\(\int x^3 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [561]

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2} \]

[Out]

-1/8*a*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/b^2+1/10*(b^2*x^4+2*a*b*x^2+a^2)^(5/2)/b^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 654, 623} \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}-\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2} \]

[In]

Int[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

-1/8*(a*(a + b*x^2)*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2))/b^2 + (a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/(10*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]

Rule 1125

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && Integ
erQ[(m - 1)/2] && (GtQ[m, 0] || LtQ[0, 4*p, -m - 1])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2}-\frac {a \text {Subst}\left (\int \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right )}{2 b} \\ & = -\frac {a \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^2}+\frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{10 b^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.69 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x^4 \left (10 a^3+20 a^2 b x^2+15 a b^2 x^4+4 b^3 x^6\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{40 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^4*(10*a^3 + 20*a^2*b*x^2 + 15*a*b^2*x^4 + 4*b^3*x^6)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])
))/(40*(-a^2 - a*b*x^2 + Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))

Maple [C] (warning: unable to verify)

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

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(\frac {x^{4} \left (4 b^{3} x^{6}+15 b^{2} x^{4} a +20 a^{2} b \,x^{2}+10 a^{3}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{40}\)	\(46\)
gosper	\(\frac {x^{4} \left (4 b^{3} x^{6}+15 b^{2} x^{4} a +20 a^{2} b \,x^{2}+10 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{4} \left (4 b^{3} x^{6}+15 b^{2} x^{4} a +20 a^{2} b \,x^{2}+10 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{40 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} x^{10}}{10 b \,x^{2}+10 a}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \,b^{2} x^{8}}{8 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b \,x^{6}}{2 b \,x^{2}+2 a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} x^{4}}{4 b \,x^{2}+4 a}\)	\(116\)

[In]

int(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/40*x^4*(4*b^3*x^6+15*a*b^2*x^4+20*a^2*b*x^2+10*a^3)*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.52 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{10} \, b^{3} x^{10} + \frac {3}{8} \, a b^{2} x^{8} + \frac {1}{2} \, a^{2} b x^{6} + \frac {1}{4} \, a^{3} x^{4} \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/10*b^3*x^10 + 3/8*a*b^2*x^8 + 1/2*a^2*b*x^6 + 1/4*a^3*x^4

Sympy [F]

\[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**3*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.52 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{10} \, b^{3} x^{10} + \frac {3}{8} \, a b^{2} x^{8} + \frac {1}{2} \, a^{2} b x^{6} + \frac {1}{4} \, a^{3} x^{4} \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/10*b^3*x^10 + 3/8*a*b^2*x^8 + 1/2*a^2*b*x^6 + 1/4*a^3*x^4

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{40} \, {\left (4 \, b^{3} x^{10} + 15 \, a b^{2} x^{8} + 20 \, a^{2} b x^{6} + 10 \, a^{3} x^{4}\right )} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^3*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/40*(4*b^3*x^10 + 15*a*b^2*x^8 + 20*a^2*b*x^6 + 10*a^3*x^4)*sgn(b*x^2 + a)

Mupad [B] (verification not implemented)

Time = 13.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.69 \[ \int x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}\,\left (-a^2+3\,a\,b\,x^2+4\,b^2\,x^4\right )}{40\,b^2} \]

[In]

int(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

((a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)*(4*b^2*x^4 - a^2 + 3*a*b*x^2))/(40*b^2)