\(\int (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3})^p x \, dx\) [103]

Optimal result

Integrand size = 26, antiderivative size = 292 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=-\frac {3 a^5 \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (1+2 p)}+\frac {15 a^4 \left (a+b \sqrt [3]{x}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (1+p)}-\frac {30 a^3 \left (a+b \sqrt [3]{x}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (3+2 p)}+\frac {15 a^2 \left (a+b \sqrt [3]{x}\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2+p)}-\frac {15 a \left (a+b \sqrt [3]{x}\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (5+2 p)}+\frac {3 \left (a+b \sqrt [3]{x}\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (3+p)} \] Output:

-3*a^5*(a+b*x^(1/3))*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/b^6/(1+2*p)+15/2*a^ 
4*(a+b*x^(1/3))^2*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/b^6/(p+1)-30*a^3*(a+b* 
x^(1/3))^3*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/b^6/(3+2*p)+15*a^2*(a+b*x^(1/ 
3))^4*(a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/b^6/(2+p)-15*a*(a+b*x^(1/3))^5*(a^ 
2+2*a*b*x^(1/3)+b^2*x^(2/3))^p/b^6/(5+2*p)+3/2*(a+b*x^(1/3))^6*(a^2+2*a*b* 
x^(1/3)+b^2*x^(2/3))^p/b^6/(3+p)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.49 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \left (-\frac {2 a^5}{1+2 p}+\frac {5 a^4 \left (a+b \sqrt [3]{x}\right )}{1+p}-\frac {20 a^3 \left (a+b \sqrt [3]{x}\right )^2}{3+2 p}+\frac {10 a^2 \left (a+b \sqrt [3]{x}\right )^3}{2+p}-\frac {10 a \left (a+b \sqrt [3]{x}\right )^4}{5+2 p}+\frac {\left (a+b \sqrt [3]{x}\right )^5}{3+p}\right ) \left (a+b \sqrt [3]{x}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p}{2 b^6} \] Input:

Integrate[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*x,x]
 

Output:

(3*((-2*a^5)/(1 + 2*p) + (5*a^4*(a + b*x^(1/3)))/(1 + p) - (20*a^3*(a + b* 
x^(1/3))^2)/(3 + 2*p) + (10*a^2*(a + b*x^(1/3))^3)/(2 + p) - (10*a*(a + b* 
x^(1/3))^4)/(5 + 2*p) + (a + b*x^(1/3))^5/(3 + p))*(a + b*x^(1/3))*((a + b 
*x^(1/3))^2)^p)/(2*b^6)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.82, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1385, 798, 53, 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.

\(\displaystyle \int x \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \, dx\)

\(\Big \downarrow \) 1385

\(\displaystyle \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \int \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p} xdx\)

\(\Big \downarrow \) 798

\(\displaystyle 3 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \int \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p} x^{5/3}d\sqrt [3]{x}\)

\(\Big \downarrow \) 53

\(\displaystyle 3 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p \int \left (-\frac {a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p}}{b^5}+\frac {5 a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p+1}}{b^5}-\frac {10 a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p+2}}{b^5}+\frac {10 a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p+3}}{b^5}-\frac {5 a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p+4}}{b^5}+\frac {a^5 \left (\frac {\sqrt [3]{x} b}{a}+1\right )^{2 p+5}}{b^5}\right )d\sqrt [3]{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (\frac {5 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 (p+1)}}{2 b^6 (p+1)}+\frac {5 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 (p+2)}}{b^6 (p+2)}+\frac {a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 (p+3)}}{2 b^6 (p+3)}-\frac {a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 p+1}}{b^6 (2 p+1)}-\frac {10 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 p+3}}{b^6 (2 p+3)}-\frac {5 a^6 \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{2 p+5}}{b^6 (2 p+5)}\right ) \left (\frac {b \sqrt [3]{x}}{a}+1\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\)

Input:

Int[(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p*x,x]
 

Output:

(3*((5*a^6*(1 + (b*x^(1/3))/a)^(2*(1 + p)))/(2*b^6*(1 + p)) + (5*a^6*(1 + 
(b*x^(1/3))/a)^(2*(2 + p)))/(b^6*(2 + p)) + (a^6*(1 + (b*x^(1/3))/a)^(2*(3 
 + p)))/(2*b^6*(3 + p)) - (a^6*(1 + (b*x^(1/3))/a)^(1 + 2*p))/(b^6*(1 + 2* 
p)) - (10*a^6*(1 + (b*x^(1/3))/a)^(3 + 2*p))/(b^6*(3 + 2*p)) - (5*a^6*(1 + 
 (b*x^(1/3))/a)^(5 + 2*p))/(b^6*(5 + 2*p)))*(a^2 + 2*a*b*x^(1/3) + b^2*x^( 
2/3))^p)/(1 + (b*x^(1/3))/a)^(2*p)
 

