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

Optimal. Leaf size=468 \[ \frac{3 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^9 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+9)}-\frac{12 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^8 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+4)}+\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^7 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+7)}-\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+3)}+\frac{210 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+5)}-\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+2)}+\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+3)}-\frac{12 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+1)}+\frac{3 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+1)} \]

[Out]

(3*a^9*(1 + (b*x^(1/3))/a)*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(1 + 2*p)) - (12*a^9*(1 + (b*x^(1/3))/a
)^2*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(1 + p)) + (84*a^9*(1 + (b*x^(1/3))/a)^3*(a^2 + 2*a*b*x^(1/3)
+ b^2*x^(2/3))^p)/(b^9*(3 + 2*p)) - (84*a^9*(1 + (b*x^(1/3))/a)^4*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*
(2 + p)) + (210*a^9*(1 + (b*x^(1/3))/a)^5*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(5 + 2*p)) - (84*a^9*(1
+ (b*x^(1/3))/a)^6*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(3 + p)) + (84*a^9*(1 + (b*x^(1/3))/a)^7*(a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(7 + 2*p)) - (12*a^9*(1 + (b*x^(1/3))/a)^8*(a^2 + 2*a*b*x^(1/3) + b^2*x^
(2/3))^p)/(b^9*(4 + p)) + (3*a^9*(1 + (b*x^(1/3))/a)^9*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(9 + 2*p))

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Rubi [A]  time = 0.224245, antiderivative size = 468, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {1356, 266, 43} \[ \frac{3 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^9 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+9)}-\frac{12 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^8 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+4)}+\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^7 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+7)}-\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+3)}+\frac{210 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+5)}-\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+2)}+\frac{84 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+3)}-\frac{12 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (p+1)}+\frac{3 a^9 \left (\frac{b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^9*(1 + (b*x^(1/3))/a)*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(1 + 2*p)) - (12*a^9*(1 + (b*x^(1/3))/a
)^2*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(1 + p)) + (84*a^9*(1 + (b*x^(1/3))/a)^3*(a^2 + 2*a*b*x^(1/3)
+ b^2*x^(2/3))^p)/(b^9*(3 + 2*p)) - (84*a^9*(1 + (b*x^(1/3))/a)^4*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*
(2 + p)) + (210*a^9*(1 + (b*x^(1/3))/a)^5*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(5 + 2*p)) - (84*a^9*(1
+ (b*x^(1/3))/a)^6*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(3 + p)) + (84*a^9*(1 + (b*x^(1/3))/a)^7*(a^2 +
2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(7 + 2*p)) - (12*a^9*(1 + (b*x^(1/3))/a)^8*(a^2 + 2*a*b*x^(1/3) + b^2*x^
(2/3))^p)/(b^9*(4 + p)) + (3*a^9*(1 + (b*x^(1/3))/a)^9*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^9*(9 + 2*p))

Rule 1356

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 43

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] && NeQ[b*c - a*d, 0] && 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])

Rubi steps

\begin{align*} \int \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p x^2 \, dx &=\left (\left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \int \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{2 p} x^2 \, dx\\ &=\left (3 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \operatorname{Subst}\left (\int x^8 \left (1+\frac{b x}{a}\right )^{2 p} \, dx,x,\sqrt [3]{x}\right )\\ &=\left (3 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^{-2 p} \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p\right ) \operatorname{Subst}\left (\int \left (\frac{a^8 \left (1+\frac{b x}{a}\right )^{2 p}}{b^8}-\frac{8 a^8 \left (1+\frac{b x}{a}\right )^{1+2 p}}{b^8}+\frac{28 a^8 \left (1+\frac{b x}{a}\right )^{2+2 p}}{b^8}-\frac{56 a^8 \left (1+\frac{b x}{a}\right )^{3+2 p}}{b^8}+\frac{70 a^8 \left (1+\frac{b x}{a}\right )^{4+2 p}}{b^8}-\frac{56 a^8 \left (1+\frac{b x}{a}\right )^{5+2 p}}{b^8}+\frac{28 a^8 \left (1+\frac{b x}{a}\right )^{6+2 p}}{b^8}-\frac{8 a^8 \left (1+\frac{b x}{a}\right )^{7+2 p}}{b^8}+\frac{a^8 \left (1+\frac{b x}{a}\right )^{8+2 p}}{b^8}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{3 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (1+2 p)}-\frac{12 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (1+p)}+\frac{84 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (3+2 p)}-\frac{84 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (2+p)}+\frac{210 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (5+2 p)}-\frac{84 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (3+p)}+\frac{84 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^7 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (7+2 p)}-\frac{12 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^8 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (4+p)}+\frac{3 a^9 \left (1+\frac{b \sqrt [3]{x}}{a}\right )^9 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^9 (9+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.224451, size = 207, normalized size = 0.44 \[ \frac{3 \left (a+b \sqrt [3]{x}\right ) \left (-\frac{4 a^7 \left (a+b \sqrt [3]{x}\right )}{p+1}+\frac{28 a^6 \left (a+b \sqrt [3]{x}\right )^2}{2 p+3}-\frac{28 a^5 \left (a+b \sqrt [3]{x}\right )^3}{p+2}+\frac{70 a^4 \left (a+b \sqrt [3]{x}\right )^4}{2 p+5}-\frac{28 a^3 \left (a+b \sqrt [3]{x}\right )^5}{p+3}+\frac{28 a^2 \left (a+b \sqrt [3]{x}\right )^6}{2 p+7}+\frac{a^8}{2 p+1}-\frac{4 a \left (a+b \sqrt [3]{x}\right )^7}{p+4}+\frac{\left (a+b \sqrt [3]{x}\right )^8}{2 p+9}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p}{b^9} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a^8/(1 + 2*p) - (4*a^7*(a + b*x^(1/3)))/(1 + p) + (28*a^6*(a + b*x^(1/3))^2)/(3 + 2*p) - (28*a^5*(a + b*x^
(1/3))^3)/(2 + p) + (70*a^4*(a + b*x^(1/3))^4)/(5 + 2*p) - (28*a^3*(a + b*x^(1/3))^5)/(3 + p) + (28*a^2*(a + b
*x^(1/3))^6)/(7 + 2*p) - (4*a*(a + b*x^(1/3))^7)/(4 + p) + (a + b*x^(1/3))^8/(9 + 2*p))*(a + b*x^(1/3))*((a +
b*x^(1/3))^2)^p)/b^9

