Integrand size = 24, antiderivative size = 24 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx =\text {Too large to display} \] Output:
16/729*b^3*n^3/x^3-3088/27783*b^3*d*n^3/e/x^(7/3)+221344/496125*b^3*d^2*n^ 3/e^2/x^(5/3)-637984/297675*b^3*d^3*n^3/e^3/x+3475504/99225*b^3*d^4*n^3/e^ 4/x^(1/3)+3475504/99225*b^3*d^(9/2)*n^3*arctan(d^(1/2)*x^(1/3)/e^(1/2))/e^ (9/2)+4504/315*I*b^3*d^(9/2)*n^3*polylog(2,-1+2*e^(1/2)/(e^(1/2)-I*d^(1/2) *x^(1/3)))/e^(9/2)-9008/315*b^3*d^(9/2)*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^(9/2)-8/81*b^2*n^2*(a+b*ln (c*(d+e/x^(2/3))^n))/x^3+128/441*b^2*d*n^2*(a+b*ln(c*(d+e/x^(2/3))^n))/e/x ^(7/3)-1144/1575*b^2*d^2*n^2*(a+b*ln(c*(d+e/x^(2/3))^n))/e^2/x^(5/3)+1984/ 945*b^2*d^3*n^2*(a+b*ln(c*(d+e/x^(2/3))^n))/e^3/x-4504/315*b^2*d^4*n^2*(a+ b*ln(c*(d+e/x^(2/3))^n))/e^4/x^(1/3)-4504/315*b^2*d^(9/2)*n^2*arctan(d^(1/ 2)*x^(1/3)/e^(1/2))*(a+b*ln(c*(d+e/x^(2/3))^n))/e^(9/2)+2/9*b*n*(a+b*ln(c* (d+e/x^(2/3))^n))^2/x^3-2/7*b*d*n*(a+b*ln(c*(d+e/x^(2/3))^n))^2/e/x^(7/3)+ 2/5*b*d^2*n*(a+b*ln(c*(d+e/x^(2/3))^n))^2/e^2/x^(5/3)-2/3*b*d^3*n*(a+b*ln( c*(d+e/x^(2/3))^n))^2/e^3/x+2*b*d^4*n*(a+b*ln(c*(d+e/x^(2/3))^n))^2/e^4/x^ (1/3)-1/3*(a+b*ln(c*(d+e/x^(2/3))^n))^3/x^3+4504/315*I*b^3*d^(9/2)*n^3*arc tan(d^(1/2)*x^(1/3)/e^(1/2))^2/e^(9/2)+2/3*b*d^5*n*Defer(Int)((a+b*ln(c*(d +e/x^(2/3))^n))^2/(e+d*x^(2/3))/x^(2/3),x)/e^4
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(6328\) vs. \(2(784)=1568\).
Time = 27.19 (sec) , antiderivative size = 6328, normalized size of antiderivative = 263.67 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\text {Result too large to show} \] Input:
Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^4,x]
Output:
Result too large to show
Not integrable
Time = 4.67 (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.
\(\displaystyle \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx\) |
\(\Big \downarrow \) 2908 |
\(\displaystyle 3 \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^{10/3}}d\sqrt [3]{x}\) |
\(\Big \downarrow \) 2907 |
\(\displaystyle 3 \left (-\frac {2}{3} b e n \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{\left (d+\frac {e}{x^{2/3}}\right ) x^4}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{9 x^3}\right )\) |
\(\Big \downarrow \) 2005 |
\(\displaystyle 3 \left (-\frac {2}{3} b e n \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{\left (x^{2/3} d+e\right ) x^{10/3}}d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{9 x^3}\right )\) |
\(\Big \downarrow \) 2926 |
\(\displaystyle 3 \left (-\frac {2}{3} b e n \int \left (-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 d^5}{e^5 \left (x^{2/3} d+e\right )}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 d^4}{e^5 x^{2/3}}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 d^3}{e^4 x^{4/3}}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 d^2}{e^3 x^2}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2 d}{e^2 x^{8/3}}+\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e x^{10/3}}\right )d\sqrt [3]{x}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{9 x^3}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 3 \left (-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{9 x^3}-\frac {2}{3} b e n \left (-\frac {d^5 \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^5}+\frac {2252 b d^{9/2} 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 )}{315 e^{11/2}}-\frac {d^4 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{e^5 \sqrt [3]{x}}+\frac {2252 b d^4 n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{315 e^5 \sqrt [3]{x}}+\frac {d^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{3 e^4 x}-\frac {992 b d^3 n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{945 e^4 x}-\frac {d^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{5 e^3 x^{5/3}}+\frac {572 b d^2 n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{1575 e^3 