Defintions of rubi rules used

rule 53

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, 
x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) 
|| LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
 

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1385

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[a^IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/(1 + 2*c*(x^n/b))^(2* 
FracPart[p]))   Int[u*(1 + 2*c*(x^n/b))^(2*p), x], x] /; FreeQ[{a, b, c, n, 
 p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p] && NeQ[u, 
 x^(n - 1)] && NeQ[u, x^(2*n - 1)]
 

rule 2009

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

Maple [F]

\[\int \left (a^{2}+2 a b \,x^{\frac {1}{3}}+b^{2} x^{\frac {2}{3}}\right )^{p} x d x\]

Input:

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x)
 

Output:

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x)
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.02 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=-\frac {3 \, {\left (30 \, a^{6} - {\left (8 \, b^{6} p^{5} + 60 \, b^{6} p^{4} + 170 \, b^{6} p^{3} + 225 \, b^{6} p^{2} + 137 \, b^{6} p + 30 \, b^{6}\right )} x^{2} - 20 \, {\left (2 \, a^{3} b^{3} p^{3} + 3 \, a^{3} b^{3} p^{2} + a^{3} b^{3} p\right )} x + 2 \, {\left (30 \, a^{4} b^{2} p^{2} + 15 \, a^{4} b^{2} p - {\left (4 \, a b^{5} p^{5} + 20 \, a b^{5} p^{4} + 35 \, a b^{5} p^{3} + 25 \, a b^{5} p^{2} + 6 \, a b^{5} p\right )} x\right )} x^{\frac {2}{3}} - 5 \, {\left (12 \, a^{5} b p - {\left (4 \, a^{2} b^{4} p^{4} + 12 \, a^{2} b^{4} p^{3} + 11 \, a^{2} b^{4} p^{2} + 3 \, a^{2} b^{4} p\right )} x\right )} x^{\frac {1}{3}}\right )} {\left (b^{2} x^{\frac {2}{3}} + 2 \, a b x^{\frac {1}{3}} + a^{2}\right )}^{p}}{2 \, {\left (8 \, b^{6} p^{6} + 84 \, b^{6} p^{5} + 350 \, b^{6} p^{4} + 735 \, b^{6} p^{3} + 812 \, b^{6} p^{2} + 441 \, b^{6} p + 90 \, b^{6}\right )}} \] Input:

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x, algorithm="fricas")
 

Output:

-3/2*(30*a^6 - (8*b^6*p^5 + 60*b^6*p^4 + 170*b^6*p^3 + 225*b^6*p^2 + 137*b 
^6*p + 30*b^6)*x^2 - 20*(2*a^3*b^3*p^3 + 3*a^3*b^3*p^2 + a^3*b^3*p)*x + 2* 
(30*a^4*b^2*p^2 + 15*a^4*b^2*p - (4*a*b^5*p^5 + 20*a*b^5*p^4 + 35*a*b^5*p^ 
3 + 25*a*b^5*p^2 + 6*a*b^5*p)*x)*x^(2/3) - 5*(12*a^5*b*p - (4*a^2*b^4*p^4 
+ 12*a^2*b^4*p^3 + 11*a^2*b^4*p^2 + 3*a^2*b^4*p)*x)*x^(1/3))*(b^2*x^(2/3) 
+ 2*a*b*x^(1/3) + a^2)^p/(8*b^6*p^6 + 84*b^6*p^5 + 350*b^6*p^4 + 735*b^6*p 
^3 + 812*b^6*p^2 + 441*b^6*p + 90*b^6)
 

Sympy [F]

\[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\int x \left (\left (a + b \sqrt [3]{x}\right )^{2}\right )^{p}\, dx \] Input:

integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**p*x,x)
 

Output:

Integral(x*((a + b*x**(1/3))**2)**p, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.68 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \, {\left ({\left (8 \, p^{5} + 60 \, p^{4} + 170 \, p^{3} + 225 \, p^{2} + 137 \, p + 30\right )} b^{6} x^{2} + 2 \, {\left (4 \, p^{5} + 20 \, p^{4} + 35 \, p^{3} + 25 \, p^{2} + 6 \, p\right )} a b^{5} x^{\frac {5}{3}} - 5 \, {\left (4 \, p^{4} + 12 \, p^{3} + 11 \, p^{2} + 3 \, p\right )} a^{2} b^{4} x^{\frac {4}{3}} + 20 \, {\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a^{3} b^{3} x - 30 \, {\left (2 \, p^{2} + p\right )} a^{4} b^{2} x^{\frac {2}{3}} + 60 \, a^{5} b p x^{\frac {1}{3}} - 30 \, a^{6}\right )} {\left (b x^{\frac {1}{3}} + a\right )}^{2 \, p}}{2 \, {\left (8 \, p^{6} + 84 \, p^{5} + 350 \, p^{4} + 735 \, p^{3} + 812 \, p^{2} + 441 \, p + 90\right )} b^{6}} \] Input:

