3.561 \(\int x^3 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\)

Optimal. Leaf size=67 \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0509264, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1111, 640, 609} \[ \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1111

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rubi steps

\begin{align*} \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right )\\ &=\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}-\frac{a \operatorname{Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right )}{2 b}\\ &=-\frac{a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2}+\frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0153101, size = 61, normalized size = 0.91 \[ \frac{x^4 \sqrt{\left (a+b x^2\right )^2} \left (20 a^2 b x^2+10 a^3+15 a b^2 x^4+4 b^3 x^6\right )}{40 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[(a + b*x^2)^2]*(10*a^3 + 20*a^2*b*x^2 + 15*a*b^2*x^4 + 4*b^3*x^6))/(40*(a + b*x^2))

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Maple [A]  time = 0.173, size = 58, normalized size = 0.9 \begin{align*}{\frac{{x}^{4} \left ( 4\,{b}^{3}{x}^{6}+15\,a{b}^{2}{x}^{4}+20\,{a}^{2}b{x}^{2}+10\,{a}^{3} \right ) }{40\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/40*x^4*(4*b^3*x^6+15*a*b^2*x^4+20*a^2*b*x^2+10*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47925, size = 82, normalized size = 1.22 \begin{align*} \frac{1}{10} \, b^{3} x^{10} + \frac{3}{8} \, a b^{2} x^{8} + \frac{1}{2} \, a^{2} b x^{6} + \frac{1}{4} \, a^{3} 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)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{3}{2}}\, dx \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)**(3/2),x)

[Out]

Integral(x**3*((a + b*x**2)**2)**(3/2), x)

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Giac [A]  time = 1.12088, size = 61, normalized size = 0.91 \begin{align*} \frac{1}{40} \,{\left (4 \, b^{3} x^{10} + 15 \, a b^{2} x^{8} + 20 \, a^{2} b x^{6} + 10 \, a^{3} x^{4}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \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)^(3/2),x, algorithm="giac")

[Out]

1/40*(4*b^3*x^10 + 15*a*b^2*x^8 + 20*a^2*b*x^6 + 10*a^3*x^4)*sgn(b*x^2 + a)