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

Optimal. Leaf size=363 \[ \frac{b e^2 k n x^{-2 m} (g x)^{2 m} \text{PolyLog}\left (2,\frac{f x^m}{e}+1\right )}{2 f^2 g m^2}+\frac{(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac{e^2 k x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^2 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}-\frac{b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac{b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac{b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac{f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac{3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}+\frac{b k n (g x)^{2 m}}{4 g m^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.41775, antiderivative size = 363, 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, 43, 2376, 16, 32, 30, 19, 2454, 2394, 2315} \[ \frac{b e^2 k n x^{-2 m} (g x)^{2 m} \text{PolyLog}\left (2,\frac{f x^m}{e}+1\right )}{2 f^2 g m^2}+\frac{(g x)^{2 m} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f x^m\right )^k\right )}{2 g m}-\frac{e^2 k x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right ) \left (a+b \log \left (c x^n\right )\right )}{2 f^2 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}-\frac{b n (g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{4 g m^2}+\frac{b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac{b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (-\frac{f x^m}{e}\right ) \log \left (e+f x^m\right )}{2 f^2 g m^2}-\frac{3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}+\frac{b k n (g x)^{2 m}}{4 g m^2} \]

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]

(b*k*n*(g*x)^(2*m))/(4*g*m^2) - (3*b*e*k*n*(g*x)^(2*m))/(4*f*g*m^2*x^m) - (k*(g*x)^(2*m)*(a + b*Log[c*x^n]))/(
4*g*m) + (e*k*(g*x)^(2*m)*(a + b*Log[c*x^n]))/(2*f*g*m*x^m) + (b*e^2*k*n*(g*x)^(2*m)*Log[e + f*x^m])/(4*f^2*g*
m^2*x^(2*m)) + (b*e^2*k*n*(g*x)^(2*m)*Log[-((f*x^m)/e)]*Log[e + f*x^m])/(2*f^2*g*m^2*x^(2*m)) - (e^2*k*(g*x)^(
2*m)*(a + b*Log[c*x^n])*Log[e + f*x^m])/(2*f^2*g*m*x^(2*m)) - (b*n*(g*x)^(2*m)*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) + (b*e^2*k*n*(g*x)^(2*m)*PolyLog[2, 1 + (f*x^
m)/e])/(2*f^2*g*m^2*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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -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 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]

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{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{e^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 f^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{k (g x)^{2 m}}{4 g m x}+\frac{e k x^{-1-m} (g x)^{2 m}}{2 f g m}-\frac{e^2 k x^{-1-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{2 f^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{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{e^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 f^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 \frac{(g x)^{2 m} \log \left (d \left (e+f x^m\right )^k\right )}{x} \, dx}{2 g m}+\frac{(b k n) \int \frac{(g x)^{2 m}}{x} \, dx}{4 g m}+\frac{\left (b e^2 k n\right ) \int x^{-1-2 m} (g x)^{2 m} \log \left (e+f x^m\right ) \, dx}{2 f^2 g m}-\frac{(b e k n) \int x^{-1-m} (g x)^{2 m} \, dx}{2 f g m}\\ &=-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{e^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 f^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{(b k n) \int (g x)^{-1+2 m} \, dx}{4 m}+\frac{\left (b e^2 k n x^{-2 m} (g x)^{2 m}\right ) \int \frac{\log \left (e+f x^m\right )}{x} \, dx}{2 f^2 g m}-\frac{\left (b e k n x^{-2 m} (g x)^{2 m}\right ) \int x^{-1+m} \, dx}{2 f g m}\\ &=\frac{b k n (g x)^{2 m}}{8 g m^2}-\frac{b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}-\frac{e^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 f^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 e^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 f^2 g m^2}\\ &=\frac{b k n (g x)^{2 m}}{8 g m^2}-\frac{b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac{b e^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 f^2 g m^2}-\frac{e^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 f^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 e^2 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 f g m^2}+\frac{\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \int \frac{x^{-1+3 m}}{e+f x^m} \, dx}{4 g m}\\ &=\frac{b k n (g x)^{2 m}}{8 g m^2}-\frac{b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac{b e^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 f^2 g m^2}-\frac{e^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 f^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 e^2 k n x^{-2 m} (g x)^{2 m} \text{Li}_2\left (1+\frac{f x^m}{e}\right )}{2 f^2 g m^2}+\frac{\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \operatorname{Subst}\left (\int \frac{x^2}{e+f x} \, dx,x,x^m\right )}{4 g m^2}\\ &=\frac{b k n (g x)^{2 m}}{8 g m^2}-\frac{b e k n x^{-m} (g x)^{2 m}}{2 f g m^2}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac{b e^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 f^2 g m^2}-\frac{e^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 f^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 e^2 k n x^{-2 m} (g x)^{2 m} \text{Li}_2\left (1+\frac{f x^m}{e}\right )}{2 f^2 g m^2}+\frac{\left (b f k n x^{-2 m} (g x)^{2 m}\right ) \operatorname{Subst}\left (\int \left (-\frac{e}{f^2}+\frac{x}{f}+\frac{e^2}{f^2 (e+f x)}\right ) \, dx,x,x^m\right )}{4 g m^2}\\ &=\frac{b k n (g x)^{2 m}}{4 g m^2}-\frac{3 b e k n x^{-m} (g x)^{2 m}}{4 f g m^2}-\frac{k (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{4 g m}+\frac{e k x^{-m} (g x)^{2 m} \left (a+b \log \left (c x^n\right )\right )}{2 f g m}+\frac{b e^2 k n x^{-2 m} (g x)^{2 m} \log \left (e+f x^m\right )}{4 f^2 g m^2}+\frac{b e^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 f^2 g m^2}-\frac{e^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 f^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 e^2 k n x^{-2 m} (g x)^{2 m} \text{Li}_2\left (1+\frac{f x^m}{e}\right )}{2 f^2 g m^2}\\ \end{align*}