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x, algorithm="maxima")
 

Output:

3/2*((8*p^5 + 60*p^4 + 170*p^3 + 225*p^2 + 137*p + 30)*b^6*x^2 + 2*(4*p^5 
+ 20*p^4 + 35*p^3 + 25*p^2 + 6*p)*a*b^5*x^(5/3) - 5*(4*p^4 + 12*p^3 + 11*p 
^2 + 3*p)*a^2*b^4*x^(4/3) + 20*(2*p^3 + 3*p^2 + p)*a^3*b^3*x - 30*(2*p^2 + 
 p)*a^4*b^2*x^(2/3) + 60*a^5*b*p*x^(1/3) - 30*a^6)*(b*x^(1/3) + a)^(2*p)/( 
(8*p^6 + 84*p^5 + 350*p^4 + 735*p^3 + 812*p^2 + 441*p + 90)*b^6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 745 vs. \(2 (252) = 504\).

Time = 0.14 (sec) , antiderivative size = 745, normalized size of antiderivative = 2.55 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx =\text {Too large to display} \] Input:

integrate((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x, algorithm="giac")
 

Output:

3/2*(8*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^6*p^5*x^2 + 8*(b^2*x^(2/3) 
+ 2*a*b*x^(1/3) + a^2)^p*a*b^5*p^5*x^(5/3) + 60*(b^2*x^(2/3) + 2*a*b*x^(1/ 
3) + a^2)^p*b^6*p^4*x^2 + 40*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^5*p 
^4*x^(5/3) - 20*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^4*p^4*x^(4/3) 
+ 170*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^6*p^3*x^2 + 70*(b^2*x^(2/3) 
+ 2*a*b*x^(1/3) + a^2)^p*a*b^5*p^3*x^(5/3) - 60*(b^2*x^(2/3) + 2*a*b*x^(1/ 
3) + a^2)^p*a^2*b^4*p^3*x^(4/3) + 40*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p 
*a^3*b^3*p^3*x + 225*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^6*p^2*x^2 + 5 
0*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^5*p^2*x^(5/3) - 55*(b^2*x^(2/3 
) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^4*p^2*x^(4/3) + 60*(b^2*x^(2/3) + 2*a*b*x 
^(1/3) + a^2)^p*a^3*b^3*p^2*x + 137*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p* 
b^6*p*x^2 - 60*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^2*p^2*x^(2/3) + 
 12*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^5*p*x^(5/3) - 15*(b^2*x^(2/3 
) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^4*p*x^(4/3) + 20*(b^2*x^(2/3) + 2*a*b*x^( 
1/3) + a^2)^p*a^3*b^3*p*x + 30*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^6*x 
^2 - 30*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^2*p*x^(2/3) + 60*(b^2* 
x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^5*b*p*x^(1/3) - 30*(b^2*x^(2/3) + 2*a*b 
*x^(1/3) + a^2)^p*a^6)/(8*b^6*p^6 + 84*b^6*p^5 + 350*b^6*p^4 + 735*b^6*p^3 
 + 812*b^6*p^2 + 441*b^6*p + 90*b^6)
 

Mupad [B] (verification not implemented)

Time = 11.94 (sec) , antiderivative size = 390, normalized size of antiderivative = 1.34 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx={\left (a^2+b^2\,x^{2/3}+2\,a\,b\,x^{1/3}\right )}^p\,\left (\frac {3\,x^2\,\left (8\,p^5+60\,p^4+170\,p^3+225\,p^2+137\,p+30\right )}{2\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {45\,a^6}{b^6\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {90\,a^5\,p\,x^{1/3}}{b^5\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {15\,a^2\,p\,x^{4/3}\,\left (4\,p^3+12\,p^2+11\,p+3\right )}{2\,b^2\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {30\,a^3\,p\,x\,\left (2\,p^2+3\,p+1\right )}{b^3\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}-\frac {45\,a^4\,p\,x^{2/3}\,\left (2\,p+1\right )}{b^4\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}+\frac {3\,a\,p\,x^{5/3}\,\left (4\,p^4+20\,p^3+35\,p^2+25\,p+6\right )}{b\,\left (8\,p^6+84\,p^5+350\,p^4+735\,p^3+812\,p^2+441\,p+90\right )}\right ) \] Input:

int(x*(a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^p,x)
 

Output:

