\(\int \frac {(f+g x^{-n})^2 \log (c (d+e x^n)^p)}{x} \, dx\) [368]

Optimal result

Integrand size = 27, antiderivative size = 193 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {e g^2 p x^{-n}}{2 d n}+\frac {2 e f g p \log (x)}{d}-\frac {e^2 g^2 p \log (x)}{2 d^2}-\frac {2 e f g p \log \left (d+e x^n\right )}{d n}+\frac {e^2 g^2 p \log \left (d+e x^n\right )}{2 d^2 n}-\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )}{n} \]

[Out]

-1/2*e*g^2*p/d/n/(x^n)+2*e*f*g*p*ln(x)/d-1/2*e^2*g^2*p*ln(x)/d^2-2*e*f*g*p*ln(d+e*x^n)/d/n+1/2*e^2*g^2*p*ln(d+
e*x^n)/d^2/n-1/2*g^2*ln(c*(d+e*x^n)^p)/n/(x^(2*n))-2*f*g*ln(c*(d+e*x^n)^p)/n/(x^n)+f^2*ln(-e*x^n/d)*ln(c*(d+e*
x^n)^p)/n+f^2*p*polylog(2,1+e*x^n/d)/n

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.407, Rules used = {2525, 269, 45, 2463, 2442, 46, 36, 29, 31, 2441, 2352} \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {e^2 g^2 p \log \left (d+e x^n\right )}{2 d^2 n}-\frac {e^2 g^2 p \log (x)}{2 d^2}+\frac {f^2 p \operatorname {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}-\frac {2 e f g p \log \left (d+e x^n\right )}{d n}+\frac {2 e f g p \log (x)}{d}-\frac {e g^2 p x^{-n}}{2 d n} \]

[In]

Int[((f + g/x^n)^2*Log[c*(d + e*x^n)^p])/x,x]

[Out]

-1/2*(e*g^2*p)/(d*n*x^n) + (2*e*f*g*p*Log[x])/d - (e^2*g^2*p*Log[x])/(2*d^2) - (2*e*f*g*p*Log[d + e*x^n])/(d*n
) + (e^2*g^2*p*Log[d + e*x^n])/(2*d^2*n) - (g^2*Log[c*(d + e*x^n)^p])/(2*n*x^(2*n)) - (2*f*g*Log[c*(d + e*x^n)
^p])/(n*x^n) + (f^2*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (f^2*p*PolyLog[2, 1 + (e*x^n)/d])/n

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 45

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 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 269

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

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 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]

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2525

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (f+\frac {g}{x}\right )^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n} \\ & = \frac {\text {Subst}\left (\int \left (\frac {g^2 \log \left (c (d+e x)^p\right )}{x^3}+\frac {2 f g \log \left (c (d+e x)^p\right )}{x^2}+\frac {f^2 \log \left (c (d+e x)^p\right )}{x}\right ) \, dx,x,x^n\right )}{n} \\ & = \frac {f^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac {(2 f g) \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^2} \, dx,x,x^n\right )}{n}+\frac {g^2 \text {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x^3} \, dx,x,x^n\right )}{n} \\ & = -\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {\left (e f^2 p\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}+\frac {(2 e f g p) \text {Subst}\left (\int \frac {1}{x (d+e x)} \, dx,x,x^n\right )}{n}+\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \frac {1}{x^2 (d+e x)} \, dx,x,x^n\right )}{2 n} \\ & = -\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n}+\frac {(2 e f g p) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^n\right )}{d n}-\frac {\left (2 e^2 f g p\right ) \text {Subst}\left (\int \frac {1}{d+e x} \, dx,x,x^n\right )}{d n}+\frac {\left (e g^2 p\right ) \text {Subst}\left (\int \left (\frac {1}{d x^2}-\frac {e}{d^2 x}+\frac {e^2}{d^2 (d+e x)}\right ) \, dx,x,x^n\right )}{2 n} \\ & = -\frac {e g^2 p x^{-n}}{2 d n}+\frac {2 e f g p \log (x)}{d}-\frac {e^2 g^2 p \log (x)}{2 d^2}-\frac {2 e f g p \log \left (d+e x^n\right )}{d n}+\frac {e^2 g^2 p \log \left (d+e x^n\right )}{2 d^2 n}-\frac {g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f^2 p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.83 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {-4 d e f g n p \log (x)+4 d e f g p \log \left (d+e x^n\right )+e g^2 p \left (d x^{-n}+e n \log (x)-e \log \left (d+e x^n\right )\right )+d^2 g^2 x^{-2 n} \log \left (c \left (d+e x^n\right )^p\right )+4 d^2 f g x^{-n} \log \left (c \left (d+e x^n\right )^p\right )-2 d^2 f^2 \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \operatorname {PolyLog}\left (2,1+\frac {e x^n}{d}\right )\right )}{2 d^2 n} \]

[In]

Integrate[((f + g/x^n)^2*Log[c*(d + e*x^n)^p])/x,x]

[Out]

-1/2*(-4*d*e*f*g*n*p*Log[x] + 4*d*e*f*g*p*Log[d + e*x^n] + e*g^2*p*(d/x^n + e*n*Log[x] - e*Log[d + e*x^n]) + (
d^2*g^2*Log[c*(d + e*x^n)^p])/x^(2*n) + (4*d^2*f*g*Log[c*(d + e*x^n)^p])/x^n - 2*d^2*f^2*(Log[-((e*x^n)/d)]*Lo
g[c*(d + e*x^n)^p] + p*PolyLog[2, 1 + (e*x^n)/d]))/(d^2*n)

Maple [C] (warning: unable to verify)

