Integrand size = 24, antiderivative size = 476 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=-\frac {8 b^2 e^2 n^2}{105 d^2 x^{5/3}}+\frac {32 b^2 e^3 n^2}{105 d^3 x}-\frac {568 b^2 e^4 n^2}{315 d^4 \sqrt [3]{x}}-\frac {1408 b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{315 d^{9/2}}+\frac {4 i b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{3 d^{9/2}}+\frac {8 b^2 e^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{3 d^{9/2}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{21 d x^{7/3}}+\frac {4 b e^2 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{15 d^2 x^{5/3}}-\frac {4 b e^3 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{9 d^3 x}+\frac {4 b e^4 n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^4 \sqrt [3]{x}}+\frac {4 b e^{9/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^{9/2}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 x^3}+\frac {4 i b^2 e^{9/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{3 d^{9/2}} \] Output:
-8/105*b^2*e^2*n^2/d^2/x^(5/3)+32/105*b^2*e^3*n^2/d^3/x-568/315*b^2*e^4*n^ 2/d^4/x^(1/3)-1408/315*b^2*e^(9/2)*n^2*arctan(e^(1/2)*x^(1/3)/d^(1/2))/d^( 9/2)+4/3*I*b^2*e^(9/2)*n^2*arctan(e^(1/2)*x^(1/3)/d^(1/2))^2/d^(9/2)+8/3*b ^2*e^(9/2)*n^2*arctan(e^(1/2)*x^(1/3)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+I*e^( 1/2)*x^(1/3)))/d^(9/2)-4/21*b*e*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d/x^(7/3)+4/ 15*b*e^2*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^2/x^(5/3)-4/9*b*e^3*n*(a+b*ln(c*( d+e*x^(2/3))^n))/d^3/x+4/3*b*e^4*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d^4/x^(1/3) +4/3*b*e^(9/2)*n*arctan(e^(1/2)*x^(1/3)/d^(1/2))*(a+b*ln(c*(d+e*x^(2/3))^n ))/d^(9/2)-1/3*(a+b*ln(c*(d+e*x^(2/3))^n))^2/x^3+4/3*I*b^2*e^(9/2)*n^2*pol ylog(2,1-2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x^(1/3)))/d^(9/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.63 (sec) , antiderivative size = 473, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 x^3}+\frac {4}{3} b e n \left (-\frac {2 b e^{7/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{9/2}}-\frac {2 b e n \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\frac {e x^{2/3}}{d}\right )}{35 d^2 x^{5/3}}+\frac {2 b e^2 n \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\frac {e x^{2/3}}{d}\right )}{15 d^3 x}-\frac {2 b e^3 n \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {e x^{2/3}}{d}\right )}{3 d^4 \sqrt [3]{x}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{7 d x^{7/3}}+\frac {e \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^2 x^{5/3}}-\frac {e^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^3 x}+\frac {e^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^4 \sqrt [3]{x}}+\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{9/2}}+\frac {i b e^{7/2} n \left (\arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (\arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )-2 i \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}{-i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}\right )\right )}{d^{9/2}}\right ) \] Input:
Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^4,x]
Output:
-1/3*(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^3 + (4*b*e*n*((-2*b*e^(7/2)*n*Ar cTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/d^(9/2) - (2*b*e*n*Hypergeometric2F1[-5/2 , 1, -3/2, -((e*x^(2/3))/d)])/(35*d^2*x^(5/3)) + (2*b*e^2*n*Hypergeometric 2F1[-3/2, 1, -1/2, -((e*x^(2/3))/d)])/(15*d^3*x) - (2*b*e^3*n*Hypergeometr ic2F1[-1/2, 1, 1/2, -((e*x^(2/3))/d)])/(3*d^4*x^(1/3)) - (a + b*Log[c*(d + e*x^(2/3))^n])/(7*d*x^(7/3)) + (e*(a + b*Log[c*(d + e*x^(2/3))^n]))/(5*d^ 2*x^(5/3)) - (e^2*(a + b*Log[c*(d + e*x^(2/3))^n]))/(3*d^3*x) + (e^3*(a + b*Log[c*(d + e*x^(2/3))^n]))/(d^4*x^(1/3)) + (e^(7/2)*ArcTan[(Sqrt[e]*x^(1 /3))/Sqrt[d]]*(a + b*Log[c*(d + e*x^(2/3))^n]))/d^(9/2) + (I*b*e^(7/2)*n*( ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]] - (2* I)*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3))]) + PolyLog[2, (I*Sqrt[d] + Sqrt[e]*x^(1/3))/((-I)*Sqrt[d] + Sqrt[e]*x^(1/3))]))/d^(9/2)))/3
Time = 1.35 (sec) , antiderivative size = 431, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2908, 2907, 2926, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx\) |
\(\Big \downarrow \) 2908 |
\(\displaystyle 3 \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^{10/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2907 |
\(\displaystyle 3 \left (\frac {4}{9} b e n \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{\left (d+e x^{2/3}\right ) x^{8/3}}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{9 x^3}\right )\) |
\(\Big \downarrow \) 2926 |
