3.6.22 \(\int (a+b \log (c (d+\frac {e}{x^{2/3}})^n))^2 \, dx\) [522]

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

[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 = 0.29, 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 = {2501, 2507, 2521, 2498, 269, 211, 2520, 12, 266, 6820, 5044, 4988, 2497} \begin {gather*} \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 \text {ArcTan}\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 i b^2 e^{3/2} n^2 \text {ArcTan}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{d^{3/2}}+\frac {8 b^2 e^{3/2} n^2 \text {ArcTan}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \text {ArcTan}\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} \end {gather*}

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 211

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

Rule 266

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 269

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 2497

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 2498

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 2501

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 2507

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 2520

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 2521

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 4988

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 5044

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[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 6820

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 \text {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) \text {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) \text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.70, size = 296, normalized size = 0.96 \begin {gather*} \frac {-3 b^2 e n^2 \left (e+d x^{2/3}\right ) \sqrt {-\frac {e}{d x^{2/3}}} \, _4F_3\left (1,1,1,\frac {5}{2};2,2,2;1+\frac {e}{d x^{2/3}}\right )+\frac {4 b \sqrt {d} e^2 n \tan ^{-1}\left (\frac {\sqrt {\frac {e}{x^{2/3}}}}{\sqrt {d}}\right ) \left (a-b n \log \left (d+\frac {e}{x^{2/3}}\right )+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{\sqrt {\frac {e}{x^{2/3}}}}+d x^{2/3} \left (-4 b^2 e n^2 \sqrt {-\frac {e}{d x^{2/3}}} \left (1+\log \left (\frac {1}{2} \left (1+\sqrt {-\frac {e}{d x^{2/3}}}\right )\right )\right ) \log \left (d+\frac {e}{x^{2/3}}\right )+b^2 e n^2 \sqrt {-\frac {e}{d x^{2/3}}} \log ^2\left (d+\frac {e}{x^{2/3}}\right )+\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \left (4 b e n+a d x^{2/3}+b d x^{2/3} \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\right )}{d^2 \sqrt [3]{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*e*n^2*(e + d*x^(2/3))*Sqrt[-(e/(d*x^(2/3)))]*HypergeometricPFQ[{1, 1, 1, 5/2}, {2, 2, 2}, 1 + e/(d*x^(
2/3))] + (4*b*Sqrt[d]*e^2*n*ArcTan[Sqrt[e/x^(2/3)]/Sqrt[d]]*(a - b*n*Log[d + e/x^(2/3)] + b*Log[c*(d + e/x^(2/
3))^n]))/Sqrt[e/x^(2/3)] + d*x^(2/3)*(-4*b^2*e*n^2*Sqrt[-(e/(d*x^(2/3)))]*(1 + Log[(1 + Sqrt[-(e/(d*x^(2/3)))]
)/2])*Log[d + e/x^(2/3)] + b^2*e*n^2*Sqrt[-(e/(d*x^(2/3)))]*Log[d + e/x^(2/3)]^2 + (a + b*Log[c*(d + e/x^(2/3)
)^n])*(4*b*e*n + a*d*x^(2/3) + b*d*x^(2/3)*Log[c*(d + e/x^(2/3))^n])))/(d^2*x^(1/3))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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