3.234 \(\int \frac {\log ^2(c (b x^n)^p)}{x^2} \, dx\)

Optimal. Leaf size=46 \[ -\frac {\log ^2\left (c \left (b x^n\right )^p\right )}{x}-\frac {2 n p \log \left (c \left (b x^n\right )^p\right )}{x}-\frac {2 n^2 p^2}{x} \]

[Out]

-2*n^2*p^2/x-2*n*p*ln(c*(b*x^n)^p)/x-ln(c*(b*x^n)^p)^2/x

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Rubi [A]  time = 0.07, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2305, 2304, 2445} \[ -\frac {\log ^2\left (c \left (b x^n\right )^p\right )}{x}-\frac {2 n p \log \left (c \left (b x^n\right )^p\right )}{x}-\frac {2 n^2 p^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(b*x^n)^p]^2/x^2,x]

[Out]

(-2*n^2*p^2)/x - (2*n*p*Log[c*(b*x^n)^p])/x - Log[c*(b*x^n)^p]^2/x

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 2445

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 40, normalized size = 0.87 \[ -\frac {\log ^2\left (c \left (b x^n\right )^p\right )+2 n p \log \left (c \left (b x^n\right )^p\right )+2 n^2 p^2}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(b*x^n)^p]^2/x^2,x]

[Out]

-((2*n^2*p^2 + 2*n*p*Log[c*(b*x^n)^p] + Log[c*(b*x^n)^p]^2)/x)

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fricas [A]  time = 0.64, size = 81, normalized size = 1.76 \[ -\frac {n^{2} p^{2} \log \relax (x)^{2} + 2 \, n^{2} p^{2} + 2 \, n p^{2} \log \relax (b) + p^{2} \log \relax (b)^{2} + 2 \, {\left (n p + p \log \relax (b)\right )} \log \relax (c) + \log \relax (c)^{2} + 2 \, {\left (n^{2} p^{2} + n p^{2} \log \relax (b) + n p \log \relax (c)\right )} \log \relax (x)}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^n)^p)^2/x^2,x, algorithm="fricas")

[Out]

-(n^2*p^2*log(x)^2 + 2*n^2*p^2 + 2*n*p^2*log(b) + p^2*log(b)^2 + 2*(n*p + p*log(b))*log(c) + log(c)^2 + 2*(n^2
*p^2 + n*p^2*log(b) + n*p*log(c))*log(x))/x

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giac [A]  time = 0.29, size = 90, normalized size = 1.96 \[ -\frac {n^{2} p^{2} \log \relax (x)^{2}}{x} - \frac {2 \, {\left (n^{2} p^{2} + n p^{2} \log \relax (b) + n p \log \relax (c)\right )} \log \relax (x)}{x} - \frac {2 \, n^{2} p^{2} + 2 \, n p^{2} \log \relax (b) + p^{2} \log \relax (b)^{2} + 2 \, n p \log \relax (c) + 2 \, p \log \relax (b) \log \relax (c) + \log \relax (c)^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^n)^p)^2/x^2,x, algorithm="giac")

[Out]

-n^2*p^2*log(x)^2/x - 2*(n^2*p^2 + n*p^2*log(b) + n*p*log(c))*log(x)/x - (2*n^2*p^2 + 2*n*p^2*log(b) + p^2*log
(b)^2 + 2*n*p*log(c) + 2*p*log(b)*log(c) + log(c)^2)/x

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (b \,x^{n}\right )^{p}\right )^{2}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^n)^p)^2/x^2,x)

[Out]

int(ln(c*(b*x^n)^p)^2/x^2,x)

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maxima [A]  time = 1.14, size = 46, normalized size = 1.00 \[ -\frac {2 \, n^{2} p^{2}}{x} - \frac {2 \, n p \log \left (\left (b x^{n}\right )^{p} c\right )}{x} - \frac {\log \left (\left (b x^{n}\right )^{p} c\right )^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^n)^p)^2/x^2,x, algorithm="maxima")

[Out]

-2*n^2*p^2/x - 2*n*p*log((b*x^n)^p*c)/x - log((b*x^n)^p*c)^2/x

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mupad [B]  time = 3.85, size = 40, normalized size = 0.87 \[ -\frac {2\,n^2\,p^2+2\,n\,p\,\ln \left (c\,{\left (b\,x^n\right )}^p\right )+{\ln \left (c\,{\left (b\,x^n\right )}^p\right )}^2}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(b*x^n)^p)^2/x^2,x)

[Out]

-(log(c*(b*x^n)^p)^2 + 2*n^2*p^2 + 2*n*p*log(c*(b*x^n)^p))/x

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sympy [B]  time = 1.04, size = 117, normalized size = 2.54 \[ - \frac {n^{2} p^{2} \log {\relax (x )}^{2}}{x} - \frac {2 n^{2} p^{2} \log {\relax (x )}}{x} - \frac {2 n^{2} p^{2}}{x} - \frac {2 n p^{2} \log {\relax (b )} \log {\relax (x )}}{x} - \frac {2 n p^{2} \log {\relax (b )}}{x} - \frac {2 n p \log {\relax (c )} \log {\relax (x )}}{x} - \frac {2 n p \log {\relax (c )}}{x} - \frac {p^{2} \log {\relax (b )}^{2}}{x} - \frac {2 p \log {\relax (b )} \log {\relax (c )}}{x} - \frac {\log {\relax (c )}^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**n)**p)**2/x**2,x)

[Out]

-n**2*p**2*log(x)**2/x - 2*n**2*p**2*log(x)/x - 2*n**2*p**2/x - 2*n*p**2*log(b)*log(x)/x - 2*n*p**2*log(b)/x -
2*n*p*log(c)*log(x)/x - 2*n*p*log(c)/x - p**2*log(b)**2/x - 2*p*log(b)*log(c)/x - log(c)**2/x

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