\(\int x (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [167]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 61 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \]

[Out]

-1/6*a*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/b^2+1/7*(b^2*x^2+2*a*b*x+a^2)^(7/2)/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} \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2}-\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2} \]

[In]

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

[Out]

-1/6*(a*(a + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2))/b^2 + (a^2 + 2*a*b*x + b^2*x^2)^(7/2)/(7*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 )^{7/2}}{7 b^2}-\frac {a \int \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx}{b} \\ & = -\frac {a (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^2}+\frac {\left (a^2+2 a b x+b^2 x^2\right )^{7/2}}{7 b^2} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(125\) vs. \(2(61)=122\).

Time = 1.03 (sec) , antiderivative size = 125, normalized size of antiderivative = 2.05 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^2 \left (21 a^5+70 a^4 b x+105 a^3 b^2 x^2+84 a^2 b^3 x^3+35 a b^4 x^4+6 b^5 x^5\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{42 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

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

[Out]

(x^2*(21*a^5 + 70*a^4*b*x + 105*a^3*b^2*x^2 + 84*a^2*b^3*x^3 + 35*a*b^4*x^4 + 6*b^5*x^5)*(Sqrt[a^2]*b*x + a*(S
qrt[a^2] - Sqrt[(a + b*x)^2])))/(42*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.21

method result size
gosper \(\frac {x^{2} \left (6 b^{5} x^{5}+35 a \,b^{4} x^{4}+84 a^{2} b^{3} x^{3}+105 a^{3} b^{2} x^{2}+70 a^{4} b x +21 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{42 \left (b x +a \right )^{5}}\) \(74\)
default \(\frac {x^{2} \left (6 b^{5} x^{5}+35 a \,b^{4} x^{4}+84 a^{2} b^{3} x^{3}+105 a^{3} b^{2} x^{2}+70 a^{4} b x +21 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{42 \left (b x +a \right )^{5}}\) \(74\)
risch \(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{5} x^{7}}{7 b x +7 a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a \,b^{4} x^{6}}{6 \left (b x +a \right )}+\frac {2 \sqrt {\left (b x +a \right )^{2}}\, a^{2} b^{3} x^{5}}{b x +a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a^{3} b^{2} x^{4}}{2 \left (b x +a \right )}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a^{4} b \,x^{3}}{3 \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{5} x^{2}}{2 b x +2 a}\) \(154\)

[In]

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

[Out]

1/42*x^2*(6*b^5*x^5+35*a*b^4*x^4+84*a^2*b^3*x^3+105*a^3*b^2*x^2+70*a^4*b*x+21*a^5)*((b*x+a)^2)^(5/2)/(b*x+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{7} \, b^{5} x^{7} + \frac {5}{6} \, a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{5} + \frac {5}{2} \, a^{3} b^{2} x^{4} + \frac {5}{3} \, a^{4} b x^{3} + \frac {1}{2} \, a^{5} x^{2} \]

[In]

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

[Out]

1/7*b^5*x^7 + 5/6*a*b^4*x^6 + 2*a^2*b^3*x^5 + 5/2*a^3*b^2*x^4 + 5/3*a^4*b*x^3 + 1/2*a^5*x^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (56) = 112\).

Time = 0.60 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.43 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{6}}{42 b^{2}} + \frac {a^{5} x}{42 b} + \frac {10 a^{4} x^{2}}{21} + \frac {25 a^{3} b x^{3}}{21} + \frac {55 a^{2} b^{2} x^{4}}{42} + \frac {29 a b^{3} x^{5}}{42} + \frac {b^{4} x^{6}}{7}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{2} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9}}{2 a^{2} b^{2}} & \text {for}\: a b \neq 0 \\\frac {x^{2} \left (a^{2}\right )^{\frac {5}{2}}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**6/(42*b**2) + a**5*x/(42*b) + 10*a**4*x**2/21 + 25*a**3*b*x**
3/21 + 55*a**2*b**2*x**4/42 + 29*a*b**3*x**5/42 + b**4*x**6/7), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(7/2)/
7 + (a**2 + 2*a*b*x)**(9/2)/9)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*(a**2)**(5/2)/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.23 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x}{6 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2}}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}}}{7 \, b^{2}} \]

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (53) = 106\).

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.75 \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{7} \, b^{5} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{6} \, a b^{4} x^{6} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{2} \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{4} b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a^{5} x^{2} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{7} \mathrm {sgn}\left (b x + a\right )}{42 \, b^{2}} \]

[In]

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

[Out]

1/7*b^5*x^7*sgn(b*x + a) + 5/6*a*b^4*x^6*sgn(b*x + a) + 2*a^2*b^3*x^5*sgn(b*x + a) + 5/2*a^3*b^2*x^4*sgn(b*x +
 a) + 5/3*a^4*b*x^3*sgn(b*x + a) + 1/2*a^5*x^2*sgn(b*x + a) - 1/42*a^7*sgn(b*x + a)/b^2

Mupad [F(-1)]

Timed out. \[ \int x \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\int x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2} \,d x \]

[In]

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

[Out]

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