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

Optimal. Leaf size=254 \[ \frac{f^2 p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f^2 \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac{d^2 f g p \log \left (d+e x^n\right )}{e^2 n}+\frac{d^3 g^2 p x^n}{4 e^3 n}-\frac{d^2 g^2 p x^{2 n}}{8 e^2 n}-\frac{d^4 g^2 p \log \left (d+e x^n\right )}{4 e^4 n}+\frac{d f g p x^n}{e n}+\frac{d g^2 p x^{3 n}}{12 e n}-\frac{f g p x^{2 n}}{2 n}-\frac{g^2 p x^{4 n}}{16 n} \]

[Out]

(d*f*g*p*x^n)/(e*n) + (d^3*g^2*p*x^n)/(4*e^3*n) - (f*g*p*x^(2*n))/(2*n) - (d^2*g^2*p*x^(2*n))/(8*e^2*n) + (d*g
^2*p*x^(3*n))/(12*e*n) - (g^2*p*x^(4*n))/(16*n) - (d^2*f*g*p*Log[d + e*x^n])/(e^2*n) - (d^4*g^2*p*Log[d + e*x^
n])/(4*e^4*n) + (f*g*x^(2*n)*Log[c*(d + e*x^n)^p])/n + (g^2*x^(4*n)*Log[c*(d + e*x^n)^p])/(4*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

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Rubi [A]  time = 0.267391, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2475, 266, 43, 2416, 2394, 2315, 2395} \[ \frac{f^2 p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )}{n}+\frac{f^2 \log \left (-\frac{e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 n}-\frac{d^2 f g p \log \left (d+e x^n\right )}{e^2 n}+\frac{d^3 g^2 p x^n}{4 e^3 n}-\frac{d^2 g^2 p x^{2 n}}{8 e^2 n}-\frac{d^4 g^2 p \log \left (d+e x^n\right )}{4 e^4 n}+\frac{d f g p x^n}{e n}+\frac{d g^2 p x^{3 n}}{12 e n}-\frac{f g p x^{2 n}}{2 n}-\frac{g^2 p x^{4 n}}{16 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*f*g*p*x^n)/(e*n) + (d^3*g^2*p*x^n)/(4*e^3*n) - (f*g*p*x^(2*n))/(2*n) - (d^2*g^2*p*x^(2*n))/(8*e^2*n) + (d*g
^2*p*x^(3*n))/(12*e*n) - (g^2*p*x^(4*n))/(16*n) - (d^2*f*g*p*Log[d + e*x^n])/(e^2*n) - (d^4*g^2*p*Log[d + e*x^
n])/(4*e^4*n) + (f*g*x^(2*n)*Log[c*(d + e*x^n)^p])/n + (g^2*x^(4*n)*Log[c*(d + e*x^n)^p])/(4*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 2475

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])

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 2416

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 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 2395

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]

