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

Optimal. Leaf size=139 \[ -9 b^2 n^2 \text {Li}_3\left (\frac {x^{2/3} e}{d}+1\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\frac {9}{2} b n \text {Li}_2\left (\frac {x^{2/3} e}{d}+1\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2+\frac {3}{2} \log \left (-\frac {e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3+9 b^3 n^3 \text {Li}_4\left (\frac {x^{2/3} e}{d}+1\right ) \]

[Out]

3/2*(a+b*ln(c*(d+e*x^(2/3))^n))^3*ln(-e*x^(2/3)/d)+9/2*b*n*(a+b*ln(c*(d+e*x^(2/3))^n))^2*polylog(2,1+e*x^(2/3)
/d)-9*b^2*n^2*(a+b*ln(c*(d+e*x^(2/3))^n))*polylog(3,1+e*x^(2/3)/d)+9*b^3*n^3*polylog(4,1+e*x^(2/3)/d)

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Rubi [A]  time = 0.20, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2454, 2396, 2433, 2374, 2383, 6589} \[ -9 b^2 n^2 \text {PolyLog}\left (3,\frac {e x^{2/3}}{d}+1\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\frac {9}{2} b n \text {PolyLog}\left (2,\frac {e x^{2/3}}{d}+1\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2+9 b^3 n^3 \text {PolyLog}\left (4,\frac {e x^{2/3}}{d}+1\right )+\frac {3}{2} \log \left (-\frac {e x^{2/3}}{d}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*Log[c*(d + e*x^(2/3))^n])^3*Log[-((e*x^(2/3))/d)])/2 + (9*b*n*(a + b*Log[c*(d + e*x^(2/3))^n])^2*Pol
yLog[2, 1 + (e*x^(2/3))/d])/2 - 9*b^2*n^2*(a + b*Log[c*(d + e*x^(2/3))^n])*PolyLog[3, 1 + (e*x^(2/3))/d] + 9*b
^3*n^3*PolyLog[4, 1 + (e*x^(2/3))/d]

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 {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3}{x} \, dx &=\frac {3}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^3}{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 )^3 \log \left (-\frac {e x^{2/3}}{d}\right )-\frac {1}{2} (9 b e n) \operatorname {Subst}\left (\int \frac {\log \left (-\frac {e x}{d}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{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 )^3 \log \left (-\frac {e x^{2/3}}{d}\right )-\frac {1}{2} (9 b n) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (-\frac {e \left (-\frac {d}{e}+\frac {x}{e}\right )}{d}\right )}{x} \, dx,x,d+e x^{2/3}\right )\\ &=\frac {3}{2} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \log \left (-\frac {e x^{2/3}}{d}\right )+\frac {9}{2} b n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \text {Li}_2\left (1+\frac {e x^{2/3}}{d}\right )-\left (9 b^2 n^2\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (\frac {x}{d}\right )}{x} \, dx,x,d+e x^{2/3}\right )\\ &=\frac {3}{2} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \log \left (-\frac {e x^{2/3}}{d}\right )+\frac {9}{2} b n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \text {Li}_2\left (1+\frac {e x^{2/3}}{d}\right )-9 b^2 n^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \text {Li}_3\left (1+\frac {e x^{2/3}}{d}\right )+\left (9 b^3 n^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {x}{d}\right )}{x} \, dx,x,d+e x^{2/3}\right )\\ &=\frac {3}{2} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \log \left (-\frac {e x^{2/3}}{d}\right )+\frac {9}{2} b n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \text {Li}_2\left (1+\frac {e x^{2/3}}{d}\right )-9 b^2 n^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \text {Li}_3\left (1+\frac {e x^{2/3}}{d}\right )+9 b^3 n^3 \text {Li}_4\left (1+\frac {e x^{2/3}}{d}\right )\\ \end {align*}

