\(\int x^2 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [151]

Optimal result

Integrand size = 24, antiderivative size = 96 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b^3}-\frac {2 a \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^3}+\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^3} \]

[Out]

1/4*a^2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/b^3-2/5*a*(b^2*x^2+2*a*b*x+a^2)^(5/2)/b^3+1/6*(b*x+a)*(b^2*x^2+2*a
*b*x+a^2)^(5/2)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.11, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {659} \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3}-\frac {2 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4}{5 b^3}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^3} \]

[In]

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

[Out]

(a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^3) - (2*a*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b
^3) + ((a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3)

Rule 659

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {a^2 \left (a b+b^2 x\right )^3}{b^2}-\frac {2 a \left (a b+b^2 x\right )^4}{b^3}+\frac {\left (a b+b^2 x\right )^5}{b^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}-\frac {2 a (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {(a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.07 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^3 \left (20 a^3+45 a^2 b x+36 a b^2 x^2+10 b^3 x^3\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{60 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

(x^3*(20*a^3 + 45*a^2*b*x + 36*a*b^2*x^2 + 10*b^3*x^3)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/(6
0*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.54

method	result	size
gosper	\(\frac {x^{3} \left (10 b^{3} x^{3}+36 a \,b^{2} x^{2}+45 a^{2} b x +20 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\)	\(52\)
default	\(\frac {x^{3} \left (10 b^{3} x^{3}+36 a \,b^{2} x^{2}+45 a^{2} b x +20 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{60 \left (b x +a \right )^{3}}\)	\(52\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{3} x^{3}}{3 b x +3 a}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, a^{2} b \,x^{4}}{4 \left (b x +a \right )}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, a \,b^{2} x^{5}}{5 \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{3} x^{6}}{6 b x +6 a}\)	\(100\)

[In]

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

[Out]

1/60*x^3*(10*b^3*x^3+36*a*b^2*x^2+45*a^2*b*x+20*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.36 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, b^{3} x^{6} + \frac {3}{5} \, a b^{2} x^{5} + \frac {3}{4} \, a^{2} b x^{4} + \frac {1}{3} \, a^{3} x^{3} \]

[In]

integrate(x^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/6*b^3*x^6 + 3/5*a*b^2*x^5 + 3/4*a^2*b*x^4 + 1/3*a^3*x^3

Sympy [A] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.59 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{5}}{60 b^{3}} - \frac {a^{4} x}{60 b^{2}} + \frac {a^{3} x^{2}}{60 b} + \frac {19 a^{2} x^{3}}{60} + \frac {13 a b x^{4}}{30} + \frac {b^{2} x^{5}}{6}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \left (a^{2}\right )^{\frac {3}{2}}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(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**5/(60*b**3) - a**4*x/(60*b**2) + a**3*x**2/(60*b) + 19*a**2*x*
*3/60 + 13*a*b*x**4/30 + b**2*x**5/6), Ne(b**2, 0)), ((a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 2*a**2*(a**2 + 2*a*b*x
)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(4*a**3*b**3), Ne(a*b, 0)), (x**3*(a**2)**(3/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.06 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x}{6 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a}{30 \, b^{3}} \]

[In]

integrate(x^2*(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^2*x/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3/b^3 + 1/6*(b^2*x^2 + 2
*a*b*x + a^2)^(5/2)*x/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.76 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{6} \, b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a^{2} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{6} \mathrm {sgn}\left (b x + a\right )}{60 \, b^{3}} \]

[In]

integrate(x^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/6*b^3*x^6*sgn(b*x + a) + 3/5*a*b^2*x^5*sgn(b*x + a) + 3/4*a^2*b*x^4*sgn(b*x + a) + 1/3*a^3*x^3*sgn(b*x + a)
+ 1/60*a^6*sgn(b*x + a)/b^3

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

[In]

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

[Out]

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