x^{5/3}}+\frac {d \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{7 e^2 x^{7/3}}-\frac {64 b d n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{441 e^2 x^{7/3}}-\frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^2}{9 e x^3}+\frac {4 b n \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )}{81 e x^3}-\frac {2252 i b^2 d^{9/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )^2}{315 e^{11/2}}-\frac {1737752 b^2 d^{9/2} n^2 \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{99225 e^{11/2}}+\frac {4504 b^2 d^{9/2} 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 )}{315 e^{11/2}}-\frac {2252 i b^2 d^{9/2} n^2 \operatorname {PolyLog}\left (2,\frac {2 \sqrt {e}}{\sqrt {e}-i \sqrt {d} \sqrt [3]{x}}-1\right )}{315 e^{11/2}}-\frac {1737752 b^2 d^4 n^2}{99225 e^5 \sqrt [3]{x}}+\frac {318992 b^2 d^3 n^2}{297675 e^4 x}-\frac {110672 b^2 d^2 n^2}{496125 e^3 x^{5/3}}+\frac {1544 b^2 d n^2}{27783 e^2 x^{7/3}}-\frac {8 b^2 n^2}{729 e x^3}\right )\right )\) |
Input:
Int[(a + b*Log[c*(d + e/x^(2/3))^n])^3/x^4,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
\[\int \frac {{\left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )}^{3}}{x^{4}}d x\]
Input:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^4,x)
Output:
int((a+b*ln(c*(d+e/x^(2/3))^n))^3/x^4,x)
Not integrable
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 3.50 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^4,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^4, x)
Timed out. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e/x**(2/3))**n))**3/x**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/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
Not integrable
Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*log(c*(d+e/x^(2/3))^n))^3/x^4,x, algorithm="giac")
Output:
integrate((b*log(c*(d + e/x^(2/3))^n) + a)^3/x^4, x)
Not integrable
Time = 22.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )\right )}^3}{x^4} \,d x \] Input:
int((a + b*log(c*(d + e/x^(2/3))^n))^3/x^4,x)
Output:
int((a + b*log(c*(d + e/x^(2/3))^n))^3/x^4, x)
Not integrable
Time = 0.18 (sec) , antiderivative size = 302, normalized size of antiderivative = 12.58 \[ \int \frac {\left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )^3}{x^4} \, dx=\frac {630 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} d}{\sqrt {e}\, \sqrt {d}}\right ) a^{2} b \,d^{4} n \,x^{3}+630 x^{\frac {8}{3}} a^{2} b \,d^{4} e n -90 x^{\frac {2}{3}} a^{2} b d \,e^{4} n +126 x^{\frac {4}{3}} a^{2} b \,d^{2} e^{3} n -210 \left (\int \frac {{\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2}}{x^{\frac {14}{3}} d +e \,x^{4}}d x \right ) b^{3} e^{6} n \,x^{3}-420 \left (\int \frac {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}{x^{\frac {14}{3}} d +e \,x^{4}}d x \right ) a \,b^{2} e^{6} n \,x^{3}-105 {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{3} b^{3} e^{5}-315 {\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right )}^{2} a \,b^{2} e^{5}-315 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) a^{2} b \,e^{5}-105 a^{3} e^{5}-210 a^{2} b \,d^{3} e^{2} n \,x^{2}+70 a^{2} b \,e^{5} n}{315 e^{5} x^{3}} \] Input:
int((a+b*log(c*(d+e/x^(2/3))^n))^3/x^4,x)
Output:
(630*sqrt(e)*sqrt(d)*atan((x**(1/3)*d)/(sqrt(e)*sqrt(d)))*a**2*b*d**4*n*x* *3 + 630*x**(2/3)*a**2*b*d**4*e*n*x**2 - 90*x**(2/3)*a**2*b*d*e**4*n + 126 *x**(1/3)*a**2*b*d**2*e**3*n*x - 210*int(log(((x**(2/3)*d + e)**n*c)/x**(( 2*n)/3))**2/(x**(2/3)*d*x**4 + e*x**4),x)*b**3*e**6*n*x**3 - 420*int(log(( (x**(2/3)*d + e)**n*c)/x**((2*n)/3))/(x**(2/3)*d*x**4 + e*x**4),x)*a*b**2* e**6*n*x**3 - 105*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**3*b**3*e**5 - 315*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))**2*a*b**2*e**5 - 315*log((( x**(2/3)*d + e)**n*c)/x**((2*n)/3))*a**2*b*e**5 - 105*a**3*e**5 - 210*a**2 *b*d**3*e**2*n*x**2 + 70*a**2*b*e**5*n)/(315*e**5*x**3)