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

Optimal. Leaf size=23 \[ -\frac {\log \left (c \left (b x^n\right )^p\right )}{x}-\frac {n p}{x} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2304, 2445} \[ -\frac {\log \left (c \left (b x^n\right )^p\right )}{x}-\frac {n p}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((n*p)/x) - Log[c*(b*x^n)^p]/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 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 \left (c \left (b x^n\right )^p\right )}{x^2} \, dx &=\operatorname {Subst}\left (\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 {n p}{x}-\frac {\log \left (c \left (b x^n\right )^p\right )}{x}\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.00 \[ -\frac {\log \left (c \left (b x^n\right )^p\right )}{x}-\frac {n p}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((n*p)/x) - Log[c*(b*x^n)^p]/x

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fricas [A]  time = 0.66, size = 20, normalized size = 0.87 \[ -\frac {n p \log \relax (x) + n p + p \log \relax (b) + \log \relax (c)}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 25, normalized size = 1.09 \[ -\frac {n p \log \relax (x)}{x} - \frac {n p + p \log \relax (b) + \log \relax (c)}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.05, size = 23, normalized size = 1.00 \[ -\frac {n p}{x} - \frac {\log \left (\left (b x^{n}\right )^{p} c\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-n*p/x - log((b*x^n)^p*c)/x

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mupad [B]  time = 3.88, size = 19, normalized size = 0.83 \[ -\frac {\ln \left (c\,{\left (b\,x^n\right )}^p\right )+n\,p}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(c*(b*x^n)^p) + n*p)/x

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sympy [A]  time = 1.00, size = 26, normalized size = 1.13 \[ - \frac {n p \log {\relax (x )}}{x} - \frac {n p}{x} - \frac {p \log {\relax (b )}}{x} - \frac {\log {\relax (c )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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