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

Optimal. Leaf size=176 \[ \frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {f^2 p \text {Li}_2\left (\frac {e x^n}{d}+1\right )}{n}+\frac {d g^2 p x^n}{2 e n}-\frac {2 f g p x^n}{n}-\frac {g^2 p x^{2 n}}{4 n} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2475, 43, 2416, 2389, 2295, 2394, 2315, 2395} \[ \frac {f^2 p \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}+\frac {f^2 \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^2 n}+\frac {d g^2 p x^n}{2 e n}-\frac {2 f g p x^n}{n}-\frac {g^2 p x^{2 n}}{4 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*f*g*p*x^n)/n + (d*g^2*p*x^n)/(2*e*n) - (g^2*p*x^(2*n))/(4*n) - (d^2*g^2*p*Log[d + e*x^n])/(2*e^2*n) + (g^2
*x^(2*n)*Log[c*(d + e*x^n)^p])/(2*n) + (2*f*g*(d + e*x^n)*Log[c*(d + e*x^n)^p])/(e*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 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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 2389

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

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 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]

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 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])

Rubi steps

\begin {align*} \int \frac {\left (f+g x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(f+g x)^2 \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (2 f g \log \left (c (d+e x)^p\right )+\frac {f^2 \log \left (c (d+e x)^p\right )}{x}+g^2 x \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 \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}+\frac {g^2 \operatorname {Subst}\left (\int x \log \left (c (d+e x)^p\right ) \, 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 {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) \operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^n\right )}{e 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 {\left (e g^2 p\right ) \operatorname {Subst}\left (\int \frac {x^2}{d+e x} \, dx,x,x^n\right )}{2 n}\\ &=-\frac {2 f g p x^n}{n}+\frac {g^2 x^{2 n} \log \left (c \left (d+e x^n\right )^p\right )}{2 n}+\frac {2 f g \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e 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 {\left (e g^2 p\right ) \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 )}{2 n}\\ &=-\frac {2 f g p x^n}{n}+\frac {d g^2 p x^n}{2 e n}-\frac {g^2 p x^{2 n}}{4 n}-\frac {d^2 g^2 p \log \left (d+e x^n\right )}{2 e^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 \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e 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*}

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Mathematica [A]  time = 0.19, size = 124, normalized size = 0.70 \[ \frac {2 e \log \left (c \left (d+e x^n\right )^p\right ) \left (2 e f^2 \log \left (-\frac {e x^n}{d}\right )+4 d f g+e g x^n \left (4 f+g x^n\right )\right )-2 d^2 g^2 p \log \left (d+e x^n\right )+4 e^2 f^2 p \text {Li}_2\left (\frac {e x^n}{d}+1\right )-e g p x^n \left (-2 d g+8 e f+e g x^n\right )}{4 e^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 192, normalized size = 1.09 \[ -\frac {4 \, e^{2} f^{2} n p \log \relax (x) \log \left (\frac {e x^{n} + d}{d}\right ) - 4 \, e^{2} f^{2} n \log \relax (c) \log \relax (x) + 4 \, e^{2} f^{2} p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + {\left (e^{2} g^{2} p - 2 \, e^{2} g^{2} \log \relax (c)\right )} x^{2 \, n} - 2 \, {\left (4 \, e^{2} f g \log \relax (c) - {\left (4 \, e^{2} f g - d e g^{2}\right )} p\right )} x^{n} - 2 \, {\left (2 \, e^{2} f^{2} n p \log \relax (x) + e^{2} g^{2} p x^{2 \, n} + 4 \, e^{2} f g p x^{n} + {\left (4 \, d e f g - d^{2} g^{2}\right )} p\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (g x^{n} + f\right )}^{2} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 3.80, size = 665, normalized size = 3.78 \[ -\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right ) \ln \left (x^{n}\right )}{2 n}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \left (x^{n}\right )}{2 n}+\frac {i \pi \,f^{2} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \left (x^{n}\right )}{2 n}-\frac {i \pi \,f^{2} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3} \ln \left (x^{n}\right )}{2 n}-f^{2} p \ln \relax (x ) \ln \left (\frac {e \,x^{n}+d}{d}\right )-\frac {i \pi f g \,x^{n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )}{n}+\frac {i \pi f g \,x^{n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{n}+\frac {i \pi f g \,x^{n} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{n}-\frac {i \pi f g \,x^{n} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{n}-\frac {i \pi \,g^{2} x^{2 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )}{4 n}+\frac {i \pi \,g^{2} x^{2 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{4 n}+\frac {i \pi \,g^{2} x^{2 n} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{4 n}-\frac {i \pi \,g^{2} x^{2 n} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{4 n}-\frac {d^{2} g^{2} p \ln \left (e \,x^{n}+d \right )}{2 e^{2} n}+\frac {2 d f g p \ln \left (e \,x^{n}+d \right )}{e n}+\frac {d \,g^{2} p \,x^{n}}{2 e n}-\frac {f^{2} p \dilog \left (\frac {e \,x^{n}+d}{d}\right )}{n}+\frac {f^{2} \ln \relax (c ) \ln \left (x^{n}\right )}{n}-\frac {2 f g p \,x^{n}}{n}+\frac {2 f g \,x^{n} \ln \relax (c )}{n}-\frac {g^{2} p \,x^{2 n}}{4 n}+\frac {g^{2} x^{2 n} \ln \relax (c )}{2 n}+\frac {\left (2 f^{2} n \ln \relax (x )+4 f g \,x^{n}+g^{2} x^{2 n}\right ) \ln \left (\left (e \,x^{n}+d \right )^{p}\right )}{2 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*e^2*f^2*n^2*p*log(x)^2 + (e^2*g^2*p - 2*e^2*g^2*log(c))*x^(2*n) + 2*(4*e^2*f*g*p - d*e*g^2*p - 4*e^2*f
*g*log(c))*x^n - 2*(2*e^2*f^2*n*log(x) + e^2*g^2*x^(2*n) + 4*e^2*f*g*x^n)*log((e*x^n + d)^p) - 2*(4*d*e*f*g*n*
p - d^2*g^2*n*p + 2*e^2*f^2*n*log(c))*log(x))/(e^2*n) + integrate(1/2*(2*d*e^2*f^2*n*p*log(x) - 4*d^2*e*f*g*p
+ d^3*g^2*p)/(e^3*x*x^n + d*e^2*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,{\left (f+g\,x^n\right )}^2}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (f + g x^{n}\right )^{2} \log {\left (c \left (d + e x^{n}\right )^{p} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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