3.447 \(\int x^5 (a^2+2 a b x^2+b^2 x^4)^3 \, dx\)

Optimal. Leaf size=53 \[ \frac{a^2 \left (a+b x^2\right )^7}{14 b^3}+\frac{\left (a+b x^2\right )^9}{18 b^3}-\frac{a \left (a+b x^2\right )^8}{8 b^3} \]

[Out]

(a^2*(a + b*x^2)^7)/(14*b^3) - (a*(a + b*x^2)^8)/(8*b^3) + (a + b*x^2)^9/(18*b^3)

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Rubi [A]  time = 0.0853567, antiderivative size = 53, 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, Rules used = {28, 266, 43} \[ \frac{a^2 \left (a+b x^2\right )^7}{14 b^3}+\frac{\left (a+b x^2\right )^9}{18 b^3}-\frac{a \left (a+b x^2\right )^8}{8 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

(a^2*(a + b*x^2)^7)/(14*b^3) - (a*(a + b*x^2)^8)/(8*b^3) + (a + b*x^2)^9/(18*b^3)

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 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])

Rubi steps

\begin{align*} \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx &=\frac{\int x^5 \left (a b+b^2 x^2\right )^6 \, dx}{b^6}\\ &=\frac{\operatorname{Subst}\left (\int x^2 \left (a b+b^2 x\right )^6 \, dx,x,x^2\right )}{2 b^6}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 \left (a b+b^2 x\right )^6}{b^2}-\frac{2 a \left (a b+b^2 x\right )^7}{b^3}+\frac{\left (a b+b^2 x\right )^8}{b^4}\right ) \, dx,x,x^2\right )}{2 b^6}\\ &=\frac{a^2 \left (a+b x^2\right )^7}{14 b^3}-\frac{a \left (a+b x^2\right )^8}{8 b^3}+\frac{\left (a+b x^2\right )^9}{18 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0026915, size = 82, normalized size = 1.55 \[ \frac{15}{14} a^2 b^4 x^{14}+\frac{5}{3} a^3 b^3 x^{12}+\frac{3}{2} a^4 b^2 x^{10}+\frac{3}{4} a^5 b x^8+\frac{a^6 x^6}{6}+\frac{3}{8} a b^5 x^{16}+\frac{b^6 x^{18}}{18} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

(a^6*x^6)/6 + (3*a^5*b*x^8)/4 + (3*a^4*b^2*x^10)/2 + (5*a^3*b^3*x^12)/3 + (15*a^2*b^4*x^14)/14 + (3*a*b^5*x^16
)/8 + (b^6*x^18)/18

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Maple [A]  time = 0.042, size = 69, normalized size = 1.3 \begin{align*}{\frac{{b}^{6}{x}^{18}}{18}}+{\frac{3\,a{b}^{5}{x}^{16}}{8}}+{\frac{15\,{a}^{2}{b}^{4}{x}^{14}}{14}}+{\frac{5\,{a}^{3}{b}^{3}{x}^{12}}{3}}+{\frac{3\,{a}^{4}{b}^{2}{x}^{10}}{2}}+{\frac{3\,{a}^{5}b{x}^{8}}{4}}+{\frac{{a}^{6}{x}^{6}}{6}} \end{align*}

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)^3,x)

[Out]

1/18*b^6*x^18+3/8*a*b^5*x^16+15/14*a^2*b^4*x^14+5/3*a^3*b^3*x^12+3/2*a^4*b^2*x^10+3/4*a^5*b*x^8+1/6*a^6*x^6

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Maxima [A]  time = 0.980579, size = 92, normalized size = 1.74 \begin{align*} \frac{1}{18} \, b^{6} x^{18} + \frac{3}{8} \, a b^{5} x^{16} + \frac{15}{14} \, a^{2} b^{4} x^{14} + \frac{5}{3} \, a^{3} b^{3} x^{12} + \frac{3}{2} \, a^{4} b^{2} x^{10} + \frac{3}{4} \, a^{5} b x^{8} + \frac{1}{6} \, a^{6} x^{6} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

1/18*b^6*x^18 + 3/8*a*b^5*x^16 + 15/14*a^2*b^4*x^14 + 5/3*a^3*b^3*x^12 + 3/2*a^4*b^2*x^10 + 3/4*a^5*b*x^8 + 1/
6*a^6*x^6

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Fricas [A]  time = 1.51299, size = 163, normalized size = 3.08 \begin{align*} \frac{1}{18} x^{18} b^{6} + \frac{3}{8} x^{16} b^{5} a + \frac{15}{14} x^{14} b^{4} a^{2} + \frac{5}{3} x^{12} b^{3} a^{3} + \frac{3}{2} x^{10} b^{2} a^{4} + \frac{3}{4} x^{8} b a^{5} + \frac{1}{6} x^{6} a^{6} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

1/18*x^18*b^6 + 3/8*x^16*b^5*a + 15/14*x^14*b^4*a^2 + 5/3*x^12*b^3*a^3 + 3/2*x^10*b^2*a^4 + 3/4*x^8*b*a^5 + 1/
6*x^6*a^6

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Sympy [A]  time = 0.081546, size = 80, normalized size = 1.51 \begin{align*} \frac{a^{6} x^{6}}{6} + \frac{3 a^{5} b x^{8}}{4} + \frac{3 a^{4} b^{2} x^{10}}{2} + \frac{5 a^{3} b^{3} x^{12}}{3} + \frac{15 a^{2} b^{4} x^{14}}{14} + \frac{3 a b^{5} x^{16}}{8} + \frac{b^{6} x^{18}}{18} \end{align*}

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)**3,x)

[Out]

a**6*x**6/6 + 3*a**5*b*x**8/4 + 3*a**4*b**2*x**10/2 + 5*a**3*b**3*x**12/3 + 15*a**2*b**4*x**14/14 + 3*a*b**5*x
**16/8 + b**6*x**18/18

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Giac [A]  time = 1.1859, size = 92, normalized size = 1.74 \begin{align*} \frac{1}{18} \, b^{6} x^{18} + \frac{3}{8} \, a b^{5} x^{16} + \frac{15}{14} \, a^{2} b^{4} x^{14} + \frac{5}{3} \, a^{3} b^{3} x^{12} + \frac{3}{2} \, a^{4} b^{2} x^{10} + \frac{3}{4} \, a^{5} b x^{8} + \frac{1}{6} \, a^{6} x^{6} \end{align*}

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)^3,x, algorithm="giac")

[Out]

1/18*b^6*x^18 + 3/8*a*b^5*x^16 + 15/14*a^2*b^4*x^14 + 5/3*a^3*b^3*x^12 + 3/2*a^4*b^2*x^10 + 3/4*a^5*b*x^8 + 1/
6*a^6*x^6