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

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

[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)

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Rubi [A]  time = 0.0152343, 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} \[ \frac{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2}-\frac{a (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}{2 b^2} \]

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 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 \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*}

Mathematica [A]  time = 0.0083166, size = 33, normalized size = 0.54 \[ \frac{x^2 \sqrt{(a+b x)^2} (3 a+2 b x)}{6 (a+b x)} \]

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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Maple [A]  time = 0.042, size = 30, normalized size = 0.5 \begin{align*}{\frac{{x}^{2} \left ( 2\,bx+3\,a \right ) }{6\,bx+6\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.84533, size = 31, normalized size = 0.51 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} \end{align*}

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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Sympy [A]  time = 0.148866, size = 12, normalized size = 0.2 \begin{align*} \frac{a x^{2}}{2} + \frac{b x^{3}}{3} \end{align*}

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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Giac [A]  time = 1.24059, size = 53, normalized size = 0.87 \begin{align*} \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{align*}

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