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

Optimal. Leaf size=142 \[ \frac{3 \left (a+b \sqrt [3]{x}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+3)}-\frac{3 a \left (a+b \sqrt [3]{x}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (p+1)}+\frac{3 a^2 \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.0677279, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {1341, 646, 43} \[ \frac{3 \left (a+b \sqrt [3]{x}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+3)}-\frac{3 a \left (a+b \sqrt [3]{x}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (p+1)}+\frac{3 a^2 \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1341

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[I
nt[x^(k - 1)*(a + b*x^(k*n) + c*x^(2*k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] &
& FractionQ[n]

Rule 646

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

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 \, dx &=3 \operatorname{Subst}\left (\int x^2 \left (a^2+2 a b x+b^2 x^2\right )^p \, dx,x,\sqrt [3]{x}\right )\\ &=\left (3 \left (b \left (a+b \sqrt [3]{x}\right )\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^2 \left (a b+b^2 x\right )^{2 p} \, dx,x,\sqrt [3]{x}\right )\\ &=\left (3 \left (b \left (a+b \sqrt [3]{x}\right )\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^2 \left (a b+b^2 x\right )^{2 p}}{b^2}-\frac{2 a \left (a b+b^2 x\right )^{1+2 p}}{b^3}+\frac{\left (a b+b^2 x\right )^{2+2 p}}{b^4}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac{3 a^2 \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (1+2 p)}-\frac{3 a \left (a+b \sqrt [3]{x}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (1+p)}+\frac{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^3 (3+2 p)}\\ \end{align*}

Mathematica [A]  time = 0.0489338, size = 83, normalized size = 0.58 \[ \frac{3 \left (a+b \sqrt [3]{x}\right ) \left (\left (a+b \sqrt [3]{x}\right )^2\right )^p \left (a^2-a b (2 p+1) \sqrt [3]{x}+b^2 \left (2 p^2+3 p+1\right ) x^{2/3}\right )}{b^3 (p+1) (2 p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.005, size = 0, normalized size = 0. \begin{align*} \int \left ({a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}} \right ) ^{p}\, 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)

[Out]

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

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Maxima [A]  time = 0.972562, size = 104, normalized size = 0.73 \begin{align*} \frac{3 \,{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x +{\left (2 \, p^{2} + p\right )} a b^{2} x^{\frac{2}{3}} - 2 \, a^{2} b p x^{\frac{1}{3}} + a^{3}\right )}{\left (b x^{\frac{1}{3}} + a\right )}^{2 \, p}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \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, algorithm="maxima")

[Out]

3*((2*p^2 + 3*p + 1)*b^3*x + (2*p^2 + p)*a*b^2*x^(2/3) - 2*a^2*b*p*x^(1/3) + a^3)*(b*x^(1/3) + a)^(2*p)/((4*p^
3 + 12*p^2 + 11*p + 3)*b^3)

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Fricas [A]  time = 2.12405, size = 240, normalized size = 1.69 \begin{align*} -\frac{3 \,{\left (2 \, a^{2} b p x^{\frac{1}{3}} - a^{3} -{\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x -{\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{\frac{2}{3}}\right )}{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \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, algorithm="fricas")

[Out]

-3*(2*a^2*b*p*x^(1/3) - a^3 - (2*b^3*p^2 + 3*b^3*p + b^3)*x - (2*a*b^2*p^2 + a*b^2*p)*x^(2/3))*(b^2*x^(2/3) +
2*a*b*x^(1/3) + a^2)^p/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)

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

[Out]

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

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Giac [A]  time = 1.16262, size = 309, normalized size = 2.18 \begin{align*} \frac{3 \,{\left (2 \,{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} b^{3} p^{2} x + 2 \,{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} a b^{2} p^{2} x^{\frac{2}{3}} + 3 \,{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} b^{3} p x +{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} a b^{2} p x^{\frac{2}{3}} - 2 \,{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} a^{2} b p x^{\frac{1}{3}} +{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} b^{3} x +{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p} a^{3}\right )}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \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, algorithm="giac")

[Out]

3*(2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^3*p^2*x + 2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^2*p^2*x^(2/
3) + 3*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^3*p*x + (b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a*b^2*p*x^(2/3) -
2*(b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^2*b*p*x^(1/3) + (b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*b^3*x + (b^2*
x^(2/3) + 2*a*b*x^(1/3) + a^2)^p*a^3)/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)