3.2.67 \(\int (f x)^{-1-n} \log ^2(c (d+e x^n)^p) \, dx\) [167]

Optimal. Leaf size=124 \[ \frac {2 e p x^{1+n} (f x)^{-1-n} \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{d n}-\frac {x (f x)^{-1-n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{d n}+\frac {2 e p^2 x^{1+n} (f x)^{-1-n} \text {Li}_2\left (1+\frac {e x^n}{d}\right )}{d n} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {2506, 2504, 2444, 2441, 2352} \begin {gather*} \frac {2 e p^2 x^{n+1} (f x)^{-n-1} \text {PolyLog}\left (2,\frac {e x^n}{d}+1\right )}{d n}-\frac {x (f x)^{-n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{d n}+\frac {2 e p x^{n+1} (f x)^{-n-1} \log \left (-\frac {e x^n}{d}\right ) \log \left (c \left (d+e x^n\right )^p\right )}{d n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*p*x^(1 + n)*(f*x)^(-1 - n)*Log[-((e*x^n)/d)]*Log[c*(d + e*x^n)^p])/(d*n) - (x*(f*x)^(-1 - n)*(d + e*x^n)*
Log[c*(d + e*x^n)^p]^2)/(d*n) + (2*e*p^2*x^(1 + n)*(f*x)^(-1 - n)*PolyLog[2, 1 + (e*x^n)/d])/(d*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 2444

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

Rule 2504

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 2506

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

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Mathematica [A]
time = 0.07, size = 148, normalized size = 1.19 \begin {gather*} -\frac {(f x)^{-n} \left (2 e p^2 x^n \log \left (-\frac {d x^{-n}}{e}\right ) \log \left (-e-d x^{-n}\right )-e p^2 x^n \log ^2\left (-e-d x^{-n}\right )+2 e p x^n \log \left (-e-d x^{-n}\right ) \log \left (c \left (d+e x^n\right )^p\right )+d \log ^2\left (c \left (d+e x^n\right )^p\right )+2 e p^2 x^n \text {Li}_2\left (1+\frac {d x^{-n}}{e}\right )\right )}{d f n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \left (f x \right )^{-1-n} \ln \left (c \left (d +e \,x^{n}\right )^{p}\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume((-n)-1>0)', see `assume?` for
more details

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Fricas [A]
time = 0.37, size = 207, normalized size = 1.67 \begin {gather*} -\frac {2 \, f^{-n - 1} n p^{2} x^{n} e \log \left (x\right ) \log \left (\frac {x^{n} e + d}{d}\right ) - 2 \, f^{-n - 1} n p x^{n} e \log \left (c\right ) \log \left (x\right ) + 2 \, f^{-n - 1} p^{2} x^{n} {\rm Li}_2\left (-\frac {x^{n} e + d}{d} + 1\right ) e + d f^{-n - 1} \log \left (c\right )^{2} + {\left (f^{-n - 1} p^{2} x^{n} e + d f^{-n - 1} p^{2}\right )} \log \left (x^{n} e + d\right )^{2} + 2 \, {\left (d f^{-n - 1} p \log \left (c\right ) - {\left (n p^{2} e \log \left (x\right ) - p e \log \left (c\right )\right )} f^{-n - 1} x^{n}\right )} \log \left (x^{n} e + d\right )}{d n x^{n}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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