\(\int (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [153]

Optimal result

Integrand size = 20, antiderivative size = 32 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {623} \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b} \]

[In]

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

[Out]

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

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)]

Rubi steps \begin{align*} \text {integral}& = \frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^{3/2}}{4 b} \]

[In]

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

[Out]

((a + b*x)*((a + b*x)^2)^(3/2))/(4*b)

Maple [A] (verified)

Time = 2.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{4 b}\)	\(20\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b x +a \right )^{3}}{4 b}\)	\(22\)
gosper	\(\frac {x \left (b^{3} x^{3}+4 a \,b^{2} x^{2}+6 a^{2} b x +4 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{4 \left (b x +a \right )^{3}}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, b^{3} x^{4} + a b^{2} x^{3} + \frac {3}{2} \, a^{2} b x^{2} + a^{3} x \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (29) = 58\).

Time = 0.80 (sec) , antiderivative size = 294, normalized size of antiderivative = 9.19 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=a^{2} \left (\begin {cases} \left (\frac {a}{2 b} + \frac {x}{2}\right ) \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} & \text {for}\: b^{2} \neq 0 \\\frac {\left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3 a b} & \text {for}\: a b \neq 0 \\x \sqrt {a^{2}} & \text {otherwise} \end {cases}\right ) + 2 a b \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{2}}{6 b^{2}} + \frac {a x}{6 b} + \frac {x^{2}}{3}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \sqrt {a^{2}}}{2} & \text {otherwise} \end {cases}\right ) + b^{2} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{3}}{12 b^{3}} - \frac {a^{2} x}{12 b^{2}} + \frac {a x^{2}}{12 b} + \frac {x^{3}}{4}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \sqrt {a^{2}}}{3} & \text {otherwise} \end {cases}\right ) \]

[In]

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

[Out]

a**2*Piecewise(((a/(2*b) + x/2)*sqrt(a**2 + 2*a*b*x + b**2*x**2), Ne(b**2, 0)), ((a**2 + 2*a*b*x)**(3/2)/(3*a*
b), Ne(a*b, 0)), (x*sqrt(a**2), True)) + 2*a*b*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + a
*x/(6*b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2*a*b*x)**(5/2)/5)/(2*a**2*b**2)
, Ne(a*b, 0)), (x**2*sqrt(a**2)/2, True)) + b**2*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 [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a}{4 \, b} \]

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (28) = 56\).

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.78 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} a^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{4} \mathrm {sgn}\left (b x + a\right )}{4 \, b} + \frac {1}{4} \, {\left (b x^{2} + 2 \, a x\right )}^{2} b \mathrm {sgn}\left (b x + a\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (x\,b^2+a\,b\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{4\,b^2} \]

[In]

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

[Out]

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