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Maple [F]  time = 0.007, size = 0, normalized size = 0. \begin{align*} \int \left ({a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}} \right ) ^{p}{x}^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06306, size = 489, normalized size = 1.04 \begin{align*} \frac{3 \,{\left ({\left (16 \, p^{8} + 288 \, p^{7} + 2184 \, p^{6} + 9072 \, p^{5} + 22449 \, p^{4} + 33642 \, p^{3} + 29531 \, p^{2} + 13698 \, p + 2520\right )} b^{9} x^{3} +{\left (16 \, p^{8} + 224 \, p^{7} + 1288 \, p^{6} + 3920 \, p^{5} + 6769 \, p^{4} + 6566 \, p^{3} + 3267 \, p^{2} + 630 \, p\right )} a b^{8} x^{\frac{8}{3}} - 8 \,{\left (8 \, p^{7} + 84 \, p^{6} + 350 \, p^{5} + 735 \, p^{4} + 812 \, p^{3} + 441 \, p^{2} + 90 \, p\right )} a^{2} b^{7} x^{\frac{7}{3}} + 28 \,{\left (8 \, p^{6} + 60 \, p^{5} + 170 \, p^{4} + 225 \, p^{3} + 137 \, p^{2} + 30 \, p\right )} a^{3} b^{6} x^{2} - 168 \,{\left (4 \, p^{5} + 20 \, p^{4} + 35 \, p^{3} + 25 \, p^{2} + 6 \, p\right )} a^{4} b^{5} x^{\frac{5}{3}} + 420 \,{\left (4 \, p^{4} + 12 \, p^{3} + 11 \, p^{2} + 3 \, p\right )} a^{5} b^{4} x^{\frac{4}{3}} - 1680 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a^{6} b^{3} x + 2520 \,{\left (2 \, p^{2} + p\right )} a^{7} b^{2} x^{\frac{2}{3}} - 5040 \, a^{8} b p x^{\frac{1}{3}} + 2520 \, a^{9}\right )}{\left (b x^{\frac{1}{3}} + a\right )}^{2 \, p}}{{\left (32 \, p^{9} + 720 \, p^{8} + 6960 \, p^{7} + 37800 \, p^{6} + 126546 \, p^{5} + 269325 \, p^{4} + 361840 \, p^{3} + 293175 \, p^{2} + 128322 \, p + 22680\right )} b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*((16*p^8 + 288*p^7 + 2184*p^6 + 9072*p^5 + 22449*p^4 + 33642*p^3 + 29531*p^2 + 13698*p + 2520)*b^9*x^3 + (16
*p^8 + 224*p^7 + 1288*p^6 + 3920*p^5 + 6769*p^4 + 6566*p^3 + 3267*p^2 + 630*p)*a*b^8*x^(8/3) - 8*(8*p^7 + 84*p
^6 + 350*p^5 + 735*p^4 + 812*p^3 + 441*p^2 + 90*p)*a^2*b^7*x^(7/3) + 28*(8*p^6 + 60*p^5 + 170*p^4 + 225*p^3 +
137*p^2 + 30*p)*a^3*b^6*x^2 - 168*(4*p^5 + 20*p^4 + 35*p^3 + 25*p^2 + 6*p)*a^4*b^5*x^(5/3) + 420*(4*p^4 + 12*p
^3 + 11*p^2 + 3*p)*a^5*b^4*x^(4/3) - 1680*(2*p^3 + 3*p^2 + p)*a^6*b^3*x + 2520*(2*p^2 + p)*a^7*b^2*x^(2/3) - 5
040*a^8*b*p*x^(1/3) + 2520*a^9)*(b*x^(1/3) + a)^(2*p)/((32*p^9 + 720*p^8 + 6960*p^7 + 37800*p^6 + 126546*p^5 +
269325*p^4 + 361840*p^3 + 293175*p^2 + 128322*p + 22680)*b^9)

