\(\int (a+b x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [1975]

Optimal result

Integrand size = 26, antiderivative size = 27 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {643} \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b} \]

[In]

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

[Out]

(a^2 + 2*a*b*x + b^2*x^2)^(5/2)/(5*b)

Rule 643

Int[((d_) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*((a + b*x + c*x^2)^(p +
1)/(b*(p + 1))), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {(a+b x)^4 \sqrt {(a+b x)^2}}{5 b} \]

[In]

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

[Out]

((a + b*x)^4*Sqrt[(a + b*x)^2])/(5*b)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 2.

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{5}}{5 b}\)	\(19\)
default	\(\frac {\left (b x +a \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 b}\)	\(22\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b x +a \right )^{4}}{5 b}\)	\(22\)
gosper	\(\frac {x \left (x^{4} b^{4}+5 a \,b^{3} x^{3}+10 x^{2} b^{2} a^{2}+10 b \,a^{3} x +5 a^{4}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{5 \left (b x +a \right )^{3}}\)	\(60\)

[In]

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

[Out]

1/5*csgn(b*x+a)*(b*x+a)^5/b

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.56 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{4} x^{5} + a b^{3} x^{4} + 2 \, a^{2} b^{2} x^{3} + 2 \, a^{3} b x^{2} + a^{4} x \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (22) = 44\).

Time = 0.16 (sec) , antiderivative size = 158, normalized size of antiderivative = 5.85 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {a^{4} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{5 b} + \frac {4 a^{3} x \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac {6 a^{2} b x^{2} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac {4 a b^{2} x^{3} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{5} + \frac {b^{3} x^{4} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}}}{5} & \text {for}\: b \neq 0 \\a x \left (a^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**4*sqrt(a**2 + 2*a*b*x + b**2*x**2)/(5*b) + 4*a**3*x*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + 6*a**2*
b*x**2*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + 4*a*b**2*x**3*sqrt(a**2 + 2*a*b*x + b**2*x**2)/5 + b**3*x**4*sqrt(
a**2 + 2*a*b*x + b**2*x**2)/5, Ne(b, 0)), (a*x*(a**2)**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}{5 \, b} \]

[In]

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

[Out]

1/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)/b

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (23) = 46\).

Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.19 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{5} \, b^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + a b^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b x^{2} \mathrm {sgn}\left (b x + a\right ) + a^{4} x \mathrm {sgn}\left (b x + a\right ) + \frac {a^{5} \mathrm {sgn}\left (b x + a\right )}{5 \, b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 10.75 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (a+b\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b} \]

[In]

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

[Out]

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