3.2.18 \(\int x \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

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

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Rubi [A]  time = 0.02, 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 )^{3/2}}{3 b^2}-\frac {a (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 33, normalized size = 0.54 \begin {gather*} \frac {x^2 \sqrt {(a+b x)^2} (3 a+2 b x)}{6 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x^2*Sqrt[(a + b*x)^2]*(3*a + 2*b*x))/(6*(a + b*x))

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IntegrateAlgebraic [F]  time = 0.22, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][x*Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.40, size = 13, normalized size = 0.21 \begin {gather*} \frac {1}{3} \, b x^{3} + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/3*b*x^3 + 1/2*a*x^2

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giac [A]  time = 0.16, size = 39, normalized size = 0.64 \begin {gather*} \frac {1}{3} \, b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a x^{2} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{3} \mathrm {sgn}\left (b x + a\right )}{6 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/3*b*x^3*sgn(b*x + a) + 1/2*a*x^2*sgn(b*x + a) - 1/6*a^3*sgn(b*x + a)/b^2

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maple [A]  time = 0.05, size = 30, normalized size = 0.49 \begin {gather*} \frac {\left (2 b x +3 a \right ) \sqrt {\left (b x +a \right )^{2}}\, x^{2}}{6 b x +6 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((b*x+a)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 1.40, size = 75, normalized size = 1.23 \begin {gather*} -\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a x}{2 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.18, size = 55, normalized size = 0.90 \begin {gather*} \frac {\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((a + b*x)^2)^(1/2),x)

[Out]

((8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^4)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.20 \begin {gather*} \frac {a x^{2}}{2} + \frac {b x^{3}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((b*x+a)**2)**(1/2),x)

[Out]

a*x**2/2 + b*x**3/3

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