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

Optimal. Leaf size=44 \[ \frac{1}{3} p \text{PolyLog}\left (2,\frac{b x^3}{a}+1\right )+\frac{1}{3} \log \left (-\frac{b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right ) \]

[Out]

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

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Rubi [A]  time = 0.0477984, antiderivative size = 44, normalized size of antiderivative = 1., 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} \[ \frac{1}{3} p \text{PolyLog}\left (2,\frac{b x^3}{a}+1\right )+\frac{1}{3} \log \left (-\frac{b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x,x]

[Out]

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

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0068243, size = 43, normalized size = 0.98 \[ \frac{1}{3} \left (p \text{PolyLog}\left (2,\frac{a+b x^3}{a}\right )+\log \left (-\frac{b x^3}{a}\right ) \log \left (c \left (a+b x^3\right )^p\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^3)^p]/x,x]

[Out]

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

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Maple [C]  time = 0.431, size = 180, normalized size = 4.1 \begin{align*} \ln \left ( x \right ) \ln \left ( \left ( b{x}^{3}+a \right ) ^{p} \right ) -p\sum _{{\it \_R1}={\it RootOf} \left ( b{{\it \_Z}}^{3}+a \right ) }\ln \left ( x \right ) \ln \left ({\frac{{\it \_R1}-x}{{\it \_R1}}} \right ) +{\it dilog} \left ({\frac{{\it \_R1}-x}{{\it \_R1}}} \right ) +{\frac{i}{2}}\ln \left ( x \right ) \pi \,{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}-{\frac{i}{2}}\ln \left ( x \right ) \pi \,{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{2}}\ln \left ( x \right ) \pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{3}+{\frac{i}{2}}\ln \left ( x \right ) \pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +\ln \left ( c \right ) \ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)*ln((b*x^3+a)^p)-p*sum(ln(x)*ln((_R1-x)/_R1)+dilog((_R1-x)/_R1),_R1=RootOf(_Z^3*b+a))+1/2*I*ln(x)*Pi*csgn
(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-1/2*I*ln(x)*Pi*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)-1/2
*I*ln(x)*Pi*csgn(I*c*(b*x^3+a)^p)^3+1/2*I*ln(x)*Pi*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)+ln(c)*ln(x)

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Maxima [B]  time = 1.07101, size = 108, normalized size = 2.45 \begin{align*} \frac{1}{3} \, b p{\left (\frac{3 \, \log \left (b x^{3} + a\right ) \log \left (x\right )}{b} - \frac{3 \, \log \left (\frac{b x^{3}}{a} + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-\frac{b x^{3}}{a}\right )}{b}\right )} - p \log \left (b x^{3} + a\right ) \log \left (x\right ) + \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*p*(3*log(b*x^3 + a)*log(x)/b - (3*log(b*x^3/a + 1)*log(x) + dilog(-b*x^3/a))/b) - p*log(b*x^3 + a)*log(x
) + log((b*x^3 + a)^p*c)*log(x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log((b*x^3 + a)^p*c)/x, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((b*x^3 + a)^p*c)/x, x)