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

Time = 6.57 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.73

method	result	size
risch	\(\frac {\left (2 f^{2} \ln \left (x \right ) n \,x^{2 n}-4 f g \,x^{n}-g^{2}\right ) x^{-2 n} \ln \left (\left (d +e \,x^{n}\right )^{p}\right )}{2 n}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (d +e \,x^{n}\right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (-2 f g \,x^{-n}+f^{2} \ln \left (x^{n}\right )-\frac {g^{2} x^{-2 n}}{2}\right )}{n}-\frac {2 e f g p \ln \left (d +e \,x^{n}\right )}{d n}+\frac {2 p e f g \ln \left (x^{n}\right )}{n d}+\frac {e^{2} g^{2} p \ln \left (d +e \,x^{n}\right )}{2 d^{2} n}-\frac {e \,g^{2} p \,x^{-n}}{2 d n}-\frac {p \,e^{2} g^{2} \ln \left (x^{n}\right )}{2 n \,d^{2}}-\frac {p \,f^{2} \operatorname {dilog}\left (\frac {d +e \,x^{n}}{d}\right )}{n}-p \,f^{2} \ln \left (x \right ) \ln \left (\frac {d +e \,x^{n}}{d}\right )\)	\(334\)

[In]

int((f+g/(x^n))^2*ln(c*(d+e*x^n)^p)/x,x,method=_RETURNVERBOSE)

[Out]

1/2*(2*f^2*ln(x)*n*(x^n)^2-4*f*g*x^n-g^2)/n/(x^n)^2*ln((d+e*x^n)^p)+(1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+
e*x^n)^p)^2-1/2*I*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*csgn(I*c)-1/2*I*Pi*csgn(I*c*(d+e*x^n)^p)^3+1/2*
I*Pi*csgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)+ln(c))/n*(-2*f*g/(x^n)+f^2*ln(x^n)-1/2*g^2/(x^n)^2)-2*e*f*g*p*ln(d+e*x^
n)/d/n+2*p*e/n*f*g/d*ln(x^n)+1/2*e^2*g^2*p*ln(d+e*x^n)/d^2/n-1/2*e*g^2*p/d/n/(x^n)-1/2*p*e^2/n*g^2/d^2*ln(x^n)
-p/n*f^2*dilog((d+e*x^n)/d)-p*f^2*ln(x)*ln((d+e*x^n)/d)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.09 \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=-\frac {2 \, d^{2} f^{2} n p x^{2 \, n} \log \left (x\right ) \log \left (\frac {e x^{n} + d}{d}\right ) + 2 \, d^{2} f^{2} p x^{2 \, n} {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + d^{2} g^{2} \log \left (c\right ) - {\left (2 \, d^{2} f^{2} n \log \left (c\right ) + {\left (4 \, d e f g - e^{2} g^{2}\right )} n p\right )} x^{2 \, n} \log \left (x\right ) + {\left (d e g^{2} p + 4 \, d^{2} f g \log \left (c\right )\right )} x^{n} + {\left (4 \, d^{2} f g p x^{n} + d^{2} g^{2} p - {\left (2 \, d^{2} f^{2} n p \log \left (x\right ) - {\left (4 \, d e f g - e^{2} g^{2}\right )} p\right )} x^{2 \, n}\right )} \log \left (e x^{n} + d\right )}{2 \, d^{2} n x^{2 \, n}} \]

[In]

integrate((f+g/(x^n))^2*log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")

[Out]

-1/2*(2*d^2*f^2*n*p*x^(2*n)*log(x)*log((e*x^n + d)/d) + 2*d^2*f^2*p*x^(2*n)*dilog(-(e*x^n + d)/d + 1) + d^2*g^
2*log(c) - (2*d^2*f^2*n*log(c) + (4*d*e*f*g - e^2*g^2)*n*p)*x^(2*n)*log(x) + (d*e*g^2*p + 4*d^2*f*g*log(c))*x^
n + (4*d^2*f*g*p*x^n + d^2*g^2*p - (2*d^2*f^2*n*p*log(x) - (4*d*e*f*g - e^2*g^2)*p)*x^(2*n))*log(e*x^n + d))/(
d^2*n*x^(2*n))

Sympy [F]

\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {x^{- 2 n} \left (f x^{n} + g\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]

[In]

integrate((f+g/(x**n))**2*ln(c*(d+e*x**n)**p)/x,x)

[Out]

Integral((f*x**n + g)**2*log(c*(d + e*x**n)**p)/(x*x**(2*n)), x)

Maxima [F]

\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate((f+g/(x^n))^2*log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")

[Out]

-1/2*(d*g^2*log(c) + (d*f^2*n^2*p*log(x)^2 - 2*d*f^2*n*log(c)*log(x))*x^(2*n) + (e*g^2*p + 4*d*f*g*log(c))*x^n
- (2*d*f^2*n*x^(2*n)*log(x) - 4*d*f*g*x^n - d*g^2)*log((e*x^n + d)^p))/(d*n*x^(2*n)) + integrate(1/2*(2*d^2*f
^2*n*p*log(x) + 4*d*e*f*g*p - e^2*g^2*p)/(d*e*x*x^n + d^2*x), x)

Giac [F]

\[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int { \frac {{\left (f + \frac {g}{x^{n}}\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate((f+g/(x^n))^2*log(c*(d+e*x^n)^p)/x,x, algorithm="giac")

[Out]

integrate((f + g/x^n)^2*log((e*x^n + d)^p*c)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (f+g x^{-n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+\frac {g}{x^n}\right )}^2}{x} \,d x \]

[In]

int((log(c*(d + e*x^n)^p)*(f + g/x^n)^2)/x,x)

[Out]

int((log(c*(d + e*x^n)^p)*(f + g/x^n)^2)/x, x)