3.5.78 \(\int \frac {(a+b \log (c (d+e x^{2/3})^n))^2}{x^2} \, dx\) [478]

Optimal. Leaf size=298 \[ \frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {4 i b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d \sqrt [3]{x}}-\frac {4 b e^{3/2} n \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{3/2}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x}-\frac {4 i b^2 e^{3/2} n^2 \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.27, antiderivative size = 298, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {2508, 2507, 2526, 2505, 211, 2520, 12, 5040, 4964, 2449, 2352} \begin {gather*} -\frac {4 i b^2 e^{3/2} n^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}}-\frac {4 b e^{3/2} n \text {ArcTan}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{3/2}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d \sqrt [3]{x}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x}-\frac {4 i b^2 e^{3/2} n^2 \text {ArcTan}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{d^{3/2}}+\frac {8 b^2 e^{3/2} n^2 \text {ArcTan}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \text {ArcTan}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^2*e^(3/2)*n^2*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/d^(3/2) - ((4*I)*b^2*e^(3/2)*n^2*ArcTan[(Sqrt[e]*x^(1/3)
)/Sqrt[d]]^2)/d^(3/2) - (8*b^2*e^(3/2)*n^2*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt
[e]*x^(1/3))])/d^(3/2) - (4*b*e*n*(a + b*Log[c*(d + e*x^(2/3))^n]))/(d*x^(1/3)) - (4*b*e^(3/2)*n*ArcTan[(Sqrt[
e]*x^(1/3))/Sqrt[d]]*(a + b*Log[c*(d + e*x^(2/3))^n]))/d^(3/2) - (a + b*Log[c*(d + e*x^(2/3))^n])^2/x - ((4*I)
*b^2*e^(3/2)*n^2*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*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 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2505

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

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 2508

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

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 2526

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c,
d, e, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] && IntegerQ[s]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[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 5040

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*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 247, normalized size = 0.83 \begin {gather*} \frac {-4 i b^2 e^{3/2} n^2 x \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2-4 b e^{3/2} n x \tan ^{-1}\left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a-2 b n+2 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\sqrt {d} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \left (a d+4 b e n x^{2/3}+b d \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-4 i b^2 e^{3/2} n^2 x \text {Li}_2\left (\frac {i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}{-i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )^{2}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

-b^2*n^2*log(x^(2/3)*e + d)^2/x + integrate(1/3*(3*(b^2*log(c)^2 + 2*a*b*log(c) + a^2)*x*e + 2*(2*b^2*n*x*e +
3*(b^2*log(c) + a*b)*x*e + 3*(b^2*d*log(c) + a*b*d)*x^(1/3))*n*log(x^(2/3)*e + d) + 3*(b^2*d*log(c)^2 + 2*a*b*
d*log(c) + a^2*d)*x^(1/3))/(x^3*e + d*x^(7/3)), 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^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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**2,x)

[Out]

Timed out

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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^2,x, algorithm="giac")

[Out]

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

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

[Out]

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

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