Integrand size = 24, antiderivative size = 24 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\frac {16 b^3 n^3}{9 x}-\frac {208 b^3 d n^3}{3 e \sqrt [3]{x}}-\frac {208 b^3 d^{3/2} n^3 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 e^{3/2}}-\frac {32 i b^3 d^{3/2} n^3 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{e^{3/2}}+\frac {64 b^3 d^{3/2} n^3 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{e^{3/2}}-\frac {8 b^2 n^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 x}+\frac {32 b^2 d n^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{e \sqrt [3]{x}}+\frac {32 b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{e^{3/2}}+\frac {2 b n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x}-\frac {6 b d n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e \sqrt [3]{x}}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x}-\frac {32 i b^3 d^{3/2} n^3 \operatorname {PolyLog}\left (2,-1+\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{e^{3/2}}-\frac {2 b d^2 n \text {Int}\left (\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{\left (e+d x^{2/3}\right ) x^{2/3}},x\right )}{e} \] Output:
16/9*b^3*n^3/x-208/3*b^3*d*n^3/e/x^(1/3)-208/3*b^3*d^(3/2)*n^3*arctan(d^(1 /2)*x^(1/3)/e^(1/2))/e^(3/2)-32*I*b^3*d^(3/2)*n^3*arctan(d^(1/2)*x^(1/3)/e ^(1/2))^2/e^(3/2)+64*b^3*d^(3/2)*n^3*arctan(d^(1/2)*x^(1/3)/e^(1/2))*ln(2- 2*e^(1/2)/(e^(1/2)-I*d^(1/2)*x^(1/3)))/e^(3/2)-8/3*b^2*n^2*(a+b*ln(c*(d+e/ x^(2/3))^n))/x+32*b^2*d*n^2*(a+b*ln(c*(d+e/x^(2/3))^n))/e/x^(1/3)+32*b^2*d ^(3/2)*n^2*arctan(d^(1/2)*x^(1/3)/e^(1/2))*(a+b*ln(c*(d+e/x^(2/3))^n))/e^( 3/2)+2*b*n*(a+b*ln(c*(d+e/x^(2/3))^n))^2/x-6*b*d*n*(a+b*ln(c*(d+e/x^(2/3)) ^n))^2/e/x^(1/3)-(a+b*ln(c*(d+e/x^(2/3))^n))^3/x-32*I*b^3*d^(3/2)*n^3*poly log(2,-1+2*e^(1/2)/(e^(1/2)-I*d^(1/2)*x^(1/3)))/e^(3/2)-2*b*d^2*n*Defer(In t)((a+b*ln(c*(d+e/x^(2/3))^n))^2/(e+d*x^(2/3))/x^(2/3),x)/e
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(5502\) vs. \(2(483)=966\).
Time = 19.56 (sec) , antiderivative size = 5502, normalized size of antiderivative = 229.25 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\text {Result too large to show} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^2,x]
Output:
Result too large to show
Not integrable
Time = 2.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {2908, 2907, 2005, 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.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx\) |
\(\Big \downarrow \) 2908 |
\(\displaystyle 3 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^{4/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2907 |
\(\displaystyle 3 \left (-2 b e n \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{\left (d+\frac {e}{x^{2/3}}\right ) x^2}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{3 x}\right )\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle 3 \left (-2 b e n \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{\left (x^{2/3} d+e\right ) x^{4/3}}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{3 x}\right )\) |
\(\Big \downarrow \) 2926 |
\(\displaystyle 3 \left (-2 b e n \int \left (-\frac {d \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e^2 x^{2/3}}+\frac {d^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e^2 \left (x^{2/3} d+e\right )}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e x^{4/3}}\right )d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{3 x}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{3 x}-2 b e n \left (\frac {d^2 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{x^{2/3} d+e}d\sqrt [3]{x}}{e^2}-\frac {16 b d^{3/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 e^{5/2}}+\frac {d \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e^2 \sqrt [3]{x}}-\frac {16 b d n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{3 e^2 \sqrt [3]{x}}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 e x}+\frac {4 b n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{9 e x}+\frac {16 i b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{3 e^{5/2}}+\frac {104 b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{9 e^{5/2}}-\frac {32 b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right ) \log \left (2-\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}\right )}{3 e^{5/2}}+\frac {16 i b^2 d^{3/2} n^2 \operatorname {PolyLog}\left (2,\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}-1\right )}{3 e^{5/2}}+\frac {104 b^2 d n^2}{9 e^2 \sqrt [3]{x}}-\frac {8 b^2 n^2}{27 e x}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^2,x]
Output:
$Aborted
Int[(Fx_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p*Fx, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && Neg Q[n]
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]
Not integrable
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )}^{3}}{x^{2}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^2,x)
Output:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^2,x)
Not integrable
Time = 0.08 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.50 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{3}}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^2,x, algorithm="fricas")
Output:
integral((b^3*log(c*((d*x + e*x^(1/3))/x)^n)^3 + 3*a*b^2*log(c*((d*x + e*x ^(1/3))/x)^n)^2 + 3*a^2*b*log(c*((d*x + e*x^(1/3))/x)^n) + a^3)/x^2, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(2/3))**n))**3/x**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^2,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
Not integrable
Time = 0.53 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{3}}{x^{2}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^2,x, algorithm="giac")
Output:
integrate((b*log(c*(d + e/x^(2/3))^n) + a)^3/x^2, x)
Not integrable
Time = 25.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )\right )}^3}{x^2} \,d x \] Input:
int((a + b*log(c*(d + e/x^(2/3))^n))^3/x^2,x)
Output:
int((a + b*log(c*(d + e/x^(2/3))^n))^3/x^2, x)
Not integrable
Time = 0.18 (sec) , antiderivative size = 245, normalized size of antiderivative = 10.21 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^2} \, dx=\frac {-6 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} d}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} b d n x -6 x^{\frac {2}{3}} a^{2} b d e n -2 \left (\int \frac {{\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2}}{x^{\frac {8}{3}} d +e \,x^{2}}d x \right ) b^{3} e^{3} n x -4 \left (\int \frac {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}{x^{\frac {8}{3}} d +e \,x^{2}}d x \right ) a \,b^{2} e^{3} n x -{\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{3} b^{3} e^{2}-3 {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2} a \,b^{2} e^{2}-3 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) a^{2} b \,e^{2}-a^{3} e^{2}+2 a^{2} b \,e^{2} n}{e^{2} x} \] Input:
int((a+b*log(c*(d+e/x^(2/3))^n))^3/x^2,x)
Output:
( - 6*sqrt(e)*sqrt(d)*atan((x**(1/3)*d)/(sqrt(e)*sqrt(d)))*a**2*b*d*n*x - 6*x**(2/3)*a**2*b*d*e*n - 2*int(log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))* *2/(x**(2/3)*d*x**2 + e*x**2),x)*b**3*e**3*n*x - 4*int(log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))/(x**(2/3)*d*x**2 + e*x**2),x)*a*b**2*e**3*n*x - log (((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**3*b**3*e**2 - 3*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2*a*b**2*e**2 - 3*log(((x**(2/3)*d + e)**n*c)/x** ((2*n)/3))*a**2*b*e**2 - a**3*e**2 + 2*a**2*b*e**2*n)/(e**2*x)