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

Optimal. Leaf size=315 \[ \frac{3 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+3)}-\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+5)}+\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (p+2)}-\frac{30 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+3)}+\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+1)}-\frac{3 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.139651, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1356, 266, 43} \[ \frac{3 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^6 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+3)}-\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^5 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+5)}+\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^4 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (p+2)}-\frac{30 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+3)}+\frac{15 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{2 b^6 (p+1)}-\frac{3 a^6 \left (\frac{b \sqrt [3]{x}}{a}+1\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^6 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^6*(1 + (b*x^(1/3))/a)*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^6*(1 + 2*p)) + (15*a^6*(1 + (b*x^(1/3))/
a)^2*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(2*b^6*(1 + p)) - (30*a^6*(1 + (b*x^(1/3))/a)^3*(a^2 + 2*a*b*x^(1/
3) + b^2*x^(2/3))^p)/(b^6*(3 + 2*p)) + (15*a^6*(1 + (b*x^(1/3))/a)^4*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b
^6*(2 + p)) - (15*a^6*(1 + (b*x^(1/3))/a)^5*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(b^6*(5 + 2*p)) + (3*a^6*(1
+ (b*x^(1/3))/a)^6*(a^2 + 2*a*b*x^(1/3) + b^2*x^(2/3))^p)/(2*b^6*(3 + 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 \, 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 \, 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^5 \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^5 \left (1+\frac{b x}{a}\right )^{2 p}}{b^5}+\frac{5 a^5 \left (1+\frac{b x}{a}\right )^{1+2 p}}{b^5}-\frac{10 a^5 \left (1+\frac{b x}{a}\right )^{2+2 p}}{b^5}+\frac{10 a^5 \left (1+\frac{b x}{a}\right )^{3+2 p}}{b^5}-\frac{5 a^5 \left (1+\frac{b x}{a}\right )^{4+2 p}}{b^5}+\frac{a^5 \left (1+\frac{b x}{a}\right )^{5+2 p}}{b^5}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=-\frac{3 a^6 \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^6 (1+2 p)}+\frac{15 a^6 \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}{2 b^6 (1+p)}-\frac{30 a^6 \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^6 (3+2 p)}+\frac{15 a^6 \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^6 (2+p)}-\frac{15 a^6 \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^6 (5+2 p)}+\frac{3 a^6 \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}{2 b^6 (3+p)}\\ \end{align*}

Mathematica [A]  time = 0.185468, size = 143, normalized size = 0.45 \[ \frac{3 \left (a+b \sqrt [3]{x}\right ) \left (\frac{5 a^4 \left (a+b \sqrt [3]{x}\right )}{p+1}-\frac{20 a^3 \left (a+b \sqrt [3]{x}\right )^2}{2 p+3}+\frac{10 a^2 \left (a+b \sqrt [3]{x}\right )^3}{p+2}-\frac{2 a^5}{2 p+1}-\frac{10 a \left (a+b \sqrt [3]{x}\right )^4}{2 p+5}+\frac{\left (a+b \sqrt [3]{x}\right )^5}{p+3}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p}{2 b^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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)

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Maple [F]  time = 0.006, 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\, 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,x)

[Out]

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

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Maxima [A]  time = 0.994342, size = 267, normalized size = 0.85 \begin{align*} \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}} \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,x, algorithm="maxima")

[Out]

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)

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Fricas [A]  time = 2.5916, size = 653, normalized size = 2.07 \begin{align*} -\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 )}} \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,x, algorithm="fricas")

[Out]

-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)

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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,x)

[Out]

Timed out

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Giac [B]  time = 1.14164, size = 1006, normalized size = 3.19 \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,x, algorithm="giac")

[Out]

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 + 50*(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)