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

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

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Rubi [A]  time = 0.01, antiderivative size = 61, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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]

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*}

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Mathematica [A]  time = 0.01, size = 55, normalized size = 0.90 \begin {gather*} \frac {x^2 \sqrt {(a+b x)^2} \left (10 a^3+20 a^2 b x+15 a b^2 x^2+4 b^3 x^3\right )}{20 (a+b x)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.45, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 34, normalized size = 0.56 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.18, size = 72, normalized size = 1.18 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.04, size = 52, normalized size = 0.85 \begin {gather*} \frac {\left (4 b^{3} x^{3}+15 a \,b^{2} x^{2}+20 a^{2} b x +10 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x^{2}}{20 \left (b x +a \right )^{3}} \end {gather*}

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 [A]  time = 1.39, size = 75, normalized size = 1.23 \begin {gather*} -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{5 \, b^{2}} \end {gather*}

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]

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

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mupad [B]  time = 0.16, size = 42, normalized size = 0.69 \begin {gather*} \frac {\left (-a^2+3\,a\,b\,x+4\,b^2\,x^2\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{20\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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