\(\int x (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [152]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {654, 623} \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2}-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^2} \]

[In]

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

[Out]

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

Rule 623

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 654

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.69 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^2 \left (10 a^3+20 a^2 b x+15 a b^2 x^2+4 b^3 x^3\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{20 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

(x^2*(10*a^3 + 20*a^2*b*x + 15*a*b^2*x^2 + 4*b^3*x^3)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(20
*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85

method	result	size
gosper	\(\frac {x^{2} \left (4 b^{3} x^{3}+15 a \,b^{2} x^{2}+20 a^{2} b x +10 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\)	\(52\)
default	\(\frac {x^{2} \left (4 b^{3} x^{3}+15 a \,b^{2} x^{2}+20 a^{2} b x +10 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{20 \left (b x +a \right )^{3}}\)	\(52\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} x^{5}}{5 b x +5 a}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, a \,b^{2} x^{4}}{4 \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{2} b \,x^{3}}{b x +a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{2}}{2 b x +2 a}\)	\(99\)

[In]

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

[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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.56 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} x^{5} + \frac {3}{4} \, a b^{2} x^{4} + a^{2} b x^{3} + \frac {1}{2} \, a^{3} x^{2} \]

[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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (56) = 112\).

Time = 0.49 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.98 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{2}} + \frac {a^{3} x}{20 b} + \frac {9 a^{2} x^{2}}{20} + \frac {11 a b x^{3}}{20} + \frac {b^{2} x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \left (a^{2}\right )^{\frac {3}{2}}}{2} & \text {otherwise} \end {cases} \]

[In]

integrate(x*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**2) + a**3*x/(20*b) + 9*a**2*x**2/20 + 11*a*b*x**3/20
+ b**2*x**4/5), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/(2*a**2*b**2), N
e(a*b, 0)), (x**2*(a**2)**(3/2)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a x}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{5 \, b^{2}} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.18 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + a^{2} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{5} \mathrm {sgn}\left (b x + a\right )}{20 \, b^{2}} \]

[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

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (-a^2+3\,a\,b\,x+4\,b^2\,x^2\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{20\,b^2} \]

[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)