3.163 \(\int (f x)^{-1+3 n} \log ^2(c (d+e x^n)^p) \, dx\)

Optimal. Leaf size=372 \[ \frac {2 d^3 p x^{1-3 n} (f x)^{3 n-1} \log \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{3 e^3 n}-\frac {2 d^2 p x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^3 n}-\frac {2 p x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^3 \log \left (c \left (d+e x^n\right )^p\right )}{9 e^3 n}+\frac {d p x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{e^3 n}+\frac {x (f x)^{3 n-1} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}-\frac {d^3 p^2 x^{1-3 n} (f x)^{3 n-1} \log ^2\left (d+e x^n\right )}{3 e^3 n}+\frac {2 d^2 p^2 x^{1-2 n} (f x)^{3 n-1}}{e^2 n}+\frac {2 p^2 x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^3}{27 e^3 n}-\frac {d p^2 x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^2}{2 e^3 n} \]

[Out]

2*d^2*p^2*x^(1-2*n)*(f*x)^(-1+3*n)/e^2/n-1/2*d*p^2*x^(1-3*n)*(f*x)^(-1+3*n)*(d+e*x^n)^2/e^3/n+2/27*p^2*x^(1-3*
n)*(f*x)^(-1+3*n)*(d+e*x^n)^3/e^3/n-1/3*d^3*p^2*x^(1-3*n)*(f*x)^(-1+3*n)*ln(d+e*x^n)^2/e^3/n-2*d^2*p*x^(1-3*n)
*(f*x)^(-1+3*n)*(d+e*x^n)*ln(c*(d+e*x^n)^p)/e^3/n+d*p*x^(1-3*n)*(f*x)^(-1+3*n)*(d+e*x^n)^2*ln(c*(d+e*x^n)^p)/e
^3/n-2/9*p*x^(1-3*n)*(f*x)^(-1+3*n)*(d+e*x^n)^3*ln(c*(d+e*x^n)^p)/e^3/n+2/3*d^3*p*x^(1-3*n)*(f*x)^(-1+3*n)*ln(
d+e*x^n)*ln(c*(d+e*x^n)^p)/e^3/n+1/3*x*(f*x)^(-1+3*n)*ln(c*(d+e*x^n)^p)^2/n

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Rubi [A]  time = 0.32, antiderivative size = 278, normalized size of antiderivative = 0.75, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2456, 2454, 2398, 2411, 43, 2334, 12, 14, 2301} \[ -\frac {p x^{1-3 n} (f x)^{3 n-1} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{3 n-1} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {2 d^2 p^2 x^{1-2 n} (f x)^{3 n-1}}{e^2 n}-\frac {d^3 p^2 x^{1-3 n} (f x)^{3 n-1} \log ^2\left (d+e x^n\right )}{3 e^3 n}+\frac {2 p^2 x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^3}{27 e^3 n}-\frac {d p^2 x^{1-3 n} (f x)^{3 n-1} \left (d+e x^n\right )^2}{2 e^3 n} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p]^2,x]

[Out]

(2*d^2*p^2*x^(1 - 2*n)*(f*x)^(-1 + 3*n))/(e^2*n) - (d*p^2*x^(1 - 3*n)*(f*x)^(-1 + 3*n)*(d + e*x^n)^2)/(2*e^3*n
) + (2*p^2*x^(1 - 3*n)*(f*x)^(-1 + 3*n)*(d + e*x^n)^3)/(27*e^3*n) - (d^3*p^2*x^(1 - 3*n)*(f*x)^(-1 + 3*n)*Log[
d + e*x^n]^2)/(3*e^3*n) - (p*x^(1 - 3*n)*(f*x)^(-1 + 3*n)*((18*d^2*(d + e*x^n))/e^3 - (9*d*(d + e*x^n)^2)/e^3
+ (2*(d + e*x^n)^3)/e^3 - (6*d^3*Log[d + e*x^n])/e^3)*Log[c*(d + e*x^n)^p])/(9*n) + (x*(f*x)^(-1 + 3*n)*Log[c*
(d + e*x^n)^p]^2)/(3*n)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 2456