(a^2 + b^2*x^(2/3) + 2*a*b*x^(1/3))^p*((3*x^2*(137*p + 225*p^2 + 170*p^3 + 
 60*p^4 + 8*p^5 + 30))/(2*(441*p + 812*p^2 + 735*p^3 + 350*p^4 + 84*p^5 + 
8*p^6 + 90)) - (45*a^6)/(b^6*(441*p + 812*p^2 + 735*p^3 + 350*p^4 + 84*p^5 
 + 8*p^6 + 90)) + (90*a^5*p*x^(1/3))/(b^5*(441*p + 812*p^2 + 735*p^3 + 350 
*p^4 + 84*p^5 + 8*p^6 + 90)) - (15*a^2*p*x^(4/3)*(11*p + 12*p^2 + 4*p^3 + 
3))/(2*b^2*(441*p + 812*p^2 + 735*p^3 + 350*p^4 + 84*p^5 + 8*p^6 + 90)) + 
(30*a^3*p*x*(3*p + 2*p^2 + 1))/(b^3*(441*p + 812*p^2 + 735*p^3 + 350*p^4 + 
 84*p^5 + 8*p^6 + 90)) - (45*a^4*p*x^(2/3)*(2*p + 1))/(b^4*(441*p + 812*p^ 
2 + 735*p^3 + 350*p^4 + 84*p^5 + 8*p^6 + 90)) + (3*a*p*x^(5/3)*(25*p + 35* 
p^2 + 20*p^3 + 4*p^4 + 6))/(b*(441*p + 812*p^2 + 735*p^3 + 350*p^4 + 84*p^ 
5 + 8*p^6 + 90)))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.05 \[ \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x \, dx=\frac {3 \left (x^{\frac {2}{3}} b^{2}+2 x^{\frac {1}{3}} a b +a^{2}\right )^{p} \left (-60 x^{\frac {2}{3}} a^{4} b^{2} p^{2}-30 x^{\frac {2}{3}} a^{4} b^{2} p +8 x^{\frac {5}{3}} a \,b^{5} p^{5}+40 x^{\frac {5}{3}} a \,b^{5} p^{4}+70 x^{\frac {5}{3}} a \,b^{5} p^{3}+50 x^{\frac {5}{3}} a \,b^{5} p^{2}+12 x^{\frac {5}{3}} a \,b^{5} p +60 x^{\frac {1}{3}} a^{5} b p -20 x^{\frac {4}{3}} a^{2} b^{4} p^{4}-60 x^{\frac {4}{3}} a^{2} b^{4} p^{3}-55 x^{\frac {4}{3}} a^{2} b^{4} p^{2}-15 x^{\frac {4}{3}} a^{2} b^{4} p -30 a^{6}+40 a^{3} b^{3} p^{3} x +60 a^{3} b^{3} p^{2} x +20 a^{3} b^{3} p x +8 b^{6} p^{5} x^{2}+60 b^{6} p^{4} x^{2}+170 b^{6} p^{3} x^{2}+225 b^{6} p^{2} x^{2}+137 b^{6} p \,x^{2}+30 b^{6} x^{2}\right )}{2 b^{6} \left (8 p^{6}+84 p^{5}+350 p^{4}+735 p^{3}+812 p^{2}+441 p +90\right )} \] Input:

int((a^2+2*a*b*x^(1/3)+b^2*x^(2/3))^p*x,x)
 

Output:

(3*(x**(2/3)*b**2 + 2*x**(1/3)*a*b + a**2)**p*( - 60*x**(2/3)*a**4*b**2*p* 
*2 - 30*x**(2/3)*a**4*b**2*p + 8*x**(2/3)*a*b**5*p**5*x + 40*x**(2/3)*a*b* 
*5*p**4*x + 70*x**(2/3)*a*b**5*p**3*x + 50*x**(2/3)*a*b**5*p**2*x + 12*x** 
(2/3)*a*b**5*p*x + 60*x**(1/3)*a**5*b*p - 20*x**(1/3)*a**2*b**4*p**4*x - 6 
0*x**(1/3)*a**2*b**4*p**3*x - 55*x**(1/3)*a**2*b**4*p**2*x - 15*x**(1/3)*a 
**2*b**4*p*x - 30*a**6 + 40*a**3*b**3*p**3*x + 60*a**3*b**3*p**2*x + 20*a* 
*3*b**3*p*x + 8*b**6*p**5*x**2 + 60*b**6*p**4*x**2 + 170*b**6*p**3*x**2 + 
225*b**6*p**2*x**2 + 137*b**6*p*x**2 + 30*b**6*x**2))/(2*b**6*(8*p**6 + 84 
*p**5 + 350*p**4 + 735*p**3 + 812*p**2 + 441*p + 90))