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

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

[Out]

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

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Rubi [A]  time = 0.05, 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} \[ \frac {1}{2} p \text {PolyLog}\left (2,\frac {b x^2}{a}+1\right )+\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x,x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 0.98 \[ \frac {1}{2} \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \text {Li}_2\left (\frac {b x^2+a}{a}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x,x]

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x^2 + a)^p*c)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((b*x^2 + a)^p*c)/x, x)

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maple [C]  time = 0.25, size = 232, normalized size = 5.27 \[ -\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \ln \relax (x )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2} \ln \relax (x )}{2}-\frac {i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3} \ln \relax (x )}{2}-p \ln \relax (x ) \ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \ln \relax (x ) \ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \dilog \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \dilog \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \relax (c ) \ln \relax (x )+\ln \relax (x ) \ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*ln((b*x^2+a)^p)-p*ln(x)*ln((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-p*ln(x)*ln((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))
-p*dilog((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))-p*dilog((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+1/2*I*ln(x)*Pi*csgn(I*(b*x
^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-1/2*I*ln(x)*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-1/2*I*ln(x
)*Pi*csgn(I*c*(b*x^2+a)^p)^3+1/2*I*ln(x)*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+ln(c)*ln(x)

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maxima [B]  time = 0.70, size = 80, normalized size = 1.82 \[ \frac {1}{2} \, b p {\left (\frac {2 \, \log \left (b x^{2} + a\right ) \log \relax (x)}{b} - \frac {2 \, \log \left (\frac {b x^{2}}{a} + 1\right ) \log \relax (x) + {\rm Li}_2\left (-\frac {b x^{2}}{a}\right )}{b}\right )} - p \log \left (b x^{2} + a\right ) \log \relax (x) + \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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 \[ \int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b*x**2)**p)/x, x)

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