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

Rubi steps

\begin {align*} \int (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx &=\left (x^{1-3 n} (f x)^{-1+3 n}\right ) \int x^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx\\ &=\frac {\left (x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int x^2 \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}-\frac {\left (2 e p x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \frac {x^3 \log \left (c (d+e x)^p\right )}{d+e x} \, dx,x,x^n\right )}{3 n}\\ &=\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}-\frac {\left (2 p x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^3 \log \left (c x^p\right )}{x} \, dx,x,d+e x^n\right )}{3 n}\\ &=-\frac {p x^{1-3 n} (f x)^{-1+3 n} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {\left (2 p^2 x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \frac {18 d^2 x-9 d x^2+2 x^3-6 d^3 \log (x)}{6 e^3 x} \, dx,x,d+e x^n\right )}{3 n}\\ &=-\frac {p x^{1-3 n} (f x)^{-1+3 n} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {\left (p^2 x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \frac {18 d^2 x-9 d x^2+2 x^3-6 d^3 \log (x)}{x} \, dx,x,d+e x^n\right )}{9 e^3 n}\\ &=-\frac {p x^{1-3 n} (f x)^{-1+3 n} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}+\frac {\left (p^2 x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \left (18 d^2-9 d x+2 x^2-\frac {6 d^3 \log (x)}{x}\right ) \, dx,x,d+e x^n\right )}{9 e^3 n}\\ &=\frac {2 d^2 p^2 x^{1-2 n} (f x)^{-1+3 n}}{e^2 n}-\frac {d p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^2}{2 e^3 n}+\frac {2 p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^3}{27 e^3 n}-\frac {p x^{1-3 n} (f x)^{-1+3 n} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}-\frac {\left (2 d^3 p^2 x^{1-3 n} (f x)^{-1+3 n}\right ) \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,d+e x^n\right )}{3 e^3 n}\\ &=\frac {2 d^2 p^2 x^{1-2 n} (f x)^{-1+3 n}}{e^2 n}-\frac {d p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^2}{2 e^3 n}+\frac {2 p^2 x^{1-3 n} (f x)^{-1+3 n} \left (d+e x^n\right )^3}{27 e^3 n}-\frac {d^3 p^2 x^{1-3 n} (f x)^{-1+3 n} \log ^2\left (d+e x^n\right )}{3 e^3 n}-\frac {p x^{1-3 n} (f x)^{-1+3 n} \left (\frac {18 d^2 \left (d+e x^n\right )}{e^3}-\frac {9 d \left (d+e x^n\right )^2}{e^3}+\frac {2 \left (d+e x^n\right )^3}{e^3}-\frac {6 d^3 \log \left (d+e x^n\right )}{e^3}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{9 n}+\frac {x (f x)^{-1+3 n} \log ^2\left (c \left (d+e x^n\right )^p\right )}{3 n}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 171, normalized size = 0.46 \[ \frac {x^{-3 n} (f x)^{3 n} \left (6 d^3 p \log \left (d+e x^n\right ) \left (6 \log \left (c \left (d+e x^n\right )^p\right )-11 p\right )+e x^n \left (-6 p \left (6 d^2-3 d e x^n+2 e^2 x^{2 n}\right ) \log \left (c \left (d+e x^n\right )^p\right )+18 e^2 x^{2 n} \log ^2\left (c \left (d+e x^n\right )^p\right )+p^2 \left (66 d^2-15 d e x^n+4 e^2 x^{2 n}\right )\right )-18 d^3 p^2 \log ^2\left (d+e x^n\right )\right )}{54 e^3 f n} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^(-1 + 3*n)*Log[c*(d + e*x^n)^p]^2,x]

[Out]

((f*x)^(3*n)*(-18*d^3*p^2*Log[d + e*x^n]^2 + 6*d^3*p*Log[d + e*x^n]*(-11*p + 6*Log[c*(d + e*x^n)^p]) + e*x^n*(
p^2*(66*d^2 - 15*d*e*x^n + 4*e^2*x^(2*n)) - 6*p*(6*d^2 - 3*d*e*x^n + 2*e^2*x^(2*n))*Log[c*(d + e*x^n)^p] + 18*
e^2*x^(2*n)*Log[c*(d + e*x^n)^p]^2)))/(54*e^3*f*n*x^(3*n))

