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

Optimal. Leaf size=309 \[ -\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}+\frac {4 i b^2 e^{3/2} n^2 \text {Li}_2\left (\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}-1\right )}{d^{3/2}}+\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}+\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}} \]

[Out]

4*a*b*e*n*x^(1/3)/d+8*b^2*e^(3/2)*n^2*arctan(x^(1/3)*d^(1/2)/e^(1/2))/d^(3/2)+4*I*b^2*e^(3/2)*n^2*arctan(x^(1/
3)*d^(1/2)/e^(1/2))^2/d^(3/2)+4*b^2*e*n*x^(1/3)*ln(c*(d+e/x^(2/3))^n)/d-4*b*e^(3/2)*n*arctan(x^(1/3)*d^(1/2)/e
^(1/2))*(a+b*ln(c*(d+e/x^(2/3))^n))/d^(3/2)+x*(a+b*ln(c*(d+e/x^(2/3))^n))^2-8*b^2*e^(3/2)*n^2*arctan(x^(1/3)*d
^(1/2)/e^(1/2))*ln(2-2*e^(1/2)/(-I*x^(1/3)*d^(1/2)+e^(1/2)))/d^(3/2)+4*I*b^2*e^(3/2)*n^2*polylog(2,-1+2*e^(1/2
)/(-I*x^(1/3)*d^(1/2)+e^(1/2)))/d^(3/2)

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Rubi [A]  time = 0.44, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.650, Rules used = {2451, 2457, 2471, 2448, 263, 205, 2470, 12, 260, 6688, 4924, 4868, 2447} \[ \frac {4 i b^2 e^{3/2} n^2 \text {PolyLog}\left (2,-1+\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{d^{3/2}}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}+\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}+\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right ) \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*b*e*n*x^(1/3))/d + (8*b^2*e^(3/2)*n^2*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]])/d^(3/2) + ((4*I)*b^2*e^(3/2)*n^2
*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]^2)/d^(3/2) - (8*b^2*e^(3/2)*n^2*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]*Log[2 - (
2*Sqrt[e])/(Sqrt[e] - I*Sqrt[d]*x^(1/3))])/d^(3/2) + (4*b^2*e*n*x^(1/3)*Log[c*(d + e/x^(2/3))^n])/d - (4*b*e^(
3/2)*n*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]]*(a + b*Log[c*(d + e/x^(2/3))^n]))/d^(3/2) + x*(a + b*Log[c*(d + e/x^(
2/3))^n])^2 + ((4*I)*b^2*e^(3/2)*n^2*PolyLog[2, -1 + (2*Sqrt[e])/(Sqrt[e] - I*Sqrt[d]*x^(1/3))])/d^(3/2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 2448

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2451

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_), x_Symbol] :> With[{k = Denominator[n]}, Di
st[k, Subst[Int[x^(k - 1)*(a + b*Log[c*(d + e*x^(k*n))^p])^q, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p,
q}, x] && FractionQ[n]

Rule 2457

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

Rule 2470

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[(u*x^(n - 1))/(d + e*x^n
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 2471

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol]
:> With[{t = ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, (f + g*x^s)^r, x]}, Int[t, x] /; SumQ[t]] /; Free
Q[{a, b, c, d, e, f, g, n, p, q, r, s}, x] && IntegerQ[n] && IGtQ[q, 0] && IntegerQ[r] && IntegerQ[s] && (EqQ[
q, 1] || (GtQ[r, 0] && GtQ[s, 1]) || (LtQ[s, 0] && LtQ[r, 0]))

