\(\int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [139]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 71 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)}+\frac {b x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {b x^4 \sqrt {a^2+2 a b x+b^2 x^2}}{4 (a+b x)}+\frac {a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}{3 (a+b x)} \]

[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 45

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])

Rule 660

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.44 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.14 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x^3 (4 a+3 b x) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{12 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.42

method result size
gosper \(\frac {x^{3} \left (3 b x +4 a \right ) \sqrt {\left (b x +a \right )^{2}}}{12 b x +12 a}\) \(30\)
default \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (3 b^{2} x^{2}-2 a b x +a^{2}\right )}{12 b^{3}}\) \(36\)
risch \(\frac {a \,x^{3} \sqrt {\left (b x +a \right )^{2}}}{3 b x +3 a}+\frac {b \,x^{4} \sqrt {\left (b x +a \right )^{2}}}{4 b x +4 a}\) \(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.18 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b x^{4} + \frac {1}{3} \, a x^{3} \]

[In]

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

[Out]

1/4*b*x^4 + 1/3*a*x^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (46) = 92\).

Time = 1.01 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.80 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**3/(12*b**3) - a**2*x/(12*b**2) + a*x**2/(12*b) + x**3/4), Ne(b
**2, 0)), ((a**4*(a**2 + 2*a*b*x)**(3/2)/3 - 2*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(4*
a**3*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2)/3, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (45) = 90\).

Time = 0.21 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.44 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}}{2 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x}{4 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a}{12 \, b^{3}} \]

[In]

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

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3/b^3 + 1/4*(b^2*x^2 + 2*a*b
*x + a^2)^(3/2)*x/b^2 - 5/12*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.55 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{4} \, b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{4} \mathrm {sgn}\left (b x + a\right )}{12 \, b^{3}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.89 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{12\,b^3} \]

[In]

int(x^2*((a + b*x)^2)^(1/2),x)

[Out]

((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(12*b^3)