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

Optimal. Leaf size=34 \[ \frac{\left (a+b x^2\right )^6}{12 b^2}-\frac{a \left (a+b x^2\right )^5}{10 b^2} \]

[Out]

-(a*(a + b*x^2)^5)/(10*b^2) + (a + b*x^2)^6/(12*b^2)

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Rubi [A]  time = 0.0395182, 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} \[ \frac{\left (a+b x^2\right )^6}{12 b^2}-\frac{a \left (a+b x^2\right )^5}{10 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(a + b*x^2)^5)/(10*b^2) + (a + b*x^2)^6/(12*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 )^2 \, dx &=\frac{\int x^3 \left (a b+b^2 x^2\right )^4 \, dx}{b^4}\\ &=\frac{\operatorname{Subst}\left (\int x \left (a b+b^2 x\right )^4 \, dx,x,x^2\right )}{2 b^4}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{a \left (a b+b^2 x\right )^4}{b}+\frac{\left (a b+b^2 x\right )^5}{b^2}\right ) \, dx,x,x^2\right )}{2 b^4}\\ &=-\frac{a \left (a+b x^2\right )^5}{10 b^2}+\frac{\left (a+b x^2\right )^6}{12 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0021975, size = 56, normalized size = 1.65 \[ \frac{3}{4} a^2 b^2 x^8+\frac{2}{3} a^3 b x^6+\frac{a^4 x^4}{4}+\frac{2}{5} a b^3 x^{10}+\frac{b^4 x^{12}}{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 47, normalized size = 1.4 \begin{align*}{\frac{{b}^{4}{x}^{12}}{12}}+{\frac{2\,a{b}^{3}{x}^{10}}{5}}+{\frac{3\,{a}^{2}{b}^{2}{x}^{8}}{4}}+{\frac{2\,{a}^{3}b{x}^{6}}{3}}+{\frac{{a}^{4}{x}^{4}}{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)^2,x)

[Out]

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

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Maxima [A]  time = 0.994285, size = 62, normalized size = 1.82 \begin{align*} \frac{1}{12} \, b^{4} x^{12} + \frac{2}{5} \, a b^{3} x^{10} + \frac{3}{4} \, a^{2} b^{2} x^{8} + \frac{2}{3} \, a^{3} b x^{6} + \frac{1}{4} \, a^{4} 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.46452, size = 108, normalized size = 3.18 \begin{align*} \frac{1}{12} x^{12} b^{4} + \frac{2}{5} x^{10} b^{3} a + \frac{3}{4} x^{8} b^{2} a^{2} + \frac{2}{3} x^{6} b a^{3} + \frac{1}{4} x^{4} a^{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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.074655, size = 53, normalized size = 1.56 \begin{align*} \frac{a^{4} x^{4}}{4} + \frac{2 a^{3} b x^{6}}{3} + \frac{3 a^{2} b^{2} x^{8}}{4} + \frac{2 a b^{3} x^{10}}{5} + \frac{b^{4} x^{12}}{12} \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)**2,x)

[Out]

a**4*x**4/4 + 2*a**3*b*x**6/3 + 3*a**2*b**2*x**8/4 + 2*a*b**3*x**10/5 + b**4*x**12/12

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Giac [A]  time = 1.13718, size = 62, normalized size = 1.82 \begin{align*} \frac{1}{12} \, b^{4} x^{12} + \frac{2}{5} \, a b^{3} x^{10} + \frac{3}{4} \, a^{2} b^{2} x^{8} + \frac{2}{3} \, a^{3} b x^{6} + \frac{1}{4} \, a^{4} 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)^2,x, algorithm="giac")

[Out]

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