3.360 \(\int \frac {(f+g x^{3 n}) \log (c (d+e x^n)^p)}{x} \, dx\)

Optimal. Leaf size=144 \[ \frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {d^3 g p \log \left (d+e x^n\right )}{3 e^3 n}-\frac {d^2 g p x^n}{3 e^2 n}+\frac {f p \text {Li}_2\left (\frac {e x^n}{d}+1\right )}{n}+\frac {d g p x^{2 n}}{6 e n}-\frac {g p x^{3 n}}{9 n} \]

[Out]

-1/3*d^2*g*p*x^n/e^2/n+1/6*d*g*p*x^(2*n)/e/n-1/9*g*p*x^(3*n)/n+1/3*d^3*g*p*ln(d+e*x^n)/e^3/n+1/3*g*x^(3*n)*ln(
c*(d+e*x^n)^p)/n+f*ln(-e*x^n/d)*ln(c*(d+e*x^n)^p)/n+f*p*polylog(2,1+e*x^n/d)/n

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Rubi [A]  time = 0.17, antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2475, 14, 2416, 2394, 2315, 2395, 43} \[ \frac {f p \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 n}-\frac {d^2 g p x^n}{3 e^2 n}+\frac {d^3 g p \log \left (d+e x^n\right )}{3 e^3 n}+\frac {d g p x^{2 n}}{6 e n}-\frac {g p x^{3 n}}{9 n} \]

Antiderivative was successfully verified.

[In]

Int[((f + g*x^(3*n))*Log[c*(d + e*x^n)^p])/x,x]

[Out]

-(d^2*g*p*x^n)/(3*e^2*n) + (d*g*p*x^(2*n))/(6*e*n) - (g*p*x^(3*n))/(9*n) + (d^3*g*p*Log[d + e*x^n])/(3*e^3*n)
+ (g*x^(3*n)*Log[c*(d + e*x^n)^p])/(3*n) + (f*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/n + (f*p*PolyLog[2, 1 +
(e*x^n)/d])/n

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 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 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^{3 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (f+g x^3\right ) \log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {f \log \left (c (d+e x)^p\right )}{x}+g x^2 \log \left (c (d+e x)^p\right )\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {f \operatorname {Subst}\left (\int \frac {\log \left (c (d+e x)^p\right )}{x} \, dx,x,x^n\right )}{n}+\frac {g \operatorname {Subst}\left (\int x^2 \log \left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}-\frac {(e f p) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right )}{d+e x} \, dx,x,x^n\right )}{n}-\frac {(e g p) \operatorname {Subst}\left (\int \frac {x^3}{d+e x} \, dx,x,x^n\right )}{3 n}\\ &=\frac {g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n}-\frac {(e g p) \operatorname {Subst}\left (\int \left (\frac {d^2}{e^3}-\frac {d x}{e^2}+\frac {x^2}{e}-\frac {d^3}{e^3 (d+e x)}\right ) \, dx,x,x^n\right )}{3 n}\\ &=-\frac {d^2 g p x^n}{3 e^2 n}+\frac {d g p x^{2 n}}{6 e n}-\frac {g p x^{3 n}}{9 n}+\frac {d^3 g p \log \left (d+e x^n\right )}{3 e^3 n}+\frac {g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {f \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{n}+\frac {f p \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{n}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 118, normalized size = 0.82 \[ \frac {18 f \left (\log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )+p \text {Li}_2\left (\frac {e x^n}{d}+1\right )\right )+6 g x^{3 n} \log \left (c \left (d+e x^n\right )^p\right )-\frac {g p \left (e x^n \left (6 d^2-3 d e x^n+2 e^2 x^{2 n}\right )-6 d^3 \log \left (d+e x^n\right )\right )}{e^3}}{18 n} \]

Antiderivative was successfully verified.

[In]

Integrate[((f + g*x^(3*n))*Log[c*(d + e*x^n)^p])/x,x]

[Out]

(-((g*p*(e*x^n*(6*d^2 - 3*d*e*x^n + 2*e^2*x^(2*n)) - 6*d^3*Log[d + e*x^n]))/e^3) + 6*g*x^(3*n)*Log[c*(d + e*x^
n)^p] + 18*f*(Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p] + p*PolyLog[2, 1 + (e*x^n)/d]))/(18*n)

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fricas [A]  time = 0.81, size = 148, normalized size = 1.03 \[ -\frac {18 \, e^{3} f n p \log \relax (x) \log \left (\frac {e x^{n} + d}{d}\right ) - 18 \, e^{3} f n \log \relax (c) \log \relax (x) - 3 \, d e^{2} g p x^{2 \, n} + 6 \, d^{2} e g p x^{n} + 18 \, e^{3} f p {\rm Li}_2\left (-\frac {e x^{n} + d}{d} + 1\right ) + 2 \, {\left (e^{3} g p - 3 \, e^{3} g \log \relax (c)\right )} x^{3 \, n} - 6 \, {\left (3 \, e^{3} f n p \log \relax (x) + e^{3} g p x^{3 \, n} + d^{3} g p\right )} \log \left (e x^{n} + d\right )}{18 \, e^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(3*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="fricas")

[Out]

-1/18*(18*e^3*f*n*p*log(x)*log((e*x^n + d)/d) - 18*e^3*f*n*log(c)*log(x) - 3*d*e^2*g*p*x^(2*n) + 6*d^2*e*g*p*x
^n + 18*e^3*f*p*dilog(-(e*x^n + d)/d + 1) + 2*(e^3*g*p - 3*e^3*g*log(c))*x^(3*n) - 6*(3*e^3*f*n*p*log(x) + e^3
*g*p*x^(3*n) + d^3*g*p)*log(e*x^n + d))/(e^3*n)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(3*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="giac")

