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

Optimal. Leaf size=95 \[ -3 b n \text{PolyLog}\left (2,\frac{e}{d x^{2/3}}+1\right ) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )+3 b^2 n^2 \text{PolyLog}\left (3,\frac{e}{d x^{2/3}}+1\right )-\frac{3}{2} \log \left (-\frac{e}{d x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.131049, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {2454, 2396, 2433, 2374, 6589} \[ -3 b n \text{PolyLog}\left (2,\frac{e}{d x^{2/3}}+1\right ) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )+3 b^2 n^2 \text{PolyLog}\left (3,\frac{e}{d x^{2/3}}+1\right )-\frac{3}{2} \log \left (-\frac{e}{d x^{2/3}}\right ) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [B]  time = 0.142723, size = 199, normalized size = 2.09 \[ 2 b n \left (\frac{3}{2} \text{PolyLog}\left (2,-\frac{e}{d x^{2/3}}\right )+\log (x) \left (\log \left (d+\frac{e}{x^{2/3}}\right )-\log \left (\frac{e}{d x^{2/3}}+1\right )\right )\right ) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )-b n \log \left (d+\frac{e}{x^{2/3}}\right )\right )-\frac{3}{2} b^2 n^2 \left (-2 \text{PolyLog}\left (3,\frac{e}{d x^{2/3}}+1\right )+2 \log \left (d+\frac{e}{x^{2/3}}\right ) \text{PolyLog}\left (2,\frac{e}{d x^{2/3}}+1\right )+\log \left (-\frac{e}{d x^{2/3}}\right ) \log ^2\left (d+\frac{e}{x^{2/3}}\right )\right )+\log (x) \left (a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )-b n \log \left (d+\frac{e}{x^{2/3}}\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.352, 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 ) ^{2}}\, 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))^2/x,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b^{2} \log \left ({\left (d x^{\frac{2}{3}} + e\right )}^{n}\right )^{2} \log \left (x\right ) - \int -\frac{12 \,{\left (b^{2} d x + b^{2} e x^{\frac{1}{3}}\right )} \log \left (x^{\frac{1}{3} \, n}\right )^{2} + 3 \,{\left (b^{2} d \log \left (c\right )^{2} + 2 \, a b d \log \left (c\right ) + a^{2} d\right )} x - 2 \,{\left (2 \, b^{2} d n x \log \left (x\right ) - 3 \,{\left (b^{2} d \log \left (c\right ) + a b d\right )} x + 6 \,{\left (b^{2} d x + b^{2} e x^{\frac{1}{3}}\right )} \log \left (x^{\frac{1}{3} \, n}\right ) - 3 \,{\left (b^{2} e \log \left (c\right ) + a b e\right )} x^{\frac{1}{3}}\right )} \log \left ({\left (d x^{\frac{2}{3}} + e\right )}^{n}\right ) - 12 \,{\left ({\left (b^{2} d \log \left (c\right ) + a b d\right )} x +{\left (b^{2} e \log \left (c\right ) + a b e\right )} x^{\frac{1}{3}}\right )} \log \left (x^{\frac{1}{3} \, n}\right ) + 3 \,{\left (b^{2} e \log \left (c\right )^{2} + 2 \, a b e \log \left (c\right ) + a^{2} e\right )} x^{\frac{1}{3}}}{3 \,{\left (d x^{2} + e x^{\frac{4}{3}}\right )}}\,{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))^2/x,x, algorithm="maxima")

[Out]

b^2*log((d*x^(2/3) + e)^n)^2*log(x) - integrate(-1/3*(12*(b^2*d*x + b^2*e*x^(1/3))*log(x^(1/3*n))^2 + 3*(b^2*d
*log(c)^2 + 2*a*b*d*log(c) + a^2*d)*x - 2*(2*b^2*d*n*x*log(x) - 3*(b^2*d*log(c) + a*b*d)*x + 6*(b^2*d*x + b^2*
e*x^(1/3))*log(x^(1/3*n)) - 3*(b^2*e*log(c) + a*b*e)*x^(1/3))*log((d*x^(2/3) + e)^n) - 12*((b^2*d*log(c) + a*b
*d)*x + (b^2*e*log(c) + a*b*e)*x^(1/3))*log(x^(1/3*n)) + 3*(b^2*e*log(c)^2 + 2*a*b*e*log(c) + a^2*e)*x^(1/3))/
(d*x^2 + e*x^(4/3)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \log \left (c \left (\frac{d x + e x^{\frac{1}{3}}}{x}\right )^{n}\right )^{2} + 2 \, a b \log \left (c \left (\frac{d x + e x^{\frac{1}{3}}}{x}\right )^{n}\right ) + a^{2}}{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))^2/x,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

[Out]

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