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

Optimal. Leaf size=255 \[ \frac {x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {d x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}-\frac {p x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}+\frac {2 d p x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {p^2 x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2}{4 e^2 n}-\frac {2 d p^2 x^{1-n} (f x)^{2 n-1}}{e n} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2456, 2454, 2401, 2389, 2296, 2295, 2390, 2305, 2304} \[ \frac {x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {d x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}-\frac {p x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}+\frac {2 d p x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {p^2 x^{1-2 n} (f x)^{2 n-1} \left (d+e x^n\right )^2}{4 e^2 n}-\frac {2 d p^2 x^{1-n} (f x)^{2 n-1}}{e n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 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 2390

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

Rule 2401

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Int[Exp
andIntegrand[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[
e*f - d*g, 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 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+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx &=\left (x^{1-2 n} (f x)^{-1+2 n}\right ) \int x^{-1+2 n} \log ^2\left (c \left (d+e x^n\right )^p\right ) \, dx\\ &=\frac {\left (x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int x \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {\left (x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int \left (-\frac {d \log ^2\left (c (d+e x)^p\right )}{e}+\frac {(d+e x) \log ^2\left (c (d+e x)^p\right )}{e}\right ) \, dx,x,x^n\right )}{n}\\ &=\frac {\left (x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int (d+e x) \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{e n}-\frac {\left (d x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int \log ^2\left (c (d+e x)^p\right ) \, dx,x,x^n\right )}{e n}\\ &=\frac {\left (x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int x \log ^2\left (c x^p\right ) \, dx,x,d+e x^n\right )}{e^2 n}-\frac {\left (d x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int \log ^2\left (c x^p\right ) \, dx,x,d+e x^n\right )}{e^2 n}\\ &=-\frac {d x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {\left (p x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int x \log \left (c x^p\right ) \, dx,x,d+e x^n\right )}{e^2 n}+\frac {\left (2 d p x^{1-2 n} (f x)^{-1+2 n}\right ) \operatorname {Subst}\left (\int \log \left (c x^p\right ) \, dx,x,d+e x^n\right )}{e^2 n}\\ &=-\frac {2 d p^2 x^{1-n} (f x)^{-1+2 n}}{e n}+\frac {p^2 x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2}{4 e^2 n}+\frac {2 d p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )}{e^2 n}-\frac {p x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log \left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}-\frac {d x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right ) \log ^2\left (c \left (d+e x^n\right )^p\right )}{e^2 n}+\frac {x^{1-2 n} (f x)^{-1+2 n} \left (d+e x^n\right )^2 \log ^2\left (c \left (d+e x^n\right )^p\right )}{2 e^2 n}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 140, normalized size = 0.55 \[ \frac {x^{-2 n} (f x)^{2 n} \left (2 d^2 p \log \left (d+e x^n\right ) \left (3 p-2 \log \left (c \left (d+e x^n\right )^p\right )\right )+e x^n \left (2 e x^n \log ^2\left (c \left (d+e x^n\right )^p\right )+2 p \left (2 d-e x^n\right ) \log \left (c \left (d+e x^n\right )^p\right )+p^2 \left (e x^n-6 d\right )\right )+2 d^2 p^2 \log ^2\left (d+e x^n\right )\right )}{4 e^2 f n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 204, normalized size = 0.80 \[ \frac {{\left (e^{2} p^{2} - 2 \, e^{2} p \log \relax (c) + 2 \, e^{2} \log \relax (c)^{2}\right )} f^{2 \, n - 1} x^{2 \, n} - 2 \, {\left (3 \, d e p^{2} - 2 \, d e p \log \relax (c)\right )} f^{2 \, n - 1} x^{n} + 2 \, {\left (e^{2} f^{2 \, n - 1} p^{2} x^{2 \, n} - d^{2} f^{2 \, n - 1} p^{2}\right )} \log \left (e x^{n} + d\right )^{2} + 2 \, {\left (2 \, d e f^{2 \, n - 1} p^{2} x^{n} - {\left (e^{2} p^{2} - 2 \, e^{2} p \log \relax (c)\right )} f^{2 \, n - 1} x^{2 \, n} + {\left (3 \, d^{2} p^{2} - 2 \, d^{2} p \log \relax (c)\right )} f^{2 \, n - 1}\right )} \log \left (e x^{n} + d\right )}{4 \, e^{2} n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.72, size = 0, normalized size = 0.00 \[ \int \left (f x \right )^{2 n -1} \ln \left (c \left (e \,x^{n}+d \right )^{p}\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.57, size = 200, normalized size = 0.78 \[ -\frac {e p {\left (\frac {2 \, d^{2} f^{2 \, n} \log \left (\frac {e x^{n} + d}{e}\right )}{e^{3} n} + \frac {e f^{2 \, n} x^{2 \, n} - 2 \, d f^{2 \, n} x^{n}}{e^{2} n}\right )} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )}{2 \, f} + \frac {\left (f x\right )^{2 \, n} \log \left ({\left (e x^{n} + d\right )}^{p} c\right )^{2}}{2 \, f n} + \frac {{\left (2 \, d^{2} f^{2 \, n} \log \left (e x^{n} + d\right )^{2} + e^{2} f^{2 \, n} x^{2 \, n} - 6 \, d e f^{2 \, n} x^{n} - 2 \, {\left (2 \, f^{2 \, n} \log \relax (e) - 3 \, f^{2 \, n}\right )} d^{2} \log \left (e x^{n} + d\right )\right )} p^{2}}{4 \, e^{2} f n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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