Rubi steps

\begin{align*} \int \frac{\left (f+g x^{2 n}\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (f+g x^2\right )^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{f^2 \log \left (c (d+e x)^p\right )}{x}+2 f g x \log \left (c (d+e x)^p\right )+g^2 x^3 \log \left (c (d+e x)^p\right )\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{f^2 \operatorname{Subst}\left (\int \frac{\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac{(2 f g) \operatorname{Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}+\frac{g^2 \operatorname{Subst}\left (\int x^3 \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}\\ &=\frac{f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 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 ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}-\frac{(e f g p) \operatorname{Subst}\left (\int \frac{x^2}{d+e x} \, dx,x,x^n\right )}{n}-\frac{\left (e g^2 p\right ) \operatorname{Subst}\left (\int \frac{x^4}{d+e x} \, dx,x,x^n\right )}{4 n}\\ &=\frac{f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 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{(e f g p) \operatorname{Subst}\left (\int \left (-\frac{d}{e^2}+\frac{x}{e}+\frac{d^2}{e^2 (d+e x)}\right ) \, dx,x,x^n\right )}{n}-\frac{\left (e g^2 p\right ) \operatorname{Subst}\left (\int \left (-\frac{d^3}{e^4}+\frac{d^2 x}{e^3}-\frac{d x^2}{e^2}+\frac{x^3}{e}+\frac{d^4}{e^4 (d+e x)}\right ) \, dx,x,x^n\right )}{4 n}\\ &=\frac{d f g p x^n}{e n}+\frac{d^3 g^2 p x^n}{4 e^3 n}-\frac{f g p x^{2 n}}{2 n}-\frac{d^2 g^2 p x^{2 n}}{8 e^2 n}+\frac{d g^2 p x^{3 n}}{12 e n}-\frac{g^2 p x^{4 n}}{16 n}-\frac{d^2 f g p \log \left (d+e x^n\right )}{e^2 n}-\frac{d^4 g^2 p \log \left (d+e x^n\right )}{4 e^4 n}+\frac{f g x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac{g^2 x^{4 n} \log \left (c \left (d+e x^n\right )^p\right )}{4 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]  time = 0.259248, size = 171, normalized size = 0.67 \[ \frac{48 e^4 f^2 p \text{PolyLog}\left (2,\frac{e x^n}{d}+1\right )+12 e^4 \log \left (c \left (d+e x^n\right )^p\right ) \left (4 f^2 \log \left (-\frac{e x^n}{d}\right )+g x^{2 n} \left (4 f+g x^{2 n}\right )\right )-e g p x^n \left (6 d^2 e g x^n-12 d^3 g-4 d e^2 \left (12 f+g x^{2 n}\right )+3 e^3 x^n \left (8 f+g x^{2 n}\right )\right )-12 d^2 g p \left (d^2 g+4 e^2 f\right ) \log \left (d+e x^n\right )}{48 e^4 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(e*g*p*x^n*(-12*d^3*g + 6*d^2*e*g*x^n + 3*e^3*x^n*(8*f + g*x^(2*n)) - 4*d*e^2*(12*f + g*x^(2*n)))) - 12*d^2*
g*(4*e^2*f + d^2*g)*p*Log[d + e*x^n] + 12*e^4*(g*x^(2*n)*(4*f + g*x^(2*n)) + 4*f^2*Log[-((e*x^n)/d)])*Log[c*(d
+ e*x^n)^p] + 48*e^4*f^2*p*PolyLog[2, 1 + (e*x^n)/d])/(48*e^4*n)

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Maple [C]  time = 5.136, size = 734, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I/n*Pi*csgn(I*c*(d+e*x^n)^p)^3*f^2*ln(x^n)-d^2*f*g*p*ln(d+e*x^n)/e^2/n-p/n*f^2*dilog((d+e*x^n)/d)-1/16*p/
n*g^2*(x^n)^4+1/4/n*ln(c)*(x^n)^4*g^2+1/4*(g^2*(x^n)^4+4*f^2*ln(x)*n+4*f*g*(x^n)^2)/n*ln((d+e*x^n)^p)-1/2*I/n*
Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*csgn(I*c)*f^2*ln(x^n)+1/2*I/n*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+
e*x^n)^p)^2*(x^n)^2*f*g-1/8*I/n*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*csgn(I*c)*(x^n)^4*g^2+1/2*I/n*Pi*
csgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)*(x^n)^2*f*g-1/8*p/e^2/n*g^2*(x^n)^2*d^2-1/4*d^4*g^2*p*ln(d+e*x^n)/e^4/n-p*f^
2*ln(x)*ln((d+e*x^n)/d)+1/4*d^3*g^2*p*x^n/e^3/n-1/2*p/n*f*g*(x^n)^2+1/n*ln(c)*(x^n)^2*f*g-1/2*I/n*Pi*csgn(I*c*
(d+e*x^n)^p)^3*(x^n)^2*f*g+1/2*I/n*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)^2*f^2*ln(x^n)+1/2*I/n*Pi*csgn(
I*c*(d+e*x^n)^p)^2*csgn(I*c)*f^2*ln(x^n)+d*f*g*p*x^n/e/n+1/n*ln(c)*f^2*ln(x^n)-1/8*I/n*Pi*csgn(I*c*(d+e*x^n)^p
)^3*(x^n)^4*g^2+1/8*I/n*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)^2*(x^n)^4*g^2+1/12*p/e/n*g^2*d*(x^n)^3+1/
8*I/n*Pi*csgn(I*c*(d+e*x^n)^p)^2*csgn(I*c)*(x^n)^4*g^2-1/2*I/n*Pi*csgn(I*(d+e*x^n)^p)*csgn(I*c*(d+e*x^n)^p)*cs
gn(I*c)*(x^n)^2*f*g

