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

Optimal. Leaf size=61 \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}-\frac{a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2} \]

[Out]

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

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Rubi [A]  time = 0.014963, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {640, 609} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}-\frac{a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0140692, size = 55, normalized size = 0.9 \[ \frac{x^2 \sqrt{(a+b x)^2} \left (20 a^2 b x+10 a^3+15 a b^2 x^2+4 b^3 x^3\right )}{20 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.171, size = 52, normalized size = 0.9 \begin{align*}{\frac{{x}^{2} \left ( 4\,{x}^{3}{b}^{3}+15\,a{b}^{2}{x}^{2}+20\,{a}^{2}bx+10\,{a}^{3} \right ) }{20\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*x^2*(4*b^3*x^3+15*a*b^2*x^2+20*a^2*b*x+10*a^3)*((b*x+a)^2)^(3/2)/(b*x+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*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96812, size = 74, normalized size = 1.21 \begin{align*} \frac{1}{5} \, b^{3} x^{5} + \frac{3}{4} \, a b^{2} x^{4} + a^{2} b x^{3} + \frac{1}{2} \, a^{3} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^3*x^5 + 3/4*a*b^2*x^4 + a^2*b*x^3 + 1/2*a^3*x^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*((a + b*x)**2)**(3/2), x)

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Giac [A]  time = 1.19716, size = 97, normalized size = 1.59 \begin{align*} \frac{1}{5} \, b^{3} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, a b^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + a^{2} b x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) - \frac{a^{5} \mathrm{sgn}\left (b x + a\right )}{20 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^3*x^5*sgn(b*x + a) + 3/4*a*b^2*x^4*sgn(b*x + a) + a^2*b*x^3*sgn(b*x + a) + 1/2*a^3*x^2*sgn(b*x + a) - 1/
20*a^5*sgn(b*x + a)/b^2