 2 \, b e f k m \log \relax (c) + 2 \, a e f k m + 3 \, b e f k n\right )} g^{-2 \, m - 1} x^{m} - {\left (2 \, b e^{2} m n \log \relax (d) \log \relax (x) + {\left (2 \, b e^{2} m \log \relax (c) + 2 \, a e^{2} m + b e^{2} n\right )} \log \relax (d)\right )} g^{-2 \, m - 1} + {\left ({\left (2 \, b f^{2} k m \log \relax (c) + 2 \, a f^{2} k m + b f^{2} k n\right )} g^{-2 \, m - 1} x^{2 \, m} - {\left (2 \, b e^{2} k m n \log \relax (x) + 2 \, b e^{2} k m \log \relax (c) + 2 \, a e^{2} k m + b e^{2} k n\right )} g^{-2 \, m - 1}\right )} \log \left (f x^{m} + e\right )}{4 \, e^{2} m^{2} x^{2 \, m}} \]

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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} \left (g x\right )^{-2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k} d\right )\,{d x} \]

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)

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maple [F]  time = 0.30, size = 0, normalized size = 0.00 \[ \int \left (b \ln \left (c \,x^{n}\right )+a \right ) \left (g x \right )^{-2 m -1} \ln \left (d \left (f \,x^{m}+e \right )^{k}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (2 \, b m \log \left (x^{n}\right ) + {\left (2 \, m \log \relax (c) + n\right )} b + 2 \, a m\right )} g^{-2 \, m - 1} \log \left ({\left (f x^{m} + e\right )}^{k}\right )}{4 \, m^{2} x^{2 \, m}} + \int \frac {4 \, b e m \log \relax (c) \log \relax (d) + 4 \, a e m \log \relax (d) + {\left (2 \, {\left (f k m + 2 \, f m \log \relax (d)\right )} a + {\left (f k n + 2 \, {\left (f k m + 2 \, f m \log \relax (d)\right )} \log \relax (c)\right )} b\right )} x^{m} + 2 \, {\left (2 \, b e m \log \relax (d) + {\left (f k m + 2 \, f m \log \relax (d)\right )} b x^{m}\right )} \log \left (x^{n}\right )}{4 \, {\left (f g^{2 \, m + 1} m x x^{3 \, m} + e g^{2 \, m + 1} m x x^{2 \, m}\right )}}\,{d x} \]

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]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (d\,{\left (e+f\,x^m\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (g\,x\right )}^{2\,m+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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