Rule 4868

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

Rule 4924

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[c*x])^(p + 1))/(b*d*(p + 1)), x] + Dist[I/d, Int[(a + b*ArcTan[c*x])^p/(x*(I + c*x)), x], x] /; FreeQ[{a, b,
c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 0]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {align*} \int \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \, dx &=3 \operatorname {Subst}\left (\int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )\right )^2 \, dx,x,\sqrt [3]{x}\right )\\ &=x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+(4 b e n) \operatorname {Subst}\left (\int \frac {a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )}{d+\frac {e}{x^2}} \, dx,x,\sqrt [3]{x}\right )\\ &=x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+(4 b e n) \operatorname {Subst}\left (\int \left (\frac {a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )}{d}-\frac {e \left (a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )\right )}{d \left (e+d x^2\right )}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {(4 b e n) \operatorname {Subst}\left (\int \left (a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )\right ) \, dx,x,\sqrt [3]{x}\right )}{d}-\frac {\left (4 b e^2 n\right ) \operatorname {Subst}\left (\int \frac {a+b \log \left (c \left (d+\frac {e}{x^2}\right )^n\right )}{e+d x^2} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {\left (4 b^2 e n\right ) \operatorname {Subst}\left (\int \log \left (c \left (d+\frac {e}{x^2}\right )^n\right ) \, dx,x,\sqrt [3]{x}\right )}{d}-\frac {\left (8 b^2 e^3 n^2\right ) \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {e}}\right )}{\sqrt {d} \sqrt {e} \left (d+\frac {e}{x^2}\right ) x^3} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {\left (8 b^2 e^2 n^2\right ) \operatorname {Subst}\left (\int \frac {1}{\left (d+\frac {e}{x^2}\right ) x^2} \, dx,x,\sqrt [3]{x}\right )}{d}-\frac {\left (8 b^2 e^{5/2} n^2\right ) \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {e}}\right )}{\left (d+\frac {e}{x^2}\right ) x^3} \, dx,x,\sqrt [3]{x}\right )}{d^{3/2}}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {\left (8 b^2 e^2 n^2\right ) \operatorname {Subst}\left (\int \frac {1}{e+d x^2} \, dx,x,\sqrt [3]{x}\right )}{d}-\frac {\left (8 b^2 e^{5/2} n^2\right ) \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {e}}\right )}{x \left (e+d x^2\right )} \, dx,x,\sqrt [3]{x}\right )}{d^{3/2}}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}+\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2-\frac {\left (8 i b^2 e^{3/2} n^2\right ) \operatorname {Subst}\left (\int \frac {\tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {e}}\right )}{x \left (i+\frac {\sqrt {d} x}{\sqrt {e}}\right )} \, dx,x,\sqrt [3]{x}\right )}{d^{3/2}}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}+\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{d^{3/2}}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {\left (8 b^2 e n^2\right ) \operatorname {Subst}\left (\int \frac {\log \left (2-\frac {2}{1-\frac {i \sqrt {d} x}{\sqrt {e}}}\right )}{1+\frac {d x^2}{e}} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=\frac {4 a b e n \sqrt [3]{x}}{d}+\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}+\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{d^{3/2}}+\frac {4 b^2 e n \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{d^{3/2}}+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2+\frac {4 i b^2 e^{3/2} n^2 \text {Li}_2\left (-1+\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 523, normalized size = 1.69 \[ b e n \left (-\frac {2 \sqrt {e} \log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{(-d)^{3/2}}+\frac {2 \sqrt {e} \log \left (\sqrt {-d} \sqrt [3]{x}+\sqrt {e}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{(-d)^{3/2}}+\frac {4 a \sqrt [3]{x}}{d}+\frac {4 b \sqrt [3]{x} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{d}-\frac {8 b \sqrt {e} n \tan ^{-1}\left (\frac {\sqrt {e}}{\sqrt {d} \sqrt [3]{x}}\right )}{d^{3/2}}+\frac {b \sqrt {e} n \left (-4 \text {Li}_2\left (1-\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )+2 \text {Li}_2\left (\frac {1}{2}-\frac {\sqrt {-d} \sqrt [3]{x}}{2 \sqrt {e}}\right )+\log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right ) \left (\log \left (\sqrt {e}-\sqrt {-d} \sqrt [3]{x}\right )+2 \log \left (\frac {1}{2} \left (\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}+1\right )\right )-4 \log \left (\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )\right )}{(-d)^{3/2}}+\frac {b d \sqrt {e} n \left (2 \text {Li}_2\left (\frac {1}{2} \left (\frac {\sqrt [3]{x} \sqrt {-d}}{\sqrt {e}}+1\right )\right )-4 \text {Li}_2\left (\frac {\sqrt [3]{x} \sqrt {-d}}{\sqrt {e}}+1\right )+\log \left (\sqrt {-d} \sqrt [3]{x}+\sqrt {e}\right ) \left (\log \left (\sqrt {-d} \sqrt [3]{x}+\sqrt {e}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {-d} \sqrt [3]{x}}{2 \sqrt {e}}\right )-4 \log \left (-\frac {\sqrt {-d} \sqrt [3]{x}}{\sqrt {e}}\right )\right )\right )}{(-d)^{5/2}}\right )+x \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*(a + b*Log[c*(d + e/x^(2/3))^n])^2 + b*e*n*((4*a*x^(1/3))/d - (8*b*Sqrt[e]*n*ArcTan[Sqrt[e]/(Sqrt[d]*x^(1/3)
)])/d^(3/2) + (4*b*x^(1/3)*Log[c*(d + e/x^(2/3))^n])/d - (2*Sqrt[e]*(a + b*Log[c*(d + e/x^(2/3))^n])*Log[Sqrt[
e] - Sqrt[-d]*x^(1/3)])/(-d)^(3/2) + (2*Sqrt[e]*(a + b*Log[c*(d + e/x^(2/3))^n])*Log[Sqrt[e] + Sqrt[-d]*x^(1/3
)])/(-d)^(3/2) + (b*Sqrt[e]*n*(Log[Sqrt[e] - Sqrt[-d]*x^(1/3)]*(Log[Sqrt[e] - Sqrt[-d]*x^(1/3)] + 2*Log[(1 + (
Sqrt[-d]*x^(1/3))/Sqrt[e])/2] - 4*Log[(Sqrt[-d]*x^(1/3))/Sqrt[e]]) - 4*PolyLog[2, 1 - (Sqrt[-d]*x^(1/3))/Sqrt[
e]] + 2*PolyLog[2, 1/2 - (Sqrt[-d]*x^(1/3))/(2*Sqrt[e])]))/(-d)^(3/2) + (b*d*Sqrt[e]*n*(Log[Sqrt[e] + Sqrt[-d]
*x^(1/3)]*(Log[Sqrt[e] + Sqrt[-d]*x^(1/3)] + 2*Log[1/2 - (Sqrt[-d]*x^(1/3))/(2*Sqrt[e])] - 4*Log[-((Sqrt[-d]*x
^(1/3))/Sqrt[e])]) + 2*PolyLog[2, (1 + (Sqrt[-d]*x^(1/3))/Sqrt[e])/2] - 4*PolyLog[2, 1 + (Sqrt[-d]*x^(1/3))/Sq
rt[e]]))/(-d)^(5/2))

