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

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]

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

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Rubi [A]  time = 0.08, antiderivative size = 130, normalized size of antiderivative = 1.00, 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 = {1113, 266, 43} \[ \frac {\left (a+b x^2\right )^3 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+3)}-\frac {a \left (a+b x^2\right )^2 \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (p+1)}+\frac {a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^p}{2 b^3 (2 p+1)} \]

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))

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])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 1113

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^2 +
c*x^4)^FracPart[p])/(1 + (2*c*x^2)/b)^(2*FracPart[p]), Int[(d*x)^m*(1 + (2*c*x^2)/b)^(2*p), x], x] /; FreeQ[{
a, b, c, d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[2*p]

Rubi steps

\begin {align*} \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^p \, dx &=\left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \int x^5 \left (1+\frac {b x^2}{a}\right )^{2 p} \, dx\\ &=\frac {1}{2} \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int x^2 \left (1+\frac {b x}{a}\right )^{2 p} \, dx,x,x^2\right )\\ &=\frac {1}{2} \left (\left (1+\frac {b x^2}{a}\right )^{-2 p} \left (a^2+2 a b x^2+b^2 x^4\right )^p\right ) \operatorname {Subst}\left (\int \left (\frac {a^2 \left (1+\frac {b x}{a}\right )^{2 p}}{b^2}-\frac {2 a^2 \left (1+\frac {b x}{a}\right )^{1+2 p}}{b^2}+\frac {a^2 \left (1+\frac {b x}{a}\right )^{2+2 p}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\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 (1+2 p)}-\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 (1+p)}+\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 (3+2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 1.12, size = 108, normalized size = 0.83 \[ \frac {{\left ({\left (2 \, b^{3} p^{2} + 3 \, b^{3} p + b^{3}\right )} x^{6} - 2 \, a^{2} b p x^{2} + {\left (2 \, a b^{2} p^{2} + a b^{2} p\right )} x^{4} + a^{3}\right )} {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p}}{2 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,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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giac [A]  time = 0.19, size = 235, normalized size = 1.81 \[ \frac {2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p^{2} x^{6} + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} p x^{6} + 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p^{2} x^{4} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} b^{3} x^{6} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a b^{2} p x^{4} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{2} b p x^{2} + {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )}^{p} a^{3}}{2 \, {\left (4 \, b^{3} p^{3} + 12 \, b^{3} p^{2} + 11 \, b^{3} p + 3 \, b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 96, normalized size = 0.74 \[ \frac {\left (b \,x^{2}+a \right ) \left (2 b^{2} p^{2} x^{4}+3 b^{2} p \,x^{4}+b^{2} x^{4}-2 a b p \,x^{2}-a b \,x^{2}+a^{2}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{p}}{2 \left (4 p^{3}+12 p^{2}+11 p +3\right ) b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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 = 1.44, size = 79, normalized size = 0.61 \[ \frac {{\left ({\left (2 \, p^{2} + 3 \, p + 1\right )} b^{3} x^{6} + {\left (2 \, p^{2} + p\right )} a b^{2} x^{4} - 2 \, a^{2} b p x^{2} + a^{3}\right )} {\left (b x^{2} + a\right )}^{2 \, p}}{2 \, {\left (4 \, p^{3} + 12 \, p^{2} + 11 \, p + 3\right )} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^p,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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mupad [B]  time = 4.27, size = 137, normalized size = 1.05 \[ {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^p\,\left (\frac {x^6\,\left (p^2+\frac {3\,p}{2}+\frac {1}{2}\right )}{4\,p^3+12\,p^2+11\,p+3}+\frac {a^3}{2\,b^3\,\left (4\,p^3+12\,p^2+11\,p+3\right )}-\frac {a^2\,p\,x^2}{b^2\,\left (4\,p^3+12\,p^2+11\,p+3\right )}+\frac {a\,p\,x^4\,\left (2\,p+1\right )}{2\,b\,\left (4\,p^3+12\,p^2+11\,p+3\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \begin {cases} \frac {x^{6} \left (a^{2}\right )^{p}}{6} & \text {for}\: b = 0 \\\int \frac {x^{5}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx & \text {for}\: p = - \frac {3}{2} \\- \frac {2 a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a^{2}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} - \frac {2 a b x^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + x \right )}}{2 a b^{3} + 2 b^{4} x^{2}} + \frac {b^{2} x^{4}}{2 a b^{3} + 2 b^{4} x^{2}} & \text {for}\: p = -1 \\\int \frac {x^{5}}{\sqrt {\left (a + b x^{2}\right )^{2}}}\, dx & \text {for}\: p = - \frac {1}{2} \\\frac {a^{3} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} - \frac {2 a^{2} b p x^{2} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {2 a b^{2} p^{2} x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {a b^{2} p x^{4} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {2 b^{3} p^{2} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {3 b^{3} p x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} + \frac {b^{3} x^{6} \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{p}}{8 b^{3} p^{3} + 24 b^{3} p^{2} + 22 b^{3} p + 6 b^{3}} & \text {otherwise} \end {cases} \]

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]

Piecewise((x**6*(a**2)**p/6, Eq(b, 0)), (Integral(x**5/((a + b*x**2)**2)**(3/2), x), Eq(p, -3/2)), (-2*a**2*lo
g(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) - 2*a**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*
x**2) - 2*a**2/(2*a*b**3 + 2*b**4*x**2) - 2*a*b*x**2*log(-I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) -
2*a*b*x**2*log(I*sqrt(a)*sqrt(1/b) + x)/(2*a*b**3 + 2*b**4*x**2) + b**2*x**4/(2*a*b**3 + 2*b**4*x**2), Eq(p, -
1)), (Integral(x**5/sqrt((a + b*x**2)**2), x), Eq(p, -1/2)), (a**3*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*
p**3 + 24*b**3*p**2 + 22*b**3*p + 6*b**3) - 2*a**2*b*p*x**2*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 +
24*b**3*p**2 + 22*b**3*p + 6*b**3) + 2*a*b**2*p**2*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b
**3*p**2 + 22*b**3*p + 6*b**3) + a*b**2*p*x**4*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2
+ 22*b**3*p + 6*b**3) + 2*b**3*p**2*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b
**3*p + 6*b**3) + 3*b**3*p*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6
*b**3) + b**3*x**6*(a**2 + 2*a*b*x**2 + b**2*x**4)**p/(8*b**3*p**3 + 24*b**3*p**2 + 22*b**3*p + 6*b**3), True)
)

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