3.467 \(\int \frac{a+b \log (c (d+e x^{2/3})^n)}{x} \, dx\)

Optimal. Leaf size=55 \[ \frac{3}{2} b n \text{PolyLog}\left (2,\frac{e x^{2/3}}{d}+1\right )+\frac{3}{2} \log \left (-\frac{e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

[Out]

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

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Rubi [A]  time = 0.0507591, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2454, 2394, 2315} \[ \frac{3}{2} b n \text{PolyLog}\left (2,\frac{e x^{2/3}}{d}+1\right )+\frac{3}{2} \log \left (-\frac{e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x^(2/3))^n])/x,x]

[Out]

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

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

Mathematica [A]  time = 0.011762, size = 55, normalized size = 1. \[ \frac{3}{2} b \left (n \text{PolyLog}\left (2,\frac{d+e x^{2/3}}{d}\right )+\log \left (-\frac{e x^{2/3}}{d}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+a \log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])/x,x]

[Out]

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

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Maple [F]  time = 0.384, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{x} \left ( a+b\ln \left ( c \left ( d+e{x}^{{\frac{2}{3}}} \right ) ^{n} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d+e*x^(2/3))^n))/x,x)

[Out]

int((a+b*ln(c*(d+e*x^(2/3))^n))/x,x)

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Maxima [B]  time = 1.64034, size = 154, normalized size = 2.8 \begin{align*} -\frac{3}{2} \,{\left (2 \, \log \left (\frac{e x^{\frac{2}{3}}}{d} + 1\right ) \log \left (x^{\frac{1}{3}}\right ) +{\rm Li}_2\left (-\frac{e x^{\frac{2}{3}}}{d}\right )\right )} b n + \frac{3 \,{\left (2 \, b e n x^{\frac{2}{3}} \log \left (x^{\frac{1}{3}}\right ) - b e n x^{\frac{2}{3}}\right )}}{2 \, d} + \frac{2 \, b d \log \left ({\left (e x^{\frac{2}{3}} + d\right )}^{n}\right ) \log \left (x\right ) + 2 \,{\left (b d \log \left (c\right ) + a d\right )} \log \left (x\right ) - \frac{2 \, b e n x \log \left (x\right ) - 3 \, b e n x}{x^{\frac{1}{3}}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*(2*log(e*x^(2/3)/d + 1)*log(x^(1/3)) + dilog(-e*x^(2/3)/d))*b*n + 3/2*(2*b*e*n*x^(2/3)*log(x^(1/3)) - b*e
*n*x^(2/3))/d + 1/2*(2*b*d*log((e*x^(2/3) + d)^n)*log(x) + 2*(b*d*log(c) + a*d)*log(x) - (2*b*e*n*x*log(x) - 3
*b*e*n*x)/x^(1/3))/d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left ({\left (e x^{\frac{2}{3}} + d\right )}^{n} c\right ) + a}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log((e*x^(2/3) + d)^n*c) + a)/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((a+b*ln(c*(d+e*x**(2/3))**n))/x,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left ({\left (e x^{\frac{2}{3}} + d\right )}^{n} c\right ) + a}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*log((e*x^(2/3) + d)^n*c) + a)/x, x)