\(\int x^9 (a^2+2 a b x^2+b^2 x^4)^{3/2} \, dx\) [558]

Optimal result

Integrand size = 26, antiderivative size = 167 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {a^3 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {a^2 b x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {b^3 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )} \]

[Out]

1/10*a^3*x^10*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+1/4*a^2*b*x^12*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+3/14*a*b^2*x^14*((b*x
^2+a)^2)^(1/2)/(b*x^2+a)+1/16*b^3*x^16*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1125, 660, 45} \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {a^2 b x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {b^3 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )}+\frac {a^3 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )} \]

[In]

Int[x^9*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(a^3*x^10*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(10*(a + b*x^2)) + (a^2*b*x^12*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(4*
(a + b*x^2)) + (3*a*b^2*x^14*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(14*(a + b*x^2)) + (b^3*x^16*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])/(16*(a + b*x^2))

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]

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^4 \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 x^4 \left (a b+b^2 x\right )^3 \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \text {Subst}\left (\int \left (a^3 b^3 x^4+3 a^2 b^4 x^5+3 a b^5 x^6+b^6 x^7\right ) \, dx,x,x^2\right )}{2 b^2 \left (a b+b^2 x^2\right )} \\ & = \frac {a^3 x^{10} \sqrt {a^2+2 a b x^2+b^2 x^4}}{10 \left (a+b x^2\right )}+\frac {a^2 b x^{12} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 \left (a+b x^2\right )}+\frac {3 a b^2 x^{14} \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 \left (a+b x^2\right )}+\frac {b^3 x^{16} \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.68 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {x^{10} \left (56 a^3+140 a^2 b x^2+120 a b^2 x^4+35 b^3 x^6\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{560 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

Integrate[x^9*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]

[Out]

(x^10*(56*a^3 + 140*a^2*b*x^2 + 120*a*b^2*x^4 + 35*b^3*x^6)*(Sqrt[a^2]*b*x^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)
^2])))/(560*(-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 = 0.28

method	result	size
pseudoelliptic	\(\frac {x^{10} \left (35 b^{3} x^{6}+120 b^{2} x^{4} a +140 a^{2} b \,x^{2}+56 a^{3}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{560}\)	\(46\)
gosper	\(\frac {x^{10} \left (35 b^{3} x^{6}+120 b^{2} x^{4} a +140 a^{2} b \,x^{2}+56 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{560 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
default	\(\frac {x^{10} \left (35 b^{3} x^{6}+120 b^{2} x^{4} a +140 a^{2} b \,x^{2}+56 a^{3}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}{560 \left (b \,x^{2}+a \right )^{3}}\)	\(58\)
risch	\(\frac {a^{3} x^{10} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{10 b \,x^{2}+10 a}+\frac {a^{2} b \,x^{12} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{4 b \,x^{2}+4 a}+\frac {3 a \,b^{2} x^{14} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{14 \left (b \,x^{2}+a \right )}+\frac {b^{3} x^{16} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{16 b \,x^{2}+16 a}\)	\(116\)

[In]

int(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/560*x^10*(35*b^3*x^6+120*a*b^2*x^4+140*a^2*b*x^2+56*a^3)*csgn(b*x^2+a)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.21 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{16} \, b^{3} x^{16} + \frac {3}{14} \, a b^{2} x^{14} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{10} \, a^{3} x^{10} \]

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/16*b^3*x^16 + 3/14*a*b^2*x^14 + 1/4*a^2*b*x^12 + 1/10*a^3*x^10

Sympy [F]

\[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^{9} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate(x**9*(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(x**9*((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 = 0.21 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{16} \, b^{3} x^{16} + \frac {3}{14} \, a b^{2} x^{14} + \frac {1}{4} \, a^{2} b x^{12} + \frac {1}{10} \, a^{3} x^{10} \]

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/16*b^3*x^16 + 3/14*a*b^2*x^14 + 1/4*a^2*b*x^12 + 1/10*a^3*x^10

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.40 \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\frac {1}{16} \, b^{3} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {3}{14} \, a b^{2} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{4} \, a^{2} b x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{10} \, a^{3} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

integrate(x^9*(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

1/16*b^3*x^16*sgn(b*x^2 + a) + 3/14*a*b^2*x^14*sgn(b*x^2 + a) + 1/4*a^2*b*x^12*sgn(b*x^2 + a) + 1/10*a^3*x^10*
sgn(b*x^2 + a)

Mupad [F(-1)]

Timed out. \[ \int x^9 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx=\int x^9\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2} \,d x \]

[In]

int(x^9*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)

[Out]

int(x^9*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)