\(\int (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [168]

Optimal result

Integrand size = 20, antiderivative size = 32 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \]

[Out]

1/6*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/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} \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b} \]

[In]

Int[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/(6*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 )^{5/2}}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {(a+b x) \left ((a+b x)^2\right )^{5/2}}{6 b} \]

[In]

Integrate[(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((a + b*x)*((a + b*x)^2)^(5/2))/(6*b)

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62

method	result	size
default	\(\frac {\left (b x +a \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{6 b}\)	\(20\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (b x +a \right )^{5}}{6 b}\)	\(22\)
gosper	\(\frac {x \left (b^{5} x^{5}+6 a \,b^{4} x^{4}+15 a^{2} b^{3} x^{3}+20 a^{3} b^{2} x^{2}+15 a^{4} b x +6 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{6 \left (b x +a \right )^{5}}\)	\(71\)

[In]

int((b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.66 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{6} \, b^{5} x^{6} + a b^{4} x^{5} + \frac {5}{2} \, a^{2} b^{3} x^{4} + \frac {10}{3} \, a^{3} b^{2} x^{3} + \frac {5}{2} \, a^{4} b x^{2} + a^{5} x \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

1/6*b^5*x^6 + a*b^4*x^5 + 5/2*a^2*b^3*x^4 + 10/3*a^3*b^2*x^3 + 5/2*a^4*b*x^2 + a^5*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 = 1.43 (sec) , antiderivative size = 666, normalized size of antiderivative = 20.81 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=a^{4} \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 ) + 4 a^{3} 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 ) + 6 a^{2} 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 ) + 4 a b^{3} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{4}}{20 b^{4}} + \frac {a^{3} x}{20 b^{3}} - \frac {a^{2} x^{2}}{20 b^{2}} + \frac {a x^{3}}{20 b} + \frac {x^{4}}{5}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{6} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} + \frac {3 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} - \frac {3 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}}{8 a^{4} b^{4}} & \text {for}\: a b \neq 0 \\\frac {x^{4} \sqrt {a^{2}}}{4} & \text {otherwise} \end {cases}\right ) + b^{4} \left (\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{5}}{30 b^{5}} - \frac {a^{4} x}{30 b^{4}} + \frac {a^{3} x^{2}}{30 b^{3}} - \frac {a^{2} x^{3}}{30 b^{2}} + \frac {a x^{4}}{30 b} + \frac {x^{5}}{6}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{8} \left (a^{2} + 2 a b x\right )^{\frac {3}{2}}}{3} - \frac {4 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {6 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} - \frac {4 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11}}{16 a^{5} b^{5}} & \text {for}\: a b \neq 0 \\\frac {x^{5} \sqrt {a^{2}}}{5} & \text {otherwise} \end {cases}\right ) \]

[In]

integrate((b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

a**4*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)) + 4*a**3*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)) + 6*a**2*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))
+ 4*a*b**3*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3*x/(20*b**3) - a**2*x**2/(20*b*
*2) + a*x**3/(20*b) + x**4/5), Ne(b**2, 0)), ((-a**6*(a**2 + 2*a*b*x)**(3/2)/3 + 3*a**4*(a**2 + 2*a*b*x)**(5/2
)/5 - 3*a**2*(a**2 + 2*a*b*x)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8*a**4*b**4), Ne(a*b, 0)), (x**4*sqrt(a**
2)/4, True)) + b**4*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**5/(30*b**5) - a**4*x/(30*b**4) + a**3*x**2
/(30*b**3) - a**2*x**3/(30*b**2) + a*x**4/(30*b) + x**5/6), Ne(b**2, 0)), ((a**8*(a**2 + 2*a*b*x)**(3/2)/3 - 4
*a**6*(a**2 + 2*a*b*x)**(5/2)/5 + 6*a**4*(a**2 + 2*a*b*x)**(7/2)/7 - 4*a**2*(a**2 + 2*a*b*x)**(9/2)/9 + (a**2
+ 2*a*b*x)**(11/2)/11)/(16*a**5*b**5), Ne(a*b, 0)), (x**5*sqrt(a**2)/5, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{6} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a}{6 \, b} \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (28) = 56\).

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.59 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} a^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{6} \mathrm {sgn}\left (b x + a\right )}{6 \, b} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )}^{2} a^{2} b \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, {\left (b x^{2} + 2 \, a x\right )}^{3} b^{2} \mathrm {sgn}\left (b x + a\right ) \]

[In]

integrate((b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {\left (x\,b^2+a\,b\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}}{6\,b^2} \]

[In]

int((a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)

[Out]

((a*b + b^2*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2))/(6*b^2)