3.94 \(\int \frac {\log ^3(c (a+b x^2)^p)}{x} \, dx\)

Optimal. Leaf size=106 \[ -3 p^2 \text {Li}_3\left (\frac {b x^2}{a}+1\right ) \log \left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} p \text {Li}_2\left (\frac {b x^2}{a}+1\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )+3 p^3 \text {Li}_4\left (\frac {b x^2}{a}+1\right ) \]

[Out]

1/2*ln(-b*x^2/a)*ln(c*(b*x^2+a)^p)^3+3/2*p*ln(c*(b*x^2+a)^p)^2*polylog(2,1+b*x^2/a)-3*p^2*ln(c*(b*x^2+a)^p)*po
lylog(3,1+b*x^2/a)+3*p^3*polylog(4,1+b*x^2/a)

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Rubi [A]  time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2454, 2396, 2433, 2374, 2383, 6589} \[ -3 p^2 \text {PolyLog}\left (3,\frac {b x^2}{a}+1\right ) \log \left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} p \text {PolyLog}\left (2,\frac {b x^2}{a}+1\right ) \log ^2\left (c \left (a+b x^2\right )^p\right )+3 p^3 \text {PolyLog}\left (4,\frac {b x^2}{a}+1\right )+\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]^3/x,x]

[Out]

(Log[-((b*x^2)/a)]*Log[c*(a + b*x^2)^p]^3)/2 + (3*p*Log[c*(a + b*x^2)^p]^2*PolyLog[2, 1 + (b*x^2)/a])/2 - 3*p^
2*Log[c*(a + b*x^2)^p]*PolyLog[3, 1 + (b*x^2)/a] + 3*p^3*PolyLog[4, 1 + (b*x^2)/a]

Rule 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[k + 1, e*x^q]*(a + b*Log[c*x^n])^p)/q, x] - Dist[(b*n*p)/q, Int[(PolyLog[k + 1, e*x^q]*(a + b*Log[c*x^n])^(
p - 1))/x, x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

Rule 2396

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

Rule 2433

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*((k_.) + (l_.)*(x_))^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((k*x)/d)^r*(a + b*Log[c*x^n])^p*(f + g*Lo
g[h*((e*i - d*j)/e + (j*x)/e)^m]), x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, j, k, l, n, p, r},
x] && EqQ[e*k - d*l, 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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\log ^3\left (c \left (a+b x^2\right )^p\right )}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\log ^3\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 ^3\left (c \left (a+b x^2\right )^p\right )-\frac {1}{2} (3 b p) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right ) \log ^2\left (c (a+b x)^p\right )}{a+b x} \, dx,x,x^2\right )\\ &=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )-\frac {1}{2} (3 p) \operatorname {Subst}\left (\int \frac {\log ^2\left (c x^p\right ) \log \left (-\frac {b \left (-\frac {a}{b}+\frac {x}{b}\right )}{a}\right )}{x} \, dx,x,a+b x^2\right )\\ &=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} p \log ^2\left (c \left (a+b x^2\right )^p\right ) \text {Li}_2\left (1+\frac {b x^2}{a}\right )-\left (3 p^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (c x^p\right ) \text {Li}_2\left (\frac {x}{a}\right )}{x} \, dx,x,a+b x^2\right )\\ &=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} p \log ^2\left (c \left (a+b x^2\right )^p\right ) \text {Li}_2\left (1+\frac {b x^2}{a}\right )-3 p^2 \log \left (c \left (a+b x^2\right )^p\right ) \text {Li}_3\left (1+\frac {b x^2}{a}\right )+\left (3 p^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{a}\right )}{x} \, dx,x,a+b x^2\right )\\ &=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log ^3\left (c \left (a+b x^2\right )^p\right )+\frac {3}{2} p \log ^2\left (c \left (a+b x^2\right )^p\right ) \text {Li}_2\left (1+\frac {b x^2}{a}\right )-3 p^2 \log \left (c \left (a+b x^2\right )^p\right ) \text {Li}_3\left (1+\frac {b x^2}{a}\right )+3 p^3 \text {Li}_4\left (1+\frac {b x^2}{a}\right )\\ \end {align*}