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Fricas [A]  time = 3.18854, size = 1354, normalized size = 2.89 \begin{align*} \frac{3 \,{\left (2520 \, a^{9} +{\left (16 \, b^{9} p^{8} + 288 \, b^{9} p^{7} + 2184 \, b^{9} p^{6} + 9072 \, b^{9} p^{5} + 22449 \, b^{9} p^{4} + 33642 \, b^{9} p^{3} + 29531 \, b^{9} p^{2} + 13698 \, b^{9} p + 2520 \, b^{9}\right )} x^{3} + 28 \,{\left (8 \, a^{3} b^{6} p^{6} + 60 \, a^{3} b^{6} p^{5} + 170 \, a^{3} b^{6} p^{4} + 225 \, a^{3} b^{6} p^{3} + 137 \, a^{3} b^{6} p^{2} + 30 \, a^{3} b^{6} p\right )} x^{2} - 1680 \,{\left (2 \, a^{6} b^{3} p^{3} + 3 \, a^{6} b^{3} p^{2} + a^{6} b^{3} p\right )} x +{\left (5040 \, a^{7} b^{2} p^{2} + 2520 \, a^{7} b^{2} p +{\left (16 \, a b^{8} p^{8} + 224 \, a b^{8} p^{7} + 1288 \, a b^{8} p^{6} + 3920 \, a b^{8} p^{5} + 6769 \, a b^{8} p^{4} + 6566 \, a b^{8} p^{3} + 3267 \, a b^{8} p^{2} + 630 \, a b^{8} p\right )} x^{2} - 168 \,{\left (4 \, a^{4} b^{5} p^{5} + 20 \, a^{4} b^{5} p^{4} + 35 \, a^{4} b^{5} p^{3} + 25 \, a^{4} b^{5} p^{2} + 6 \, a^{4} b^{5} p\right )} x\right )} x^{\frac{2}{3}} - 4 \,{\left (1260 \, a^{8} b p + 2 \,{\left (8 \, a^{2} b^{7} p^{7} + 84 \, a^{2} b^{7} p^{6} + 350 \, a^{2} b^{7} p^{5} + 735 \, a^{2} b^{7} p^{4} + 812 \, a^{2} b^{7} p^{3} + 441 \, a^{2} b^{7} p^{2} + 90 \, a^{2} b^{7} p\right )} x^{2} - 105 \,{\left (4 \, a^{5} b^{4} p^{4} + 12 \, a^{5} b^{4} p^{3} + 11 \, a^{5} b^{4} p^{2} + 3 \, a^{5} 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}}{32 \, b^{9} p^{9} + 720 \, b^{9} p^{8} + 6960 \, b^{9} p^{7} + 37800 \, b^{9} p^{6} + 126546 \, b^{9} p^{5} + 269325 \, b^{9} p^{4} + 361840 \, b^{9} p^{3} + 293175 \, b^{9} p^{2} + 128322 \, b^{9} p + 22680 \, b^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(2520*a^9 + (16*b^9*p^8 + 288*b^9*p^7 + 2184*b^9*p^6 + 9072*b^9*p^5 + 22449*b^9*p^4 + 33642*b^9*p^3 + 29531*
b^9*p^2 + 13698*b^9*p + 2520*b^9)*x^3 + 28*(8*a^3*b^6*p^6 + 60*a^3*b^6*p^5 + 170*a^3*b^6*p^4 + 225*a^3*b^6*p^3
+ 137*a^3*b^6*p^2 + 30*a^3*b^6*p)*x^2 - 1680*(2*a^6*b^3*p^3 + 3*a^6*b^3*p^2 + a^6*b^3*p)*x + (5040*a^7*b^2*p^
2 + 2520*a^7*b^2*p + (16*a*b^8*p^8 + 224*a*b^8*p^7 + 1288*a*b^8*p^6 + 3920*a*b^8*p^5 + 6769*a*b^8*p^4 + 6566*a
*b^8*p^3 + 3267*a*b^8*p^2 + 630*a*b^8*p)*x^2 - 168*(4*a^4*b^5*p^5 + 20*a^4*b^5*p^4 + 35*a^4*b^5*p^3 + 25*a^4*b
^5*p^2 + 6*a^4*b^5*p)*x)*x^(2/3) - 4*(1260*a^8*b*p + 2*(8*a^2*b^7*p^7 + 84*a^2*b^7*p^6 + 350*a^2*b^7*p^5 + 735
*a^2*b^7*p^4 + 812*a^2*b^7*p^3 + 441*a^2*b^7*p^2 + 90*a^2*b^7*p)*x^2 - 105*(4*a^5*b^4*p^4 + 12*a^5*b^4*p^3 + 1
1*a^5*b^4*p^2 + 3*a^5*b^4*p)*x)*x^(1/3))*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p/(32*b^9*p^9 + 720*b^9*p^8 + 696
0*b^9*p^7 + 37800*b^9*p^6 + 126546*b^9*p^5 + 269325*b^9*p^4 + 361840*b^9*p^3 + 293175*b^9*p^2 + 128322*b^9*p +
22680*b^9)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.18422, size = 2111, normalized size = 4.51 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*(16*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^8*x^3 + 16*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p^8*x
^(8/3) + 288*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^7*x^3 + 224*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b