[Out]

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

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maple [C]  time = 3.01, size = 428, normalized size = 2.97 \[ -\frac {i \pi f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right ) \ln \relax (x )}{2}+\frac {i \pi f \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi f \,\mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2} \ln \relax (x )}{2}-\frac {i \pi f \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3} \ln \relax (x )}{2}-f p \ln \relax (x ) \ln \left (\frac {e \,x^{n}+d}{d}\right )-\frac {i \pi g \,x^{3 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )}{6 n}+\frac {i \pi g \,x^{3 n} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{6 n}+\frac {i \pi g \,x^{3 n} \mathrm {csgn}\left (i \left (e \,x^{n}+d \right )^{p}\right ) \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{2}}{6 n}-\frac {i \pi g \,x^{3 n} \mathrm {csgn}\left (i c \left (e \,x^{n}+d \right )^{p}\right )^{3}}{6 n}+f \ln \relax (c ) \ln \relax (x )+\frac {d^{3} g p \ln \left (e \,x^{n}+d \right )}{3 e^{3} n}-\frac {d^{2} g p \,x^{n}}{3 e^{2} n}+\frac {d g p \,x^{2 n}}{6 e n}-\frac {f p \dilog \left (\frac {e \,x^{n}+d}{d}\right )}{n}-\frac {g p \,x^{3 n}}{9 n}+\frac {g \,x^{3 n} \ln \relax (c )}{3 n}+\frac {\left (3 f n \ln \relax (x )+g \,x^{3 n}\right ) \ln \left (\left (e \,x^{n}+d \right )^{p}\right )}{3 n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f+g*x^(3*n))*ln(c*(e*x^n+d)^p)/x,x)

[Out]

1/3*(g*(x^n)^3+3*f*ln(x)*n)/n*ln((e*x^n+d)^p)+1/6*I*Pi*csgn(I*(e*x^n+d)^p)*csgn(I*c*(e*x^n+d)^p)^2*g*(x^n)^3/n
-1/6*I*Pi*csgn(I*c*(e*x^n+d)^p)^3*g*(x^n)^3/n+1/2*I*Pi*csgn(I*(e*x^n+d)^p)*csgn(I*c*(e*x^n+d)^p)^2*ln(x)*f-1/2
*I*Pi*csgn(I*c*(e*x^n+d)^p)^3*ln(x)*f-1/2*I*Pi*csgn(I*(e*x^n+d)^p)*csgn(I*c*(e*x^n+d)^p)*csgn(I*c)*ln(x)*f+1/2
*I*Pi*csgn(I*c*(e*x^n+d)^p)^2*csgn(I*c)*ln(x)*f+1/6*I*Pi*csgn(I*c*(e*x^n+d)^p)^2*csgn(I*c)*g*(x^n)^3/n-1/6*I*P
i*csgn(I*(e*x^n+d)^p)*csgn(I*c*(e*x^n+d)^p)*csgn(I*c)*g*(x^n)^3/n+f*ln(c)*ln(x)+1/3*ln(c)*g*(x^n)^3/n-1/9*p/n*
g*(x^n)^3+1/6*p/e/n*g*(x^n)^2*d-1/3*d^2*g*p*x^n/e^2/n+1/3*d^3*g*p*ln(e*x^n+d)/e^3/n-p/n*f*dilog((e*x^n+d)/d)-p
*f*ln(x)*ln((e*x^n+d)/d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {9 \, e^{3} f n^{2} p \log \relax (x)^{2} - 3 \, d e^{2} g p x^{2 \, n} + 6 \, d^{2} e g p x^{n} + 2 \, {\left (e^{3} g p - 3 \, e^{3} g \log \relax (c)\right )} x^{3 \, n} - 6 \, {\left (3 \, e^{3} f n \log \relax (x) + e^{3} g x^{3 \, n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p}\right ) - 6 \, {\left (d^{3} g n p + 3 \, e^{3} f n \log \relax (c)\right )} \log \relax (x)}{18 \, e^{3} n} + \int \frac {3 \, d e^{3} f n p \log \relax (x) - d^{4} g p}{3 \, {\left (e^{4} x x^{n} + d e^{3} x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x^(3*n))*log(c*(d+e*x^n)^p)/x,x, algorithm="maxima")

[Out]

-1/18*(9*e^3*f*n^2*p*log(x)^2 - 3*d*e^2*g*p*x^(2*n) + 6*d^2*e*g*p*x^n + 2*(e^3*g*p - 3*e^3*g*log(c))*x^(3*n) -
6*(3*e^3*f*n*log(x) + e^3*g*x^(3*n))*log((e*x^n + d)^p) - 6*(d^3*g*n*p + 3*e^3*f*n*log(c))*log(x))/(e^3*n) +
integrate(1/3*(3*d*e^3*f*n*p*log(x) - d^4*g*p)/(e^4*x*x^n + d*e^3*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )\,\left (f+g\,x^{3\,n}\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(c*(d + e*x^n)^p)*(f + g*x^(3*n)))/x,x)

[Out]

int((log(c*(d + e*x^n)^p)*(f + g*x^(3*n)))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f+g*x**(3*n))*ln(c*(d+e*x**n)**p)/x,x)

[Out]

Integral((f + g*x**(3*n))*log(c*(d + e*x**n)**p)/x, x)

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