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

Optimal. Leaf size=128 \[ \frac{\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+2)}-\frac{a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (2 p+3)}+\frac{a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+1)} \]

[Out]

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

_______________________________________________________________________________________

Rubi [A]  time = 0.288279, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129 \[ \frac{\left (a+b x^2\right )^4 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+2)}-\frac{a \left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{b^3 (2 p+3)}+\frac{a^2 \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{4 b^3 (p+1)} \]

Antiderivative was successfully verified.

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

[Out]

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

_______________________________________________________________________________________

Rubi in Sympy [A]  time = 43.1691, size = 114, normalized size = 0.89 \[ \frac{a^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p + 1}}{4 b^{3} \left (p + 1\right ) \left (p + 2\right ) \left (2 p + 3\right )} - \frac{a x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p + 1}}{2 b^{2} \left (p + 2\right ) \left (2 p + 3\right )} + \frac{x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p + 1}}{4 b \left (p + 2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

_______________________________________________________________________________________

Mathematica [A]  time = 0.0516488, size = 68, normalized size = 0.53 \[ \frac{\left (\left (a+b x^2\right )^2\right )^{p+1} \left (a^2-2 a b (p+1) x^2+b^2 \left (2 p^2+5 p+3\right ) x^4\right )}{4 b^3 (p+1) (p+2) (2 p+3)} \]

Antiderivative was successfully verified.

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

[Out]

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

_______________________________________________________________________________________

Maple [A]  time = 0.009, size = 99, normalized size = 0.8 \[{\frac{ \left ( 2\,{b}^{2}{p}^{2}{x}^{4}+5\,{b}^{2}p{x}^{4}+3\,{b}^{2}{x}^{4}-2\,abp{x}^{2}-2\,ab{x}^{2}+{a}^{2} \right ) \left ( b{x}^{2}+a \right ) ^{2} \left ({b}^{2}{x}^{4}+2\,ab{x}^{2}+{a}^{2} \right ) ^{p}}{4\,{b}^{3} \left ( 2\,{p}^{3}+9\,{p}^{2}+13\,p+6 \right ) }} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/4*(b*x^2+a)^2*(2*b^2*p^2*x^4+5*b^2*p*x^4+3*b^2*x^4-2*a*b*p*x^2-2*a*b*x^2+a^2)*
(b^2*x^4+2*a*b*x^2+a^2)^p/b^3/(2*p^3+9*p^2+13*p+6)

_______________________________________________________________________________________

Maxima [A]  time = 0.717756, size = 265, normalized size = 2.07 \[ \frac{{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} +{\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )}{\left (b x^{2} + a\right )}^{2 \, p} a}{2 \,{\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} + \frac{{\left ({\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{4} x^{8} + 2 \,{\left (2 \, p^{3} + 3 \, p^{2} + p\right )} a b^{3} x^{6} - 3 \,{\left (2 \, p^{2} + p\right )} a^{2} b^{2} x^{4} + 6 \, a^{3} b p x^{2} - 3 \, a^{4}\right )}{\left (b x^{2} + a\right )}^{2 \, p}}{4 \,{\left (4 \, p^{4} + 20 \, p^{3} + 35 \, p^{2} + 25 \, p + 6\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((b*x^2 + a)*(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)*a/((4*p^3 + 12*p^2 + 11*p + 3)*b^3) + 1/4*((4*p^3 + 12*p^2 + 11*
p + 3)*b^4*x^8 + 2*(2*p^3 + 3*p^2 + p)*a*b^3*x^6 - 3*(2*p^2 + p)*a^2*b^2*x^4 + 6
*a^3*b*p*x^2 - 3*a^4)*(b*x^2 + a)^(2*p)/((4*p^4 + 20*p^3 + 35*p^2 + 25*p + 6)*b^
3)

_______________________________________________________________________________________

Fricas [A]  time = 0.277798, size = 189, normalized size = 1.48 \[ \frac{{\left ({\left (2 \, b^{4} p^{2} + 5 \, b^{4} p + 3 \, b^{4}\right )} x^{8} - 2 \, a^{3} b p x^{2} + 4 \,{\left (a b^{3} p^{2} + 2 \, a b^{3} p + a b^{3}\right )} x^{6} +{\left (2 \, a^{2} b^{2} p^{2} + a^{2} b^{2} p\right )} x^{4} + a^{4}\right )}{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{4 \,{\left (2 \, b^{3} p^{3} + 9 \, b^{3} p^{2} + 13 \, b^{3} p + 6 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/4*((2*b^4*p^2 + 5*b^4*p + 3*b^4)*x^8 - 2*a^3*b*p*x^2 + 4*(a*b^3*p^2 + 2*a*b^3*
p + a*b^3)*x^6 + (2*a^2*b^2*p^2 + a^2*b^2*p)*x^4 + a^4)*(b^2*x^4 + 2*a*b*x^2 + a
^2)^p/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)

_______________________________________________________________________________________

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*x**2+a)*(b**2*x**4+2*a*b*x**2+a**2)**p,x)

[Out]

Exception raised: TypeError

_______________________________________________________________________________________

GIAC/XCAS [A]  time = 0.270142, size = 474, normalized size = 3.7 \[ \frac{2 \, b^{4} p^{2} x^{8} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 5 \, b^{4} p x^{8} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 4 \, a b^{3} p^{2} x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 3 \, b^{4} x^{8} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 8 \, a b^{3} p x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 2 \, a^{2} b^{2} p^{2} x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + 4 \, a b^{3} x^{6} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + a^{2} b^{2} p x^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} - 2 \, a^{3} b p x^{2} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )} + a^{4} e^{\left (p{\rm ln}\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )\right )}}{4 \,{\left (2 \, b^{3} p^{3} + 9 \, b^{3} p^{2} + 13 \, b^{3} p + 6 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/4*(2*b^4*p^2*x^8*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + 5*b^4*p*x^8*e^(p*ln(b^2
*x^4 + 2*a*b*x^2 + a^2)) + 4*a*b^3*p^2*x^6*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) +
3*b^4*x^8*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + 8*a*b^3*p*x^6*e^(p*ln(b^2*x^4 +
2*a*b*x^2 + a^2)) + 2*a^2*b^2*p^2*x^4*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + 4*a
*b^3*x^6*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + a^2*b^2*p*x^4*e^(p*ln(b^2*x^4 + 2
*a*b*x^2 + a^2)) - 2*a^3*b*p*x^2*e^(p*ln(b^2*x^4 + 2*a*b*x^2 + a^2)) + a^4*e^(p*
ln(b^2*x^4 + 2*a*b*x^2 + a^2)))/(2*b^3*p^3 + 9*b^3*p^2 + 13*b^3*p + 6*b^3)