^8*p^7*x^(8/3) - 64*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^7*p^7*x^(7/3) + 2184*(b^2*x^(2/3) + 2*a*b*x^(1
/3) + a^2)^p*b^9*p^6*x^3 + 1288*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p^6*x^(8/3) - 672*(b^2*x^(2/3) + 2
*a*b*x^(1/3) + a^2)^p*a^2*b^7*p^6*x^(7/3) + 224*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^3*b^6*p^6*x^2 + 9072*(
b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^5*x^3 + 3920*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p^5*x^(8/3
) - 2800*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^7*p^5*x^(7/3) + 1680*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^
p*a^3*b^6*p^5*x^2 + 22449*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^4*x^3 - 672*(b^2*x^(2/3) + 2*a*b*x^(1/3)
+ a^2)^p*a^4*b^5*p^5*x^(5/3) + 6769*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p^4*x^(8/3) - 5880*(b^2*x^(2/
3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^7*p^4*x^(7/3) + 4760*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^3*b^6*p^4*x^2 +
33642*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^3*x^3 - 3360*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^5*
p^4*x^(5/3) + 6566*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p^3*x^(8/3) + 1680*(b^2*x^(2/3) + 2*a*b*x^(1/3)
+ a^2)^p*a^5*b^4*p^4*x^(4/3) - 6496*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b^7*p^3*x^(7/3) + 6300*(b^2*x^(
2/3) + 2*a*b*x^(1/3) + a^2)^p*a^3*b^6*p^3*x^2 + 29531*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p^2*x^3 - 5880
*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^5*p^3*x^(5/3) + 3267*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*
p^2*x^(8/3) + 5040*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^5*b^4*p^3*x^(4/3) - 3528*(b^2*x^(2/3) + 2*a*b*x^(1/
3) + a^2)^p*a^2*b^7*p^2*x^(7/3) - 3360*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^6*b^3*p^3*x + 3836*(b^2*x^(2/3)
+ 2*a*b*x^(1/3) + a^2)^p*a^3*b^6*p^2*x^2 + 13698*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*p*x^3 - 4200*(b^2*
x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^5*p^2*x^(5/3) + 630*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^8*p*x^(8/
3) + 4620*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^5*b^4*p^2*x^(4/3) - 720*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^
p*a^2*b^7*p*x^(7/3) - 5040*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^6*b^3*p^2*x + 840*(b^2*x^(2/3) + 2*a*b*x^(1
/3) + a^2)^p*a^3*b^6*p*x^2 + 2520*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^9*x^3 + 5040*(b^2*x^(2/3) + 2*a*b*x^
(1/3) + a^2)^p*a^7*b^2*p^2*x^(2/3) - 1008*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^4*b^5*p*x^(5/3) + 1260*(b^2*
x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^5*b^4*p*x^(4/3) - 1680*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^6*b^3*p*x +
2520*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^7*b^2*p*x^(2/3) - 5040*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^8*
b*p*x^(1/3) + 2520*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^9)/(32*b^9*p^9 + 720*b^9*p^8 + 6960*b^9*p^7 + 37800
*b^9*p^6 + 126546*b^9*p^5 + 269325*b^9*p^4 + 361840*b^9*p^3 + 293175*b^9*p^2 + 128322*b^9*p + 22680*b^9)