\(\int x^5 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\) [148]

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

Optimal result

Integrand size = 24, antiderivative size = 151 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {a^3 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {3 a b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \]

[Out]

1/6*a^3*x^6*((b*x+a)^2)^(1/2)/(b*x+a)+3/7*a^2*b*x^7*((b*x+a)^2)^(1/2)/(b*x+a)+3/8*a*b^2*x^8*((b*x+a)^2)^(1/2)/
(b*x+a)+1/9*b^3*x^9*((b*x+a)^2)^(1/2)/(b*x+a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {660, 45} \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {3 a b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)}+\frac {a^3 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)} \]

[In]

Int[x^5*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(a^3*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*(a + b*x)) + (3*a^2*b*x^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*(a + b*
x)) + (3*a*b^2*x^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*(a + b*x)) + (b^3*x^9*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*(
a + b*x))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 660

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[(d + e*x)^m*(b/2 + c*x)^(2*p), 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^5 \left (a b+b^2 x\right )^3 \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (a^3 b^3 x^5+3 a^2 b^4 x^6+3 a b^5 x^7+b^6 x^8\right ) \, dx}{b^2 \left (a b+b^2 x\right )} \\ & = \frac {a^3 x^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 (a+b x)}+\frac {3 a^2 b x^7 \sqrt {a^2+2 a b x+b^2 x^2}}{7 (a+b x)}+\frac {3 a b^2 x^8 \sqrt {a^2+2 a b x+b^2 x^2}}{8 (a+b x)}+\frac {b^3 x^9 \sqrt {a^2+2 a b x+b^2 x^2}}{9 (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.99 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.68 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {x^6 \left (84 a^3+216 a^2 b x+189 a b^2 x^2+56 b^3 x^3\right ) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}{504 \left (-a^2-a b x+\sqrt {a^2} \sqrt {(a+b x)^2}\right )} \]

[In]

Integrate[x^5*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x^6*(84*a^3 + 216*a^2*b*x + 189*a*b^2*x^2 + 56*b^3*x^3)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))/
(504*(-a^2 - a*b*x + Sqrt[a^2]*Sqrt[(a + b*x)^2]))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.34

method result size
gosper \(\frac {x^{6} \left (56 b^{3} x^{3}+189 a \,b^{2} x^{2}+216 a^{2} b x +84 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{504 \left (b x +a \right )^{3}}\) \(52\)
default \(\frac {x^{6} \left (56 b^{3} x^{3}+189 a \,b^{2} x^{2}+216 a^{2} b x +84 a^{3}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}{504 \left (b x +a \right )^{3}}\) \(52\)
risch \(\frac {a^{3} x^{6} \sqrt {\left (b x +a \right )^{2}}}{6 b x +6 a}+\frac {3 a^{2} b \,x^{7} \sqrt {\left (b x +a \right )^{2}}}{7 \left (b x +a \right )}+\frac {3 a \,b^{2} x^{8} \sqrt {\left (b x +a \right )^{2}}}{8 \left (b x +a \right )}+\frac {b^{3} x^{9} \sqrt {\left (b x +a \right )^{2}}}{9 b x +9 a}\) \(100\)

[In]

int(x^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/504*x^6*(56*b^3*x^3+189*a*b^2*x^2+216*a^2*b*x+84*a^3)*((b*x+a)^2)^(3/2)/(b*x+a)^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.23 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{9} \, b^{3} x^{9} + \frac {3}{8} \, a b^{2} x^{8} + \frac {3}{7} \, a^{2} b x^{7} + \frac {1}{6} \, a^{3} x^{6} \]

[In]

integrate(x^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/9*b^3*x^9 + 3/8*a*b^2*x^8 + 3/7*a^2*b*x^7 + 1/6*a^3*x^6

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (107) = 214\).

Time = 0.65 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.66 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (- \frac {a^{8}}{504 b^{6}} + \frac {a^{7} x}{504 b^{5}} - \frac {a^{6} x^{2}}{504 b^{4}} + \frac {a^{5} x^{3}}{504 b^{3}} - \frac {a^{4} x^{4}}{504 b^{2}} + \frac {a^{3} x^{5}}{504 b} + \frac {83 a^{2} x^{6}}{504} + \frac {19 a b x^{7}}{72} + \frac {b^{2} x^{8}}{9}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {- \frac {a^{10} \left (a^{2} + 2 a b x\right )^{\frac {5}{2}}}{5} + \frac {5 a^{8} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} - \frac {10 a^{6} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9} + \frac {10 a^{4} \left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11} - \frac {5 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {13}{2}}}{13} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {15}{2}}}{15}}{32 a^{6} b^{6}} & \text {for}\: a b \neq 0 \\\frac {x^{6} \left (a^{2}\right )^{\frac {3}{2}}}{6} & \text {otherwise} \end {cases} \]

[In]

integrate(x**5*(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**8/(504*b**6) + a**7*x/(504*b**5) - a**6*x**2/(504*b**4) + a**
5*x**3/(504*b**3) - a**4*x**4/(504*b**2) + a**3*x**5/(504*b) + 83*a**2*x**6/504 + 19*a*b*x**7/72 + b**2*x**8/9
), Ne(b**2, 0)), ((-a**10*(a**2 + 2*a*b*x)**(5/2)/5 + 5*a**8*(a**2 + 2*a*b*x)**(7/2)/7 - 10*a**6*(a**2 + 2*a*b
*x)**(9/2)/9 + 10*a**4*(a**2 + 2*a*b*x)**(11/2)/11 - 5*a**2*(a**2 + 2*a*b*x)**(13/2)/13 + (a**2 + 2*a*b*x)**(1
5/2)/15)/(32*a**6*b**6), Ne(a*b, 0)), (x**6*(a**2)**(3/2)/6, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.25 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} x^{4}}{9 \, b^{2}} - \frac {13 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a x^{3}}{72 \, b^{3}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{5} x}{4 \, b^{5}} + \frac {37 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} x^{2}}{168 \, b^{4}} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{6}}{4 \, b^{6}} - \frac {121 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3} x}{504 \, b^{5}} + \frac {125 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{4}}{504 \, b^{6}} \]

[In]

integrate(x^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/9*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*x^4/b^2 - 13/72*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*x^3/b^3 - 1/4*(b^2*x^2 +
 2*a*b*x + a^2)^(3/2)*a^5*x/b^5 + 37/168*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^2*x^2/b^4 - 1/4*(b^2*x^2 + 2*a*b*x
+ a^2)^(3/2)*a^6/b^6 - 121/504*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3*x/b^5 + 125/504*(b^2*x^2 + 2*a*b*x + a^2)^(
5/2)*a^4/b^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.48 \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\frac {1}{9} \, b^{3} x^{9} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{8} \, a b^{2} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{7} \, a^{2} b x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{6} \, a^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) - \frac {a^{9} \mathrm {sgn}\left (b x + a\right )}{504 \, b^{6}} \]

[In]

integrate(x^5*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/9*b^3*x^9*sgn(b*x + a) + 3/8*a*b^2*x^8*sgn(b*x + a) + 3/7*a^2*b*x^7*sgn(b*x + a) + 1/6*a^3*x^6*sgn(b*x + a)
- 1/504*a^9*sgn(b*x + a)/b^6

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx=\int x^5\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

[In]

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

[Out]

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