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Mathematica [B]  time = 0.26, size = 339, normalized size = 2.44 \[ \frac {9}{2} b^2 n^2 \left (-2 \text {Li}_3\left (\frac {x^{2/3} e}{d}+1\right )+2 \text {Li}_2\left (\frac {x^{2/3} e}{d}+1\right ) \log \left (d+e x^{2/3}\right )+\log \left (-\frac {e x^{2/3}}{d}\right ) \log ^2\left (d+e x^{2/3}\right )\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )-b n \log \left (d+e x^{2/3}\right )\right )+3 b n \left (\log (x) \left (\log \left (d+e x^{2/3}\right )-\log \left (\frac {e x^{2/3}}{d}+1\right )\right )-\frac {3}{2} \text {Li}_2\left (-\frac {e x^{2/3}}{d}\right )\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )-b n \log \left (d+e x^{2/3}\right )\right )^2+\log (x) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )-b n \log \left (d+e x^{2/3}\right )\right )^3+\frac {3}{2} b^3 n^3 \left (6 \text {Li}_4\left (\frac {x^{2/3} e}{d}+1\right )+3 \text {Li}_2\left (\frac {x^{2/3} e}{d}+1\right ) \log ^2\left (d+e x^{2/3}\right )-6 \text {Li}_3\left (\frac {x^{2/3} e}{d}+1\right ) \log \left (d+e x^{2/3}\right )+\log \left (-\frac {e x^{2/3}}{d}\right ) \log ^3\left (d+e x^{2/3}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])^3*Log[x] + 3*b*n*(a - b*n*Log[d + e*x^(2/3)] + b*Log
[c*(d + e*x^(2/3))^n])^2*((Log[d + e*x^(2/3)] - Log[1 + (e*x^(2/3))/d])*Log[x] - (3*PolyLog[2, -((e*x^(2/3))/d
)])/2) + (9*b^2*n^2*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])*(Log[d + e*x^(2/3)]^2*Log[-((e*x
^(2/3))/d)] + 2*Log[d + e*x^(2/3)]*PolyLog[2, 1 + (e*x^(2/3))/d] - 2*PolyLog[3, 1 + (e*x^(2/3))/d]))/2 + (3*b^
3*n^3*(Log[d + e*x^(2/3)]^3*Log[-((e*x^(2/3))/d)] + 3*Log[d + e*x^(2/3)]^2*PolyLog[2, 1 + (e*x^(2/3))/d] - 6*L
og[d + e*x^(2/3)]*PolyLog[3, 1 + (e*x^(2/3))/d] + 6*PolyLog[4, 1 + (e*x^(2/3))/d]))/2

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{3} + 3 \, a b^{2} \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right )^{2} + 3 \, a^{2} b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \left (e \,x^{\frac {2}{3}}+d \right )^{n}\right )+a \right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.80, size = 282, normalized size = 2.03 \[ \frac {3}{2} \, {\left (\log \left (e x^{\frac {2}{3}} + d\right )^{3} \log \left (-\frac {e x^{\frac {2}{3}} + d}{d} + 1\right ) + 3 \, {\rm Li}_2\left (\frac {e x^{\frac {2}{3}} + d}{d}\right ) \log \left (e x^{\frac {2}{3}} + d\right )^{2} - 6 \, \log \left (e x^{\frac {2}{3}} + d\right ) {\rm Li}_{3}(\frac {e x^{\frac {2}{3}} + d}{d}) + 6 \, {\rm Li}_{4}(\frac {e x^{\frac {2}{3}} + d}{d})\right )} b^{3} n^{3} + a^{3} \log \relax (x) + \frac {9}{2} \, {\left (b^{3} n^{2} \log \relax (c) + a b^{2} n^{2}\right )} {\left (\log \left (e x^{\frac {2}{3}} + d\right )^{2} \log \left (-\frac {e x^{\frac {2}{3}} + d}{d} + 1\right ) + 2 \, {\rm Li}_2\left (\frac {e x^{\frac {2}{3}} + d}{d}\right ) \log \left (e x^{\frac {2}{3}} + d\right ) - 2 \, {\rm Li}_{3}(\frac {e x^{\frac {2}{3}} + d}{d})\right )} + \frac {9}{2} \, {\left (b^{3} n \log \relax (c)^{2} + 2 \, a b^{2} n \log \relax (c) + a^{2} b n\right )} {\left (\log \left (e x^{\frac {2}{3}} + d\right ) \log \left (-\frac {e x^{\frac {2}{3}} + d}{d} + 1\right ) + {\rm Li}_2\left (\frac {e x^{\frac {2}{3}} + d}{d}\right )\right )} + {\left (b^{3} \log \relax (c)^{3} + 3 \, a b^{2} \log \relax (c)^{2} + 3 \, a^{2} b \log \relax (c)\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*(log(e*x^(2/3) + d)^3*log(-(e*x^(2/3) + d)/d + 1) + 3*dilog((e*x^(2/3) + d)/d)*log(e*x^(2/3) + d)^2 - 6*lo
g(e*x^(2/3) + d)*polylog(3, (e*x^(2/3) + d)/d) + 6*polylog(4, (e*x^(2/3) + d)/d))*b^3*n^3 + a^3*log(x) + 9/2*(
b^3*n^2*log(c) + a*b^2*n^2)*(log(e*x^(2/3) + d)^2*log(-(e*x^(2/3) + d)/d + 1) + 2*dilog((e*x^(2/3) + d)/d)*log
(e*x^(2/3) + d) - 2*polylog(3, (e*x^(2/3) + d)/d)) + 9/2*(b^3*n*log(c)^2 + 2*a*b^2*n*log(c) + a^2*b*n)*(log(e*
x^(2/3) + d)*log(-(e*x^(2/3) + d)/d + 1) + dilog((e*x^(2/3) + d)/d)) + (b^3*log(c)^3 + 3*a*b^2*log(c)^2 + 3*a^
2*b*log(c))*log(x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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