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Mathematica [B]  time = 0.10, size = 279, normalized size = 2.63 \[ -\frac {3}{2} p^2 \left (-2 \text {Li}_3\left (\frac {b x^2}{a}+1\right )+2 \text {Li}_2\left (\frac {b x^2}{a}+1\right ) \log \left (a+b x^2\right )+\log \left (-\frac {b x^2}{a}\right ) \log ^2\left (a+b x^2\right )\right ) \left (p \log \left (a+b x^2\right )-\log \left (c \left (a+b x^2\right )^p\right )\right )+3 p \left (\log (x) \left (\log \left (a+b x^2\right )-\log \left (\frac {b x^2}{a}+1\right )\right )-\frac {1}{2} \text {Li}_2\left (-\frac {b x^2}{a}\right )\right ) \left (\log \left (c \left (a+b x^2\right )^p\right )-p \log \left (a+b x^2\right )\right )^2+\log (x) \left (\log \left (c \left (a+b x^2\right )^p\right )-p \log \left (a+b x^2\right )\right )^3+\frac {1}{2} p^3 \left (6 \text {Li}_4\left (\frac {b x^2}{a}+1\right )+3 \text {Li}_2\left (\frac {b x^2}{a}+1\right ) \log ^2\left (a+b x^2\right )-6 \text {Li}_3\left (\frac {b x^2}{a}+1\right ) \log \left (a+b x^2\right )+\log \left (-\frac {b x^2}{a}\right ) \log ^3\left (a+b x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]^3/x,x]

[Out]

Log[x]*(-(p*Log[a + b*x^2]) + Log[c*(a + b*x^2)^p])^3 + 3*p*(-(p*Log[a + b*x^2]) + Log[c*(a + b*x^2)^p])^2*(Lo
g[x]*(Log[a + b*x^2] - Log[1 + (b*x^2)/a]) - PolyLog[2, -((b*x^2)/a)]/2) - (3*p^2*(p*Log[a + b*x^2] - Log[c*(a
+ b*x^2)^p])*(Log[-((b*x^2)/a)]*Log[a + b*x^2]^2 + 2*Log[a + b*x^2]*PolyLog[2, 1 + (b*x^2)/a] - 2*PolyLog[3,
1 + (b*x^2)/a]))/2 + (p^3*(Log[-((b*x^2)/a)]*Log[a + b*x^2]^3 + 3*Log[a + b*x^2]^2*PolyLog[2, 1 + (b*x^2)/a] -
6*Log[a + b*x^2]*PolyLog[3, 1 + (b*x^2)/a] + 6*PolyLog[4, 1 + (b*x^2)/a]))/2

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^3/x,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)^3/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )^{3}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^3/x,x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)^3/x, x)

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maple [F]  time = 0.66, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)^3/x,x)

[Out]

int(ln(c*(b*x^2+a)^p)^3/x,x)

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maxima [B]  time = 0.65, size = 217, normalized size = 2.05 \[ \frac {1}{2} \, {\left (\log \left (b x^{2} + a\right )^{3} \log \left (-\frac {b x^{2} + a}{a} + 1\right ) + 3 \, {\rm Li}_2\left (\frac {b x^{2} + a}{a}\right ) \log \left (b x^{2} + a\right )^{2} - 6 \, \log \left (b x^{2} + a\right ) {\rm Li}_{3}(\frac {b x^{2} + a}{a}) + 6 \, {\rm Li}_{4}(\frac {b x^{2} + a}{a})\right )} p^{3} + \frac {3}{2} \, {\left (\log \left (b x^{2} + a\right )^{2} \log \left (-\frac {b x^{2} + a}{a} + 1\right ) + 2 \, {\rm Li}_2\left (\frac {b x^{2} + a}{a}\right ) \log \left (b x^{2} + a\right ) - 2 \, {\rm Li}_{3}(\frac {b x^{2} + a}{a})\right )} p^{2} \log \relax (c) + \frac {3}{2} \, {\left (\log \left (b x^{2} + a\right ) \log \left (-\frac {b x^{2} + a}{a} + 1\right ) + {\rm Li}_2\left (\frac {b x^{2} + a}{a}\right )\right )} p \log \relax (c)^{2} + \log \relax (c)^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)^3/x,x, algorithm="maxima")

[Out]

1/2*(log(b*x^2 + a)^3*log(-(b*x^2 + a)/a + 1) + 3*dilog((b*x^2 + a)/a)*log(b*x^2 + a)^2 - 6*log(b*x^2 + a)*pol
ylog(3, (b*x^2 + a)/a) + 6*polylog(4, (b*x^2 + a)/a))*p^3 + 3/2*(log(b*x^2 + a)^2*log(-(b*x^2 + a)/a + 1) + 2*
dilog((b*x^2 + a)/a)*log(b*x^2 + a) - 2*polylog(3, (b*x^2 + a)/a))*p^2*log(c) + 3/2*(log(b*x^2 + a)*log(-(b*x^
2 + a)/a + 1) + dilog((b*x^2 + a)/a))*p*log(c)^2 + log(c)^3*log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^2)^p)^3/x,x)

[Out]

int(log(c*(a + b*x^2)^p)^3/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)**3/x,x)

[Out]

Integral(log(c*(a + b*x**2)**p)**3/x, x)

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