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

Optimal. Leaf size=34 \[ \frac{\left (a+b x^2\right )^8}{16 b^2}-\frac{a \left (a+b x^2\right )^7}{14 b^2} \]

[Out]

-(a*(a + b*x^2)^7)/(14*b^2) + (a + b*x^2)^8/(16*b^2)

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Rubi [A]  time = 0.0456321, antiderivative size = 34, 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} \[ \frac{\left (a+b x^2\right )^8}{16 b^2}-\frac{a \left (a+b x^2\right )^7}{14 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*x^2)^7)/(14*b^2) + (a + b*x^2)^8/(16*b^2)

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^3 \left (a^2+2 a b x^2+b^2 x^4\right )^3 \, dx &=\frac{\int x^3 \left (a b+b^2 x^2\right )^6 \, dx}{b^6}\\ &=\frac{\operatorname{Subst}\left (\int x \left (a b+b^2 x\right )^6 \, dx,x,x^2\right )}{2 b^6}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a \left (a b+b^2 x\right )^6}{b}+\frac{\left (a b+b^2 x\right )^7}{b^2}\right ) \, dx,x,x^2\right )}{2 b^6}\\ &=-\frac{a \left (a+b x^2\right )^7}{14 b^2}+\frac{\left (a+b x^2\right )^8}{16 b^2}\\ \end{align*}

Mathematica [B]  time = 0.0025139, size = 77, normalized size = 2.26 \[ \frac{5}{4} a^2 b^4 x^{12}+2 a^3 b^3 x^{10}+\frac{15}{8} a^4 b^2 x^8+a^5 b x^6+\frac{a^6 x^4}{4}+\frac{3}{7} a b^5 x^{14}+\frac{b^6 x^{16}}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^6*x^4)/4 + a^5*b*x^6 + (15*a^4*b^2*x^8)/8 + 2*a^3*b^3*x^10 + (5*a^2*b^4*x^12)/4 + (3*a*b^5*x^14)/7 + (b^6*x
^16)/16

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Maple [B]  time = 0.042, size = 68, normalized size = 2. \begin{align*}{\frac{{b}^{6}{x}^{16}}{16}}+{\frac{3\,a{b}^{5}{x}^{14}}{7}}+{\frac{5\,{a}^{2}{b}^{4}{x}^{12}}{4}}+2\,{a}^{3}{b}^{3}{x}^{10}+{\frac{15\,{a}^{4}{b}^{2}{x}^{8}}{8}}+{a}^{5}b{x}^{6}+{\frac{{x}^{4}{a}^{6}}{4}} \end{align*}

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)^3,x)

[Out]

1/16*b^6*x^16+3/7*a*b^5*x^14+5/4*a^2*b^4*x^12+2*a^3*b^3*x^10+15/8*a^4*b^2*x^8+a^5*b*x^6+1/4*x^4*a^6

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Maxima [B]  time = 0.983868, size = 90, normalized size = 2.65 \begin{align*} \frac{1}{16} \, b^{6} x^{16} + \frac{3}{7} \, a b^{5} x^{14} + \frac{5}{4} \, a^{2} b^{4} x^{12} + 2 \, a^{3} b^{3} x^{10} + \frac{15}{8} \, a^{4} b^{2} x^{8} + a^{5} b x^{6} + \frac{1}{4} \, a^{6} x^{4} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

1/16*b^6*x^16 + 3/7*a*b^5*x^14 + 5/4*a^2*b^4*x^12 + 2*a^3*b^3*x^10 + 15/8*a^4*b^2*x^8 + a^5*b*x^6 + 1/4*a^6*x^
4

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Fricas [B]  time = 1.50752, size = 153, normalized size = 4.5 \begin{align*} \frac{1}{16} x^{16} b^{6} + \frac{3}{7} x^{14} b^{5} a + \frac{5}{4} x^{12} b^{4} a^{2} + 2 x^{10} b^{3} a^{3} + \frac{15}{8} x^{8} b^{2} a^{4} + x^{6} b a^{5} + \frac{1}{4} x^{4} a^{6} \end{align*}

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)^3,x, algorithm="fricas")

[Out]

1/16*x^16*b^6 + 3/7*x^14*b^5*a + 5/4*x^12*b^4*a^2 + 2*x^10*b^3*a^3 + 15/8*x^8*b^2*a^4 + x^6*b*a^5 + 1/4*x^4*a^
6

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Sympy [B]  time = 0.077396, size = 75, normalized size = 2.21 \begin{align*} \frac{a^{6} x^{4}}{4} + a^{5} b x^{6} + \frac{15 a^{4} b^{2} x^{8}}{8} + 2 a^{3} b^{3} x^{10} + \frac{5 a^{2} b^{4} x^{12}}{4} + \frac{3 a b^{5} x^{14}}{7} + \frac{b^{6} x^{16}}{16} \end{align*}

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)**3,x)

[Out]

a**6*x**4/4 + a**5*b*x**6 + 15*a**4*b**2*x**8/8 + 2*a**3*b**3*x**10 + 5*a**2*b**4*x**12/4 + 3*a*b**5*x**14/7 +
b**6*x**16/16

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Giac [B]  time = 1.15961, size = 90, normalized size = 2.65 \begin{align*} \frac{1}{16} \, b^{6} x^{16} + \frac{3}{7} \, a b^{5} x^{14} + \frac{5}{4} \, a^{2} b^{4} x^{12} + 2 \, a^{3} b^{3} x^{10} + \frac{15}{8} \, a^{4} b^{2} x^{8} + a^{5} b x^{6} + \frac{1}{4} \, a^{6} x^{4} \end{align*}

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)^3,x, algorithm="giac")

[Out]

1/16*b^6*x^16 + 3/7*a*b^5*x^14 + 5/4*a^2*b^4*x^12 + 2*a^3*b^3*x^10 + 15/8*a^4*b^2*x^8 + a^5*b*x^6 + 1/4*a^6*x^
4