3.41 \(\int \frac {\log (c (a+\frac {b}{x^2})^p)}{x} \, dx\)

Optimal. Leaf size=44 \[ -\frac {1}{2} \log \left (-\frac {b}{a x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-\frac {1}{2} p \text {Li}_2\left (\frac {b}{a x^2}+1\right ) \]

[Out]

-1/2*ln(c*(a+b/x^2)^p)*ln(-b/a/x^2)-1/2*p*polylog(2,1+b/a/x^2)

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Rubi [A]  time = 0.04, antiderivative size = 44, 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 = {2454, 2394, 2315} \[ -\frac {1}{2} p \text {PolyLog}\left (2,\frac {b}{a x^2}+1\right )-\frac {1}{2} \log \left (-\frac {b}{a x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x^2)^p]/x,x]

[Out]

-(Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))])/2 - (p*PolyLog[2, 1 + b/(a*x^2)])/2

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2394

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 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])

Rubi steps

\begin {align*} \int \frac {\log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{x} \, dx &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,\frac {1}{x^2}\right )\right )\\ &=-\frac {1}{2} \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )+\frac {1}{2} (b p) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {1}{2} \log \left (c \left (a+\frac {b}{x^2}\right )^p\right ) \log \left (-\frac {b}{a x^2}\right )-\frac {1}{2} p \text {Li}_2\left (1+\frac {b}{a x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 45, normalized size = 1.02 \[ -\frac {1}{2} \log \left (-\frac {b}{a x^2}\right ) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )-\frac {1}{2} p \text {Li}_2\left (\frac {a+\frac {b}{x^2}}{a}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x^2)^p]/x,x]

[Out]

-1/2*(Log[c*(a + b/x^2)^p]*Log[-(b/(a*x^2))]) - (p*PolyLog[2, (a + b/x^2)/a])/2

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (c \left (\frac {a x^{2} + b}{x^{2}}\right )^{p}\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(c*((a*x^2 + b)/x^2)^p)/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((a + b/x^2)^p*c)/x, x)

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.63, size = 89, normalized size = 2.02 \[ \frac {1}{2} \, b p {\left (\frac {2 \, \log \left (a + \frac {b}{x^{2}}\right ) \log \relax (x)}{b} + \frac {2 \, \log \relax (x)^{2}}{b} - \frac {2 \, \log \left (\frac {a x^{2}}{b} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {a x^{2}}{b}\right )}{b}\right )} - p \log \left (a + \frac {b}{x^{2}}\right ) \log \relax (x) + \log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*p*(2*log(a + b/x^2)*log(x)/b + 2*log(x)^2/b - (2*log(a*x^2/b + 1)*log(x) + dilog(-a*x^2/b))/b) - p*log(a
+ b/x^2)*log(x) + log((a + b/x^2)^p*c)*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x^2)^p)/x,x)

[Out]

int(log(c*(a + b/x^2)^p)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b/x**2)**p)/x, x)

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