\(\int x^2 (a+b \log (c (d+e x^{2/3})^n))^3 \, dx\) [485]

Optimal result
Mathematica [A] (verified)
Rubi [N/A]
Maple [N/A]
Fricas [N/A]
Sympy [F(-1)]
Maxima [F(-2)]
Giac [N/A]
Mupad [N/A]
Reduce [N/A]

Optimal result

Integrand size = 24, antiderivative size = 24 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx =\text {Too large to display} \] Output:

4504/315*a*b^2*d^4*n^2*x^(1/3)/e^4-3475504/99225*b^3*d^4*n^3*x^(1/3)/e^4+6 
37984/297675*b^3*d^3*n^3*x/e^3-221344/496125*b^3*d^2*n^3*x^(5/3)/e^2+3088/ 
27783*b^3*d*n^3*x^(7/3)/e-16/729*b^3*n^3*x^3+3475504/99225*b^3*d^(9/2)*n^3 
*arctan(e^(1/2)*x^(1/3)/d^(1/2))/e^(9/2)-4504/315*I*b^3*d^(9/2)*n^3*arctan 
(e^(1/2)*x^(1/3)/d^(1/2))^2/e^(9/2)-9008/315*b^3*d^(9/2)*n^3*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)))/e^(9/2)+4504/ 
315*b^3*d^4*n^2*x^(1/3)*ln(c*(d+e*x^(2/3))^n)/e^4-1984/945*b^2*d^3*n^2*x*( 
a+b*ln(c*(d+e*x^(2/3))^n))/e^3+1144/1575*b^2*d^2*n^2*x^(5/3)*(a+b*ln(c*(d+ 
e*x^(2/3))^n))/e^2-128/441*b^2*d*n^2*x^(7/3)*(a+b*ln(c*(d+e*x^(2/3))^n))/e 
+8/81*b^2*n^2*x^3*(a+b*ln(c*(d+e*x^(2/3))^n))-4504/315*b^2*d^(9/2)*n^2*arc 
tan(e^(1/2)*x^(1/3)/d^(1/2))*(a+b*ln(c*(d+e*x^(2/3))^n))/e^(9/2)-2*b*d^4*n 
*x^(1/3)*(a+b*ln(c*(d+e*x^(2/3))^n))^2/e^4+2/3*b*d^3*n*x*(a+b*ln(c*(d+e*x^ 
(2/3))^n))^2/e^3-2/5*b*d^2*n*x^(5/3)*(a+b*ln(c*(d+e*x^(2/3))^n))^2/e^2+2/7 
*b*d*n*x^(7/3)*(a+b*ln(c*(d+e*x^(2/3))^n))^2/e-2/9*b*n*x^3*(a+b*ln(c*(d+e* 
x^(2/3))^n))^2+1/3*x^3*(a+b*ln(c*(d+e*x^(2/3))^n))^3-4504/315*I*b^3*d^(9/2 
)*n^3*polylog(2,1-2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x^(1/3)))/e^(9/2)+2/3*b*d^5 
*n*Defer(Int)((a+b*ln(c*(d+e*x^(2/3))^n))^2/(d+e*x^(2/3))/x^(2/3),x)/e^4
 

Mathematica [A] (verified)

