\(\int x^5 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [560]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {a^2 \left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{8 b^3}-\frac {a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{5 b^3}+\frac {\left (a+b x^2\right ) \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{12 b^3} \]

[Out]

1/8*a^2*(b*x^2+a)*(b^2*x^4+2*a*b*x^2+a^2)^(3/2)/b^3-1/5*a*(b^2*x^4+2*a*b*x^2+a^2)^(5/2)/b^3+1/12*(b*x^2+a)*(b^
2*x^4+2*a*b*x^2+a^2)^(5/2)/b^3

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1125, 659} \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^3}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^4}{5 b^3}+\frac {a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^3}{8 b^3} \]

[In]

Int[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(a^2*(a + b*x^2)^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(8*b^3) - (a*(a + b*x^2)^4*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
)/(5*b^3) + ((a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^3)

Rule 659

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

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^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (\frac {a^2 \left (a b+b^2 x\right )^3}{b^2}-\frac {2 a \left (a b+b^2 x\right )^4}{b^3}+\frac {\left (a b+b^2 x\right )^5}{b^4}\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 b^3}-\frac {a \left (a+b x^2\right )^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 b^3}+\frac {\left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.07 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x^6 \left (20 a^3+45 a^2 b x^2+36 a b^2 x^4+10 b^3 x^6\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{120 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^5*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^6*(20*a^3 + 45*a^2*b*x^2 + 36*a*b^2*x^4 + 10*b^3*x^6)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2]
)))/(120*(-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 = 0.11 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.43

method	result	size
pseudoelliptic	\(\frac {x^{6} \left (10 b^{3} x^{6}+36 b^{2} x^{4} a +45 a^{2} b \,x^{2}+20 a^{3}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{120}\)	\(46\)
gosper	\(\frac {x^{6} \left (10 b^{3} x^{6}+36 b^{2} x^{4} a +45 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{120 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{6} \left (10 b^{3} x^{6}+36 b^{2} x^{4} a +45 a^{2} b \,x^{2}+20 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{120 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} x^{12}}{12 b \,x^{2}+12 a}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \,b^{2} x^{10}}{10 \left (b \,x^{2}+a \right )}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b \,x^{8}}{8 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} x^{6}}{6 b \,x^{2}+6 a}\)	\(116\)

[In]

int(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/120*x^6*(10*b^3*x^6+36*a*b^2*x^4+45*a^2*b*x^2+20*a^3)*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.33 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{12} \, b^{3} x^{12} + \frac {3}{10} \, a b^{2} x^{10} + \frac {3}{8} \, a^{2} b x^{8} + \frac {1}{6} \, a^{3} x^{6} \]

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/12*b^3*x^12 + 3/10*a*b^2*x^10 + 3/8*a^2*b*x^8 + 1/6*a^3*x^6

Sympy [F]

\[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^{5} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**5*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**5*((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 = 0.33 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{12} \, b^{3} x^{12} + \frac {3}{10} \, a b^{2} x^{10} + \frac {3}{8} \, a^{2} b x^{8} + \frac {1}{6} \, a^{3} x^{6} \]

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/12*b^3*x^12 + 3/10*a*b^2*x^10 + 3/8*a^2*b*x^8 + 1/6*a^3*x^6

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.63 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{12} \, b^{3} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{10} \, a b^{2} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{8} \, a^{2} b x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{6} \, a^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^5*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/12*b^3*x^12*sgn(b*x^2 + a) + 3/10*a*b^2*x^10*sgn(b*x^2 + a) + 3/8*a^2*b*x^8*sgn(b*x^2 + a) + 1/6*a^3*x^6*sgn
(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^5\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \]

[In]

int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)