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

Optimal. Leaf size=101 \[ \frac{x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n}-\frac{2 p x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac{2 p^2 x (f x)^{n-1}}{n} \]

[Out]

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

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Rubi [A]  time = 0.0810466, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {2456, 2454, 2389, 2296, 2295} \[ \frac{x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e n}-\frac{2 p x^{1-n} (f x)^{n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e n}+\frac{2 p^2 x (f x)^{n-1}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*p]

Rule 2295

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

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

Mathematica [A]  time = 0.0245263, size = 74, normalized size = 0.73 \[ \frac{x^{-n} (f x)^n \left (\left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )-2 p \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+2 e p^2 x^n\right )}{e f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 2.097, size = 0, normalized size = 0. \begin{align*} \int \left ( fx \right ) ^{-1+n} \left ( \ln \left ( c \left ( d+e{x}^{n} \right ) ^{p} \right ) \right ) ^{2}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 2.09927, size = 282, normalized size = 2.79 \begin{align*} \frac{{\left (2 \, e p^{2} - 2 \, e p \log \left (c\right ) + e \log \left (c\right )^{2}\right )} f^{n - 1} x^{n} +{\left (e f^{n - 1} p^{2} x^{n} + d f^{n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} - 2 \,{\left ({\left (e p^{2} - e p \log \left (c\right )\right )} f^{n - 1} x^{n} +{\left (d p^{2} - d p \log \left (c\right )\right )} f^{n - 1}\right )} \log \left (e x^{n} + d\right )}{e n} \end{align*}

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (f x\right )^{n - 1} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}\,{d x} \end{align*}

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((e*x^n + d)^p*c)^2, x)