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fricas [A]  time = 0.45, size = 266, normalized size = 0.72 \[ \frac {2 \, {\left (2 \, e^{3} p^{2} - 6 \, e^{3} p \log \relax (c) + 9 \, e^{3} \log \relax (c)^{2}\right )} f^{3 \, n - 1} x^{3 \, n} - 3 \, {\left (5 \, d e^{2} p^{2} - 6 \, d e^{2} p \log \relax (c)\right )} f^{3 \, n - 1} x^{2 \, n} + 6 \, {\left (11 \, d^{2} e p^{2} - 6 \, d^{2} e p \log \relax (c)\right )} f^{3 \, n - 1} x^{n} + 18 \, {\left (e^{3} f^{3 \, n - 1} p^{2} x^{3 \, n} + d^{3} f^{3 \, n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} + 6 \, {\left (3 \, d e^{2} f^{3 \, n - 1} p^{2} x^{2 \, n} - 6 \, d^{2} e f^{3 \, n - 1} p^{2} x^{n} - 2 \, {\left (e^{3} p^{2} - 3 \, e^{3} p \log \relax (c)\right )} f^{3 \, n - 1} x^{3 \, n} - {\left (11 \, d^{3} p^{2} - 6 \, d^{3} p \log \relax (c)\right )} f^{3 \, n - 1}\right )} \log \left (e x^{n} + d\right )}{54 \, e^{3} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="fricas")

[Out]

1/54*(2*(2*e^3*p^2 - 6*e^3*p*log(c) + 9*e^3*log(c)^2)*f^(3*n - 1)*x^(3*n) - 3*(5*d*e^2*p^2 - 6*d*e^2*p*log(c))
*f^(3*n - 1)*x^(2*n) + 6*(11*d^2*e*p^2 - 6*d^2*e*p*log(c))*f^(3*n - 1)*x^n + 18*(e^3*f^(3*n - 1)*p^2*x^(3*n) +
d^3*f^(3*n - 1)*p^2)*log(e*x^n + d)^2 + 6*(3*d*e^2*f^(3*n - 1)*p^2*x^(2*n) - 6*d^2*e*f^(3*n - 1)*p^2*x^n - 2*
(e^3*p^2 - 3*e^3*p*log(c))*f^(3*n - 1)*x^(3*n) - (11*d^3*p^2 - 6*d^3*p*log(c))*f^(3*n - 1))*log(e*x^n + d))/(e
^3*n)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (f x\right )^{3 \, n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.82, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{3 n -1} \ln \left (c \left (e \,x^{n}+d \right )^{p}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(3*n-1)*ln(c*(e*x^n+d)^p)^2,x)

[Out]

int((f*x)^(3*n-1)*ln(c*(e*x^n+d)^p)^2,x)

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maxima [A]  time = 0.56, size = 239, normalized size = 0.64 \[ \frac {e p {\left (\frac {6 \, d^{3} f^{3 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{4} n} - \frac {2 \, e^{2} f^{3 \, n} x^{3 \, n} - 3 \, d e f^{3 \, n} x^{2 \, n} + 6 \, d^{2} f^{3 \, n} x^{n}}{e^{3} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{9 \, f} + \frac {\left (f x\right )^{3 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{3 \, f n} - \frac {{\left (18 \, d^{3} f^{3 \, n} \log \left (e x^{n} + d\right )^{2} - 4 \, e^{3} f^{3 \, n} x^{3 \, n} + 15 \, d e^{2} f^{3 \, n} x^{2 \, n} - 66 \, d^{2} e f^{3 \, n} x^{n} - 6 \, {\left (6 \, f^{3 \, n} \log \relax (e) - 11 \, f^{3 \, n}\right )} d^{3} \log \left (e x^{n} + d\right )\right )} p^{2}}{54 \, e^{3} f n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+3*n)*log(c*(d+e*x^n)^p)^2,x, algorithm="maxima")

[Out]

1/9*e*p*(6*d^3*f^(3*n)*log((e*x^n + d)/e)/(e^4*n) - (2*e^2*f^(3*n)*x^(3*n) - 3*d*e*f^(3*n)*x^(2*n) + 6*d^2*f^(
3*n)*x^n)/(e^3*n))*log((e*x^n + d)^p*c)/f + 1/3*(f*x)^(3*n)*log((e*x^n + d)^p*c)^2/(f*n) - 1/54*(18*d^3*f^(3*n
)*log(e*x^n + d)^2 - 4*e^3*f^(3*n)*x^(3*n) + 15*d*e^2*f^(3*n)*x^(2*n) - 66*d^2*e*f^(3*n)*x^n - 6*(6*f^(3*n)*lo
g(e) - 11*f^(3*n))*d^3*log(e*x^n + d))*p^2/(e^3*f*n)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\ln \left (c\,{\left (d+e\,x^n\right )}^p\right )}^2\,{\left (f\,x\right )}^{3\,n-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(d + e*x^n)^p)^2*(f*x)^(3*n - 1),x)

[Out]

int(log(c*(d + e*x^n)^p)^2*(f*x)^(3*n - 1), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**(-1+3*n)*ln(c*(d+e*x**n)**p)**2,x)

[Out]

Timed out

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