Mathematica [A]  time = 0.379153, size = 352, normalized size = 0.97 \[ \frac{x^{-2 m} (g x)^{2 m} \left (2 b e^2 k n \text{PolyLog}\left (2,\frac{f x^m}{e}+1\right )+e^2 k m \log (x) \left (-2 a m-2 b m \log \left (c x^n\right )-2 b n \log \left (e+f x^m\right )+2 b n \log \left (e-e x^m\right )+b n\right )+2 a f^2 m x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )-2 a e^2 k m \log \left (e-e x^m\right )+2 a e f k m x^m-a f^2 k m x^{2 m}+2 b f^2 m x^{2 m} \log \left (c x^n\right ) \log \left (d \left (e+f x^m\right )^k\right )-2 b e^2 k m \log \left (c x^n\right ) \log \left (e-e x^m\right )+2 b e f k m x^m \log \left (c x^n\right )-b f^2 k m x^{2 m} \log \left (c x^n\right )-b f^2 n x^{2 m} \log \left (d \left (e+f x^m\right )^k\right )+2 b e^2 k n \log \left (-\frac{f x^m}{e}\right ) \log \left (e+f x^m\right )+2 b e^2 k m^2 n \log ^2(x)+b e^2 k n \log \left (e-e x^m\right )-3 b e f k n x^m+b f^2 k n x^{2 m}\right )}{4 f^2 g m^2} \]

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]

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

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Maple [F]  time = 0.24, 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)

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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

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Fricas [A]  time = 0.987284, size = 730, normalized size = 2.01 \begin{align*} -\frac{2 \, b e^{2} g^{2 \, m - 1} k m n \log \left (x\right ) \log \left (\frac{f x^{m} + e}{e}\right ) + 2 \, b e^{2} g^{2 \, m - 1} k n{\rm Li}_2\left (-\frac{f x^{m} + e}{e} + 1\right ) +{\left (b f^{2} k m \log \left (c\right ) + a f^{2} k m - b f^{2} k n -{\left (2 \, b f^{2} m \log \left (c\right ) + 2 \, a f^{2} m - b f^{2} n\right )} \log \left (d\right ) +{\left (b f^{2} k m n - 2 \, b f^{2} m n \log \left (d\right )\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 ({\left (2 \, b f^{2} k m n \log \left (x\right ) + 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 \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 \, f^{2} m^{2}} \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*e^2*g^(2*m - 1)*k*m*n*log(x)*log((f*x^m + e)/e) + 2*b*e^2*g^(2*m - 1)*k*n*dilog(-(f*x^m + e)/e + 1)
+ (b*f^2*k*m*log(c) + a*f^2*k*m - b*f^2*k*n - (2*b*f^2*m*log(c) + 2*a*f^2*m - b*f^2*n)*log(d) + (b*f^2*k*m*n -
2*b*f^2*m*n*log(d))*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*f^2*k*m*n*log(x) + 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*log(c) + 2*a*e^2*k*m - b*e^2*k*n)*g^(2*m - 1))*log(f*x^m + e))/(f^2*m^2)

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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((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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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)