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{24 \, e^{4} f^{2} n^{2} p \log \left (x\right )^{2} - 4 \, d e^{3} g^{2} p x^{3 \, n} + 3 \,{\left (e^{4} g^{2} p - 4 \, e^{4} g^{2} \log \left (c\right )\right )} x^{4 \, n} + 6 \,{\left (4 \, e^{4} f g p + d^{2} e^{2} g^{2} p - 8 \, e^{4} f g \log \left (c\right )\right )} x^{2 \, n} - 12 \,{\left (4 \, d e^{3} f g p + d^{3} e g^{2} p\right )} x^{n} - 12 \,{\left (4 \, e^{4} f^{2} n \log \left (x\right ) + e^{4} g^{2} x^{4 \, n} + 4 \, e^{4} f g x^{2 \, n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p}\right ) + 12 \,{\left (4 \, d^{2} e^{2} f g n p + d^{4} g^{2} n p - 4 \, e^{4} f^{2} n \log \left (c\right )\right )} \log \left (x\right )}{48 \, e^{4} n} + \int \frac{4 \, d e^{4} f^{2} n p \log \left (x\right ) + 4 \, d^{3} e^{2} f g p + d^{5} g^{2} p}{4 \,{\left (e^{5} x x^{n} + d e^{4} x\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(24*e^4*f^2*n^2*p*log(x)^2 - 4*d*e^3*g^2*p*x^(3*n) + 3*(e^4*g^2*p - 4*e^4*g^2*log(c))*x^(4*n) + 6*(4*e^4
*f*g*p + d^2*e^2*g^2*p - 8*e^4*f*g*log(c))*x^(2*n) - 12*(4*d*e^3*f*g*p + d^3*e*g^2*p)*x^n - 12*(4*e^4*f^2*n*lo
g(x) + e^4*g^2*x^(4*n) + 4*e^4*f*g*x^(2*n))*log((e*x^n + d)^p) + 12*(4*d^2*e^2*f*g*n*p + d^4*g^2*n*p - 4*e^4*f
^2*n*log(c))*log(x))/(e^4*n) + integrate(1/4*(4*d*e^4*f^2*n*p*log(x) + 4*d^3*e^2*f*g*p + d^5*g^2*p)/(e^5*x*x^n
+ d*e^4*x), x)

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Fricas [A]  time = 2.08815, size = 551, normalized size = 2.17 \begin{align*} -\frac{48 \, e^{4} f^{2} n p \log \left (x\right ) \log \left (\frac{e x^{n} + d}{d}\right ) - 48 \, e^{4} f^{2} n \log \left (c\right ) \log \left (x\right ) - 4 \, d e^{3} g^{2} p x^{3 \, n} + 48 \, e^{4} f^{2} p{\rm Li}_2\left (-\frac{e x^{n} + d}{d} + 1\right ) - 12 \,{\left (4 \, d e^{3} f g + d^{3} e g^{2}\right )} p x^{n} + 3 \,{\left (e^{4} g^{2} p - 4 \, e^{4} g^{2} \log \left (c\right )\right )} x^{4 \, n} - 6 \,{\left (8 \, e^{4} f g \log \left (c\right ) -{\left (4 \, e^{4} f g + d^{2} e^{2} g^{2}\right )} p\right )} x^{2 \, n} - 12 \,{\left (4 \, e^{4} f^{2} n p \log \left (x\right ) + e^{4} g^{2} p x^{4 \, n} + 4 \, e^{4} f g p x^{2 \, n} -{\left (4 \, d^{2} e^{2} f g + d^{4} g^{2}\right )} p\right )} \log \left (e x^{n} + d\right )}{48 \, e^{4} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(48*e^4*f^2*n*p*log(x)*log((e*x^n + d)/d) - 48*e^4*f^2*n*log(c)*log(x) - 4*d*e^3*g^2*p*x^(3*n) + 48*e^4*
f^2*p*dilog(-(e*x^n + d)/d + 1) - 12*(4*d*e^3*f*g + d^3*e*g^2)*p*x^n + 3*(e^4*g^2*p - 4*e^4*g^2*log(c))*x^(4*n
) - 6*(8*e^4*f*g*log(c) - (4*e^4*f*g + d^2*e^2*g^2)*p)*x^(2*n) - 12*(4*e^4*f^2*n*p*log(x) + e^4*g^2*p*x^(4*n)
+ 4*e^4*f*g*p*x^(2*n) - (4*d^2*e^2*f*g + d^4*g^2)*p)*log(e*x^n + d))/(e^4*n)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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