Time = 12.55 (sec) , antiderivative size = 1552, normalized size of antiderivative = 64.67 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*b*d^4*n*x^(1/3)*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^ 
n])^2)/e^4 + (2*b*d^3*n*x*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^( 
2/3))^n])^2)/(3*e^3) - (2*b*d^2*n*x^(5/3)*(a - b*n*Log[d + e*x^(2/3)] + b* 
Log[c*(d + e*x^(2/3))^n])^2)/(5*e^2) + (2*b*d*n*x^(7/3)*(a - b*n*Log[d + e 
*x^(2/3)] + b*Log[c*(d + e*x^(2/3))^n])^2)/(7*e) + (2*b*d^(9/2)*n*ArcTan[( 
Sqrt[e]*x^(1/3))/Sqrt[d]]*(a - b*n*Log[d + e*x^(2/3)] + b*Log[c*(d + e*x^( 
2/3))^n])^2)/e^(9/2) + b*n*x^3*Log[d + e*x^(2/3)]*(a - b*n*Log[d + e*x^(2/ 
3)] + b*Log[c*(d + e*x^(2/3))^n])^2 + (x^3*(a - b*n*Log[d + e*x^(2/3)] + b 
*Log[c*(d + e*x^(2/3))^n])^2*(3*a - 2*b*n - 3*b*n*Log[d + e*x^(2/3)] + 3*b 
*Log[c*(d + e*x^(2/3))^n]))/9 - (b^3*n^3*(1094783760*d^(9/2)*Sqrt[d + e*x^ 
(2/3)]*Sqrt[(e*x^(2/3))/(d + e*x^(2/3))]*ArcSin[Sqrt[d]/Sqrt[d + e*x^(2/3) 
]] - e*x^(2/3)*(-16*(68423985*d^4 - 4186770*d^3*e*x^(2/3) + 871542*d^2*e^2 
*x^(4/3) - 217125*d*e^3*x^2 + 42875*e^4*x^(8/3)) + 2520*(177345*d^4 - 2604 
0*d^3*e*x^(2/3) + 9009*d^2*e^2*x^(4/3) - 3600*d*e^3*x^2 + 1225*e^4*x^(8/3) 
)*Log[d + e*x^(2/3)] - 198450*(315*d^4 - 105*d^3*e*x^(2/3) + 63*d^2*e^2*x^ 
(4/3) - 45*d*e^3*x^2 + 35*e^4*x^(8/3))*Log[d + e*x^(2/3)]^2 + 10418625*e^4 
*x^(8/3)*Log[d + e*x^(2/3)]^3) + 62511750*d^(9/2)*Sqrt[(e*x^(2/3))/(d + e* 
x^(2/3))]*(8*Sqrt[d]*HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {3/2, 3/2, 3/ 
2}, d/(d + e*x^(2/3))] + Log[d + e*x^(2/3)]*(4*Sqrt[d]*HypergeometricPFQ[{ 
1/2, 1/2, 1/2}, {3/2, 3/2}, d/(d + e*x^(2/3))] + Sqrt[d + e*x^(2/3)]*Ar...
 

Rubi [N/A]

Not integrable

Time = 4.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, 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.

\(\displaystyle \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx\)

\(\Big \downarrow \) 2908

\(\displaystyle 3 \int x^{8/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3d\sqrt [3]{x}\)

\(\Big \downarrow \) 2907

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3-\frac {2}{3} b e n \int \frac {x^{10/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d+e x^{2/3}}d\sqrt [3]{x}\right )\)

\(\Big \downarrow \) 2926

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3-\frac {2}{3} b e n \int \left (-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 d^5}{e^5 \left (d+e x^{2/3}\right )}+\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 d^4}{e^5}-\frac {x^{2/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 d^3}{e^4}+\frac {x^{4/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 d^2}{e^3}-\frac {x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 d}{e^2}+\frac {x^{8/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{e}\right )d\sqrt [3]{x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (\frac {1}{9} x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3-\frac {2}{3} b e n \left (-\frac {d^5 \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{d+e x^{2/3}}d\sqrt [3]{x}}{e^5}+\frac {2252 b d^{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 )}{315 e^{11/2}}+\frac {d^4 \sqrt [3]{x} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{e^5}-\frac {d^3 x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{3 e^4}+\frac {992 b d^3 n x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{945 e^4}+\frac {d^2 x^{5/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{5 e^3}-\frac {572 b d^2 n x^{5/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{1575 e^3}-\frac {d x^{7/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{7 e^2}+\frac {64 b d n x^{7/3} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{441 e^2}+\frac {x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{9 e}-\frac {4 b n x^3 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{81 e}-\frac {2252 a b d^4 n \sqrt [3]{x}}{315 e^5}+\frac {2252 i b^2 d^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{315 e^{11/2}}-\frac {1737752 b^2 d^{9/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{99225 e^{11/2}}+\frac {4504 b^2 d^{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 )}{315 e^{11/2}}-\frac {2252 b^2 d^4 n \sqrt [3]{x} \log \left (c \left (d+e x^{2/3}\right )^n\right )}{315 e^5}+\frac {2252 i b^2 d^{9/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{315 e^{11/2}}+\frac {1737752 b^2 d^4 n^2 \sqrt [3]{x}}{99225 e^5}-\frac {318992 b^2 d^3 n^2 x}{297675 e^4}+\frac {110672 b^2 d^2 n^2 x^{5/3}}{496125 e^3}-\frac {1544 b^2 d n^2 x^{7/3}}{27783 e^2}+\frac {8 b^2 n^2 x^3}{729 e}\right )\right )\)

