3.591 \(\int x (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\)

Optimal. Leaf size=36 \[ \frac{\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b} \]

[Out]

((a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(12*b)

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Rubi [A]  time = 0.0258815, antiderivative size = 36, normalized size of antiderivative = 1., 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} \[ \frac{\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2))/(12*b)

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]

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 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, 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 )^{5/2}}{12 b}\\ \end{align*}

Mathematica [A]  time = 0.0138417, size = 27, normalized size = 0.75 \[ \frac{\left (a+b x^2\right ) \left (\left (a+b x^2\right )^2\right )^{5/2}}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.166, size = 79, normalized size = 2.2 \begin{align*}{\frac{{x}^{2} \left ({b}^{5}{x}^{10}+6\,a{b}^{4}{x}^{8}+15\,{a}^{2}{b}^{3}{x}^{6}+20\,{a}^{3}{b}^{2}{x}^{4}+15\,{a}^{4}b{x}^{2}+6\,{a}^{5} \right ) }{12\, \left ( b{x}^{2}+a \right ) ^{5}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/12*x^2*(b^5*x^10+6*a*b^4*x^8+15*a^2*b^3*x^6+20*a^3*b^2*x^4+15*a^4*b*x^2+6*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)
^5

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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*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45982, size = 132, normalized size = 3.67 \begin{align*} \frac{1}{12} \, b^{5} x^{12} + \frac{1}{2} \, a b^{4} x^{10} + \frac{5}{4} \, a^{2} b^{3} x^{8} + \frac{5}{3} \, a^{3} b^{2} x^{6} + \frac{5}{4} \, a^{4} b x^{4} + \frac{1}{2} \, a^{5} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral(x*((a + b*x**2)**2)**(5/2), x)

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Giac [B]  time = 1.17384, size = 89, normalized size = 2.47 \begin{align*} \frac{1}{12} \,{\left (b^{5} x^{12} + 6 \, a b^{4} x^{10} + 15 \, a^{2} b^{3} x^{8} + 20 \, a^{3} b^{2} x^{6} + 15 \, a^{4} b x^{4} + 6 \, a^{5} x^{2}\right )} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

1/12*(b^5*x^12 + 6*a*b^4*x^10 + 15*a^2*b^3*x^8 + 20*a^3*b^2*x^6 + 15*a^4*b*x^4 + 6*a^5*x^2)*sgn(b*x^2 + a)