\(\displaystyle 3 \left (\frac {4}{9} b e n \int \left (\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^4}{d^4 \left (d+e x^{2/3}\right )}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^3}{d^4 x^{2/3}}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e^2}{d^3 x^{4/3}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) e}{d^2 x^2}+\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{d x^{8/3}}\right )d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{9 x^3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{9 x^3}+\frac {4}{9} b e n \left (\frac {e^{7/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{9/2}}+\frac {e^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^4 \sqrt [3]{x}}-\frac {e^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 d^3 x}+\frac {e \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{5 d^2 x^{5/3}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{7 d x^{7/3}}+\frac {i b e^{7/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{d^{9/2}}-\frac {352 b e^{7/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{105 d^{9/2}}+\frac {2 b e^{7/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{9/2}}+\frac {i b e^{7/2} n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{9/2}}-\frac {142 b e^3 n}{105 d^4 \sqrt [3]{x}}+\frac {8 b e^2 n}{35 d^3 x}-\frac {2 b e n}{35 d^2 x^{5/3}}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^4,x]
Output:
3*(-1/9*(a + b*Log[c*(d + e*x^(2/3))^n])^2/x^3 + (4*b*e*n*((-2*b*e*n)/(35* d^2*x^(5/3)) + (8*b*e^2*n)/(35*d^3*x) - (142*b*e^3*n)/(105*d^4*x^(1/3)) - (352*b*e^(7/2)*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/(105*d^(9/2)) + (I*b*e ^(7/2)*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]^2)/d^(9/2) + (2*b*e^(7/2)*n*Arc Tan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3 ))])/d^(9/2) - (a + b*Log[c*(d + e*x^(2/3))^n])/(7*d*x^(7/3)) + (e*(a + b* Log[c*(d + e*x^(2/3))^n]))/(5*d^2*x^(5/3)) - (e^2*(a + b*Log[c*(d + e*x^(2 /3))^n]))/(3*d^3*x) + (e^3*(a + b*Log[c*(d + e*x^(2/3))^n]))/(d^4*x^(1/3)) + (e^(7/2)*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(a + b*Log[c*(d + e*x^(2/3)) ^n]))/d^(9/2) + (I*b*e^(7/2)*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] + I*Sqr t[e]*x^(1/3))])/d^(9/2)))/9)
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*(x_)^(m_ .), x_Symbol] :> With[{k = Denominator[n]}, Simp[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]
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]
\[\int \frac {{\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{2}}{x^{4}}d x\]
Input:
int((a+b*ln(c*(d+e*x^(2/3))^n))^2/x^4,x)
Output:
int((a+b*ln(c*(d+e*x^(2/3))^n))^2/x^4,x)
\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^2/x^4,x, algorithm="fricas")
Output:
integral((b^2*log((e*x^(2/3) + d)^n*c)^2 + 2*a*b*log((e*x^(2/3) + d)^n*c) + a^2)/x^4, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e*x**(2/3))**n))**2/x**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^2/x^4,x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\int { \frac {{\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{2}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))^2/x^4,x, algorithm="giac")
Output:
integrate((b*log((e*x^(2/3) + d)^n*c) + a)^2/x^4, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\right )}^2}{x^4} \,d x \] Input:
int((a + b*log(c*(d + e*x^(2/3))^n))^2/x^4,x)
Output:
int((a + b*log(c*(d + e*x^(2/3))^n))^2/x^4, x)
\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^4} \, dx=\frac {3780 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) a b \,e^{4} n \,x^{3}+840 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) b^{2} e^{4} n^{2} x^{3}-540 x^{\frac {2}{3}} a b \,d^{4} e n +3780 x^{\frac {8}{3}} a b d \,e^{4} n -120 x^{\frac {2}{3}} b^{2} d^{4} e \,n^{2}+840 x^{\frac {8}{3}} b^{2} d \,e^{4} n^{2}+756 x^{\frac {4}{3}} a b \,d^{3} e^{2} n +168 x^{\frac {4}{3}} b^{2} d^{3} e^{2} n^{2}-1260 \left (\int \frac {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}{x^{\frac {14}{3}} e +d \,x^{4}}d x \right ) b^{2} d^{6} n \,x^{3}-945 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b^{2} d^{5}-1890 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a b \,d^{5}-420 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{5} n -945 a^{2} d^{5}-1260 a b \,d^{2} e^{3} n \,x^{2}-280 b^{2} d^{2} e^{3} n^{2} x^{2}}{2835 d^{5} x^{3}} \] Input:
int((a+b*log(c*(d+e*x^(2/3))^n))^2/x^4,x)
Output:
(3780*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a*b*e**4*n*x**3 + 840*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*b**2*e**4*n**2 *x**3 - 540*x**(2/3)*a*b*d**4*e*n + 3780*x**(2/3)*a*b*d*e**4*n*x**2 - 120* x**(2/3)*b**2*d**4*e*n**2 + 840*x**(2/3)*b**2*d*e**4*n**2*x**2 + 756*x**(1 /3)*a*b*d**3*e**2*n*x + 168*x**(1/3)*b**2*d**3*e**2*n**2*x - 1260*int(log( (x**(2/3)*e + d)**n*c)/(x**(2/3)*e*x**4 + d*x**4),x)*b**2*d**6*n*x**3 - 94 5*log((x**(2/3)*e + d)**n*c)**2*b**2*d**5 - 1890*log((x**(2/3)*e + d)**n*c )*a*b*d**5 - 420*log((x**(2/3)*e + d)**n*c)*b**2*d**5*n - 945*a**2*d**5 - 1260*a*b*d**2*e**3*n*x**2 - 280*b**2*d**2*e**3*n**2*x**2)/(2835*d**5*x**3)