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2907
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]
 

rule 2908
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]
 

rule 2926
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]
 
Maple [N/A]

Not integrable

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

\[\int x^{2} {\left (a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )\right )}^{3}d x\]

Input:

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

Output:

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

Fricas [N/A]

Not integrable

Time = 0.09 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.12 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\int { {\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{3} x^{2} \,d x } \] Input:

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\text {Exception raised: ValueError} \] Input:

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

Giac [N/A]

Not integrable

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\int { {\left (b \log \left ({\left (e x^{\frac {2}{3}} + d\right )}^{n} c\right ) + a\right )}^{3} x^{2} \,d x } \] Input:

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

Output:

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

Mupad [N/A]

Not integrable

Time = 25.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx=\int x^2\,{\left (a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )\right )}^3 \,d x \] Input:

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

Output:

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

Reduce [N/A]

Not integrable

Time = 0.20 (sec) , antiderivative size = 929, normalized size of antiderivative = 38.71 \[ \int x^2 \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^3 \, dx =\text {Too large to display} \] Input:

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

Output:

(62511750*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a**2*b*d**4 
*n - 446909400*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a*b**2 
*d**4*n**2 + 1094783760*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d) 
))*b**3*d**4*n**3 - 12502350*x**(2/3)*log((x**(2/3)*e + d)**n*c)**2*b**3*d 
**2*e**3*n*x - 25004700*x**(2/3)*log((x**(2/3)*e + d)**n*c)*a*b**2*d**2*e* 
*3*n*x + 22702680*x**(2/3)*log((x**(2/3)*e + d)**n*c)*b**3*d**2*e**3*n**2* 
x - 12502350*x**(2/3)*a**2*b*d**2*e**3*n*x + 22702680*x**(2/3)*a*b**2*d**2 
*e**3*n**2*x - 13944672*x**(2/3)*b**3*d**2*e**3*n**3*x - 62511750*x**(1/3) 
*log((x**(2/3)*e + d)**n*c)**2*b**3*d**4*e*n + 8930250*x**(1/3)*log((x**(2 
/3)*e + d)**n*c)**2*b**3*d*e**4*n*x**2 - 125023500*x**(1/3)*log((x**(2/3)* 
e + d)**n*c)*a*b**2*d**4*e*n + 17860500*x**(1/3)*log((x**(2/3)*e + d)**n*c 
)*a*b**2*d*e**4*n*x**2 + 446909400*x**(1/3)*log((x**(2/3)*e + d)**n*c)*b** 
3*d**4*e*n**2 - 9072000*x**(1/3)*log((x**(2/3)*e + d)**n*c)*b**3*d*e**4*n* 
*2*x**2 - 62511750*x**(1/3)*a**2*b*d**4*e*n + 8930250*x**(1/3)*a**2*b*d*e* 
*4*n*x**2 + 446909400*x**(1/3)*a*b**2*d**4*e*n**2 - 9072000*x**(1/3)*a*b** 
2*d*e**4*n**2*x**2 - 1094783760*x**(1/3)*b**3*d**4*e*n**3 + 3474000*x**(1/ 
3)*b**3*d*e**4*n**3*x**2 + 20837250*int(log((x**(2/3)*e + d)**n*c)**2/(x** 
(2/3)*d + x**(1/3)*e*x),x)*b**3*d**5*e*n + 41674500*int(log((x**(2/3)*e + 
d)**n*c)/(x**(2/3)*d + x**(1/3)*e*x),x)*a*b**2*d**5*e*n - 148969800*int(lo 
g((x**(2/3)*e + d)**n*c)/(x**(2/3)*d + x**(1/3)*e*x),x)*b**3*d**5*e*n**...