\(\int x^2 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\) [166]

Optimal result

Integrand size = 24, antiderivative size = 107 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {a^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3}-\frac {2 a (a+b x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{7 b^3}+\frac {(a+b x)^7 \sqrt {a^2+2 a b x+b^2 x^2}}{8 b^3} \]

[Out]

1/6*a^2*(b*x+a)^5*((b*x+a)^2)^(1/2)/b^3-2/7*a*(b*x+a)^6*((b*x+a)^2)^(1/2)/b^3+1/8*(b*x+a)^7*((b*x+a)^2)^(1/2)/
b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 107, 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} \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^7}{8 b^3}-\frac {2 a \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^3}+\frac {a^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

[In]

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

[Out]

(a^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3) - (2*a*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b
^3) + ((a + b*x)^7*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(8*b^3)

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

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {x^3 \sqrt {(a+b x)^2} \left (56 a^5+210 a^4 b x+336 a^3 b^2 x^2+280 a^2 b^3 x^3+120 a b^4 x^4+21 b^5 x^5\right )}{168 (a+b x)} \]

[In]

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

[Out]

(x^3*Sqrt[(a + b*x)^2]*(56*a^5 + 210*a^4*b*x + 336*a^3*b^2*x^2 + 280*a^2*b^3*x^3 + 120*a*b^4*x^4 + 21*b^5*x^5)
)/(168*(a + b*x))

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {x^{3} \left (21 b^{5} x^{5}+120 a \,b^{4} x^{4}+280 a^{2} b^{3} x^{3}+336 a^{3} b^{2} x^{2}+210 a^{4} b x +56 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 \left (b x +a \right )^{5}}\)	\(74\)
default	\(\frac {x^{3} \left (21 b^{5} x^{5}+120 a \,b^{4} x^{4}+280 a^{2} b^{3} x^{3}+336 a^{3} b^{2} x^{2}+210 a^{4} b x +56 a^{5}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}{168 \left (b x +a \right )^{5}}\)	\(74\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b^{5} x^{8}}{8 b x +8 a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a \,b^{4} x^{7}}{7 \left (b x +a \right )}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a^{2} b^{3} x^{6}}{3 \left (b x +a \right )}+\frac {2 \sqrt {\left (b x +a \right )^{2}}\, a^{3} b^{2} x^{5}}{b x +a}+\frac {5 \sqrt {\left (b x +a \right )^{2}}\, a^{4} b \,x^{4}}{4 \left (b x +a \right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, a^{5} x^{3}}{3 b x +3 a}\)	\(154\)

[In]

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

[Out]

1/168*x^3*(21*b^5*x^5+120*a*b^4*x^4+280*a^2*b^3*x^3+336*a^3*b^2*x^2+210*a^4*b*x+56*a^5)*((b*x+a)^2)^(5/2)/(b*x
+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.53 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{8} \, b^{5} x^{8} + \frac {5}{7} \, a b^{4} x^{7} + \frac {5}{3} \, a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{4} \, a^{4} b x^{4} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

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

[Out]

1/8*b^5*x^8 + 5/7*a*b^4*x^7 + 5/3*a^2*b^3*x^6 + 2*a^3*b^2*x^5 + 5/4*a^4*b*x^4 + 1/3*a^5*x^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 180 vs. \(2 (71) = 142\).

Time = 0.68 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.68 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {a^{2} + 2 a b x + b^{2} x^{2}} \left (\frac {a^{7}}{168 b^{3}} - \frac {a^{6} x}{168 b^{2}} + \frac {a^{5} x^{2}}{168 b} + \frac {55 a^{4} x^{3}}{168} + \frac {155 a^{3} b x^{4}}{168} + \frac {181 a^{2} b^{2} x^{5}}{168} + \frac {33 a b^{3} x^{6}}{56} + \frac {b^{4} x^{7}}{8}\right ) & \text {for}\: b^{2} \neq 0 \\\frac {\frac {a^{4} \left (a^{2} + 2 a b x\right )^{\frac {7}{2}}}{7} - \frac {2 a^{2} \left (a^{2} + 2 a b x\right )^{\frac {9}{2}}}{9} + \frac {\left (a^{2} + 2 a b x\right )^{\frac {11}{2}}}{11}}{4 a^{3} b^{3}} & \text {for}\: a b \neq 0 \\\frac {x^{3} \left (a^{2}\right )^{\frac {5}{2}}}{3} & \text {otherwise} \end {cases} \]

[In]

integrate(x**2*(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**7/(168*b**3) - a**6*x/(168*b**2) + a**5*x**2/(168*b) + 55*a**4
*x**3/168 + 155*a**3*b*x**4/168 + 181*a**2*b**2*x**5/168 + 33*a*b**3*x**6/56 + b**4*x**7/8), Ne(b**2, 0)), ((a
**4*(a**2 + 2*a*b*x)**(7/2)/7 - 2*a**2*(a**2 + 2*a*b*x)**(9/2)/9 + (a**2 + 2*a*b*x)**(11/2)/11)/(4*a**3*b**3),
Ne(a*b, 0)), (x**3*(a**2)**(5/2)/3, True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{2} x}{6 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a^{3}}{6 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} x}{8 \, b^{2}} - \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {7}{2}} a}{56 \, b^{3}} \]

[In]

integrate(x^2*(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^2*x/b^2 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a^3/b^3 + 1/8*(b^2*x^2 + 2
*a*b*x + a^2)^(7/2)*x/b^2 - 9/56*(b^2*x^2 + 2*a*b*x + a^2)^(7/2)*a/b^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a^2+2 a b x+b^2 x^2\right )^{5/2} \, dx=\frac {1}{8} \, b^{5} x^{8} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{7} \, a b^{4} x^{7} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{3} \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {5}{4} \, a^{4} b x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a^{5} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {a^{8} \mathrm {sgn}\left (b x + a\right )}{168 \, b^{3}} \]

[In]

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

[Out]

1/8*b^5*x^8*sgn(b*x + a) + 5/7*a*b^4*x^7*sgn(b*x + a) + 5/3*a^2*b^3*x^6*sgn(b*x + a) + 2*a^3*b^2*x^5*sgn(b*x +
a) + 5/4*a^4*b*x^4*sgn(b*x + a) + 1/3*a^5*x^3*sgn(b*x + a) + 1/168*a^8*sgn(b*x + a)/b^3

Mupad [F(-1)]

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

[In]

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

[Out]

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