3.800 \(\int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\)

Optimal. Leaf size=84 \[ \frac{\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.140804, antiderivative size = 84, 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 \[ \frac{\left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^2 (p+1)}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^2 (2 p+1)} \]

Antiderivative was successfully verified.

[In]  Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

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Rubi in Sympy [A]  time = 15.8737, size = 71, normalized size = 0.85 \[ - \frac{a \left (2 a + 2 b x^{2}\right ) \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b^{2} \left (2 p + 1\right )} + \frac{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p + 1}}{4 b^{2} \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

-a*(2*a + 2*b*x**2)*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(4*b**2*(2*p + 1)) + (a**
2 + 2*a*b*x**2 + b**2*x**4)**(p + 1)/(4*b**2*(p + 1))

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Mathematica [A]  time = 0.0309718, size = 51, normalized size = 0.61 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (b (2 p+1) x^2-a\right )}{4 b^2 (p+1) (2 p+1)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

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Maple [A]  time = 0.009, size = 60, normalized size = 0.7 \[ -{\frac{ \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p} \left ( -2\,{x}^{2}pb-b{x}^{2}+a \right ) \left ( b{x}^{2}+a \right ) }{4\,{b}^{2} \left ( 2\,{p}^{2}+3\,p+1 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

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

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Maxima [A]  time = 0.70667, size = 73, normalized size = 0.87 \[ \frac{{\left (b^{2}{\left (2 \, p + 1\right )} x^{4} + 2 \, a b p x^{2} - a^{2}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{4 \,{\left (2 \, p^{2} + 3 \, p + 1\right )} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.286955, size = 95, normalized size = 1.13 \[ \frac{{\left (2 \, a b p x^{2} +{\left (2 \, b^{2} p + b^{2}\right )} x^{4} - a^{2}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Exception raised: TypeError

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GIAC/XCAS [A]  time = 0.268182, size = 189, normalized size = 2.25 \[ \frac{2 \, b^{2} p x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + b^{2} x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 2 \, a b p x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} - a^{2} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )}}{4 \,{\left (2 \, b^{2} p^{2} + 3 \, b^{2} p + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^3,x, algorithm="giac")

[Out]

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