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} \log \left (c \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )^{n}\right )^{2} + 2 \, a b \log \left (c \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right )^{n}\right ) + a^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -2 \, {\left (2 \, e n {\left (\frac {e \arctan \left (\frac {d x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} d} - \frac {x^{\frac {1}{3}}}{d}\right )} - x \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )\right )} a b + {\left (n^{2} x \log \left (d x^{\frac {2}{3}} + e\right )^{2} - \int -\frac {3 \, d x \log \relax (c)^{2} + 3 \, e x^{\frac {1}{3}} \log \relax (c)^{2} - 2 \, {\left (2 \, d n x - 3 \, d x \log \relax (c) - 3 \, e x^{\frac {1}{3}} \log \relax (c) + 6 \, {\left (d x + e x^{\frac {1}{3}}\right )} \log \left (x^{\frac {1}{3} \, n}\right )\right )} n \log \left (d x^{\frac {2}{3}} + e\right ) + 12 \, {\left (d x + e x^{\frac {1}{3}}\right )} \log \left (x^{\frac {1}{3} \, n}\right )^{2} - 12 \, {\left (d x \log \relax (c) + e x^{\frac {1}{3}} \log \relax (c)\right )} \log \left (x^{\frac {1}{3} \, n}\right )}{3 \, {\left (d x + e x^{\frac {1}{3}}\right )}}\,{d x}\right )} b^{2} + a^{2} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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