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

Optimal. Leaf size=71 \[ \frac{b x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{a x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0215479, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {646, 43} \[ \frac{b x^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac{a x^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0088511, size = 33, normalized size = 0.46 \[ \frac{x^3 \sqrt{(a+b x)^2} (4 a+3 b x)}{12 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(4*a + 3*b*x))/(12*(a + b*x))

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Maple [A]  time = 0.046, size = 30, normalized size = 0.4 \begin{align*}{\frac{{x}^{3} \left ( 3\,bx+4\,a \right ) }{12\,bx+12\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^3*(3*b*x+4*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^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.96834, size = 31, normalized size = 0.44 \begin{align*} \frac{1}{4} \, b x^{4} + \frac{1}{3} \, a x^{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

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Sympy [A]  time = 0.148424, size = 12, normalized size = 0.17 \begin{align*} \frac{a x^{3}}{3} + \frac{b x^{4}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**3/3 + b*x**4/4

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Giac [A]  time = 1.23872, size = 53, normalized size = 0.75 \begin{align*} \frac{1}{4} \, b x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{a^{4} \mathrm{sgn}\left (b x + a\right )}{12 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b*x^4*sgn(b*x + a) + 1/3*a*x^3*sgn(b*x + a) + 1/12*a^4*sgn(b*x + a)/b^3