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

Optimal. Leaf size=36 \[ \frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b} \]

[Out]

1/8*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/b

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1107, 609} \[ \frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)]

Rule 1107

Int[(x_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(a + b*x + c*x^2)^p, x],
x, x^2], x] /; FreeQ[{a, b, c, p}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 27, normalized size = 0.75 \[ \frac {\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^{3/2}}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*((a + b*x^2)^2)^(3/2))/(8*b)

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fricas [A]  time = 0.80, size = 35, normalized size = 0.97 \[ \frac {1}{8} \, b^{3} x^{8} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{2} \, a^{3} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 44, normalized size = 1.22 \[ \frac {1}{8} \, {\left (2 \, {\left (b x^{4} + 2 \, a x^{2}\right )} a^{2} + {\left (b x^{4} + 2 \, a x^{2}\right )}^{2} b\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/8*(2*(b*x^4 + 2*a*x^2)*a^2 + (b*x^4 + 2*a*x^2)^2*b)*sgn(b*x^2 + a)

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maple [A]  time = 0.00, size = 57, normalized size = 1.58 \[ \frac {\left (b^{3} x^{6}+4 a \,b^{2} x^{4}+6 a^{2} b \,x^{2}+4 a^{3}\right ) \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} x^{2}}{8 \left (b \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.36, size = 35, normalized size = 0.97 \[ \frac {1}{8} \, b^{3} x^{8} + \frac {1}{2} \, a b^{2} x^{6} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{2} \, a^{3} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.25, size = 36, normalized size = 1.00 \[ \frac {\left (b^2\,x^2+a\,b\right )\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}}{8\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

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