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

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 = 0.138828, 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 \[ \frac{3 \left (a+b \sqrt [3]{x}\right )^3 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+3)}-\frac{3 a \left (a+b \sqrt [3]{x}\right )^2 \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (p+1)}+\frac{3 a^2 \left (a+b \sqrt [3]{x}\right ) \left (a^2+2 a b \sqrt [3]{x}+b^2 x^{2/3}\right )^p}{b^3 (2 p+1)} \]

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))

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Rubi in Sympy [A]  time = 24.417, size = 144, normalized size = 1.01 \[ \frac{3 a^{2} \left (2 a + 2 b \sqrt [3]{x}\right ) \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac{2}{3}}\right )^{p}}{b^{3} \left (2 p + 1\right ) \left (2 p + 3\right )} - \frac{3 a \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac{2}{3}}\right )^{p + 1}}{b^{3} \left (p + 1\right ) \left (2 p + 3\right )} + \frac{3 x^{\frac{2}{3}} \left (2 a + 2 b \sqrt [3]{x}\right ) \left (a^{2} + 2 a b \sqrt [3]{x} + b^{2} x^{\frac{2}{3}}\right )^{p}}{2 b \left (2 p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((a**2+2*a*b*x**(1/3)+b**2*x**(2/3))**p,x)

[Out]

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

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Mathematica [A]  time = 0.0563148, 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 = 0.008, size = 0, normalized size = 0. \[ \int \left ({a}^{2}+2\,ab\sqrt [3]{x}+{b}^{2}{x}^{{\frac{2}{3}}} \right ) ^{p}\, dx \]

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 = 0.780029, size = 104, normalized size = 0.73 \[ \frac{3 \,{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x +{\left (2 \, p^{2} + p\right )} a b^{2} x^{\frac{2}{3}} - 2 \, a^{2} b p x^{\frac{1}{3}} + a^{3}\right )}{\left (b x^{\frac{1}{3}} + a\right )}^{2 \, p}}{{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^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 = 0.314024, size = 149, normalized size = 1.05 \[ -\frac{3 \,{\left (2 \, a^{2} b p x^{\frac{1}{3}} - a^{3} -{\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x -{\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{\frac{2}{3}}\right )}{\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )}^{p}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^(2/3) + 2*a*b*x^(1/3) + a^2)^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 + 1
1*b^3*p + 3*b^3)

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

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]

Timed out

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GIAC/XCAS [A]  time = 0.277844, size = 328, normalized size = 2.31 \[ \frac{3 \,{\left (2 \, b^{3} p^{2} x e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} + 2 \, a b^{2} p^{2} x^{\frac{2}{3}} e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} + 3 \, b^{3} p x e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} + a b^{2} p x^{\frac{2}{3}} e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} - 2 \, a^{2} b p x^{\frac{1}{3}} e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} + b^{3} x e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )} + a^{3} e^{\left (p{\rm ln}\left (b^{2} x^{\frac{2}{3}} + 2 \, a b x^{\frac{1}{3}} + a^{2}\right )\right )}\right )}}{4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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