3.799 \(\int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx\)

Optimal. Leaf size=130 \[ \frac{\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+3)}-\frac{a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (p+1)}+\frac{a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+1)} \]

[Out]

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

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Rubi [A]  time = 0.193112, antiderivative size = 130, 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 \[ \frac{\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+3)}-\frac{a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (p+1)}+\frac{a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+1)} \]

Antiderivative was successfully verified.

[In]  Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^p,x]

[Out]

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

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Rubi in Sympy [A]  time = 27.5382, size = 128, normalized size = 0.98 \[ \frac{a^{2} \left (2 a + 2 b x^{2}\right ) \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{2 b^{3} \left (2 p + 1\right ) \left (2 p + 3\right )} - \frac{a \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p + 1}}{2 b^{3} \left (p + 1\right ) \left (2 p + 3\right )} + \frac{x^{4} \left (2 a + 2 b x^{2}\right ) \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{4 b \left (2 p + 3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

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

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Mathematica [A]  time = 0.0501561, size = 77, normalized size = 0.59 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^p \left (a^2-a b (2 p+1) x^2+b^2 \left (2 p^2+3 p+1\right ) x^4\right )}{2 b^3 (p+1) (2 p+1) (2 p+3)} \]

Antiderivative was successfully verified.

[In]  Integrate[x^5*(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^2 - a*b*(1 + 2*p)*x^2 + b^2*(1 + 3*p + 2*p^2)*
x^4))/(2*b^3*(1 + p)*(1 + 2*p)*(3 + 2*p))

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Maple [A]  time = 0.01, size = 96, normalized size = 0.7 \[{\frac{ \left ( 2\,{b}^{2}{p}^{2}{x}^{4}+3\,{b}^{2}p{x}^{4}+{b}^{2}{x}^{4}-2\,abp{x}^{2}-ab{x}^{2}+{a}^{2} \right ) \left ( b{x}^{2}+a \right ) \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}}{2\,{b}^{3} \left ( 4\,{p}^{3}+12\,{p}^{2}+11\,p+3 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,x)

[Out]

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

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Maxima [A]  time = 0.69588, size = 107, normalized size = 0.82 \[ \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} +{\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{2 \,{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b^2*x^4 + 2*a*b*x^2 + a^2)^p*x^5,x, algorithm="maxima")

[Out]

1/2*((2*p^2 + 3*p + 1)*b^3*x^6 + (2*p^2 + p)*a*b^2*x^4 - 2*a^2*b*p*x^2 + a^3)*(b
*x^2 + a)^(2*p)/((4*p^3 + 12*p^2 + 11*p + 3)*b^3)

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Fricas [A]  time = 0.289216, size = 146, normalized size = 1.12 \[ \frac{{\left ({\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} +{\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{4} + a^{3}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{2 \,{\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

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^5,x, algorithm="fricas")

[Out]

1/2*((2*b^3*p^2 + 3*b^3*p + b^3)*x^6 - 2*a^2*b*p*x^2 + (2*a*b^2*p^2 + a*b^2*p)*x
^4 + a^3)*(b^2*x^4 + 2*a*b*x^2 + 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(-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**5*(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 = 0.268915, size = 336, normalized size = 2.58 \[ \frac{2 \, b^{3} p^{2} x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 3 \, b^{3} p x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 2 \, a b^{2} p^{2} x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + b^{3} x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + a b^{2} p x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} - 2 \, a^{2} b p x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + a^{3} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )}}{2 \,{\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

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^5,x, algorithm="giac")

[Out]

1/2*(2*b^3*p^2*x^6*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + 3*b^3*p*x^6*e^(p*ln(b^2
*x^4 + 2*a*b*x^2 + a^2)) + 2*a*b^2*p^2*x^4*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) +
b^3*x^6*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + a*b^2*p*x^4*e^(p*ln(b^2*x^4 + 2*a
*b*x^2 + a^2)) - 2*a^2*b*p*x^2*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + a^3*e^(p*ln
(b^2*x^4 + 2*a*b*x^2 + a^2)))/(4*b^3*p^3 + 12*b^3*p^2 + 11*b^3*p + 3*b^3)