\(\int x^5 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\) [589]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^2 \left (a+b x^2\right )^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{12 b^3}-\frac {a \left (a+b x^2\right )^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 b^3}+\frac {\left (a+b x^2\right )^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 b^3} \]

[Out]

1/12*a^2*(b*x^2+a)^5*((b*x^2+a)^2)^(1/2)/b^3-1/7*a*(b*x^2+a)^6*((b*x^2+a)^2)^(1/2)/b^3+1/16*(b*x^2+a)^7*((b*x^
2+a)^2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 119, 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} \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^7}{16 b^3}-\frac {a \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^6}{7 b^3}+\frac {a^2 \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )^5}{12 b^3} \]

[In]

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

[Out]

(a^2*(a + b*x^2)^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(12*b^3) - (a*(a + b*x^2)^6*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4
])/(7*b^3) + ((a + b*x^2)^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(16*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]

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

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^6 \left (56 a^5+210 a^4 b x^2+336 a^3 b^2 x^4+280 a^2 b^3 x^6+120 a b^4 x^8+21 b^5 x^{10}\right ) \left (\sqrt {a^2} b x^2+a \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{336 \left (-a^2-a b x^2+\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )} \]

[In]

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

[Out]

(x^6*(56*a^5 + 210*a^4*b*x^2 + 336*a^3*b^2*x^4 + 280*a^2*b^3*x^6 + 120*a*b^4*x^8 + 21*b^5*x^10)*(Sqrt[a^2]*b*x
^2 + a*(Sqrt[a^2] - Sqrt[(a + b*x^2)^2])))/(336*(-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 = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.55

method	result	size
pseudoelliptic	\(\frac {x^{6} \operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {3}{8} x^{10} b^{5}+\frac {15}{7} a \,x^{8} b^{4}+5 a^{2} x^{6} b^{3}+6 a^{3} x^{4} b^{2}+\frac {15}{4} x^{2} a^{4} b +a^{5}\right )}{6}\)	\(66\)
gosper	\(\frac {x^{6} \left (21 x^{10} b^{5}+120 a \,x^{8} b^{4}+280 a^{2} x^{6} b^{3}+336 a^{3} x^{4} b^{2}+210 x^{2} a^{4} b +56 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{6} \left (21 x^{10} b^{5}+120 a \,x^{8} b^{4}+280 a^{2} x^{6} b^{3}+336 a^{3} x^{4} b^{2}+210 x^{2} a^{4} b +56 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{336 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{5} x^{6}}{6 b \,x^{2}+6 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{8}}{8 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{10}}{b \,x^{2}+a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{12}}{6 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{14}}{14 \left (b \,x^{2}+a \right )}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{5} x^{16}}{16 b \,x^{2}+16 a}\)	\(177\)

[In]

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

[Out]

1/6*x^6*csgn(b*x^2+a)*(3/8*x^10*b^5+15/7*a*x^8*b^4+5*a^2*x^6*b^3+6*a^3*x^4*b^2+15/4*x^2*a^4*b+a^5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.47 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{14} \, a b^{4} x^{14} + \frac {5}{6} \, a^{2} b^{3} x^{12} + a^{3} b^{2} x^{10} + \frac {5}{8} \, a^{4} b x^{8} + \frac {1}{6} \, a^{5} x^{6} \]

[In]

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

[Out]

1/16*b^5*x^16 + 5/14*a*b^4*x^14 + 5/6*a^2*b^3*x^12 + a^3*b^2*x^10 + 5/8*a^4*b*x^8 + 1/6*a^5*x^6

Sympy [F]

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

[In]

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

[Out]

Integral(x**5*((a + b*x**2)**2)**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.47 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} + \frac {5}{14} \, a b^{4} x^{14} + \frac {5}{6} \, a^{2} b^{3} x^{12} + a^{3} b^{2} x^{10} + \frac {5}{8} \, a^{4} b x^{8} + \frac {1}{6} \, a^{5} x^{6} \]

[In]

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

[Out]

1/16*b^5*x^16 + 5/14*a*b^4*x^14 + 5/6*a^2*b^3*x^12 + a^3*b^2*x^10 + 5/8*a^4*b*x^8 + 1/6*a^5*x^6

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int x^5 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{16} \, b^{5} x^{16} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{14} \, a b^{4} x^{14} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{6} \, a^{2} b^{3} x^{12} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{3} b^{2} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{8} \, a^{4} b x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{6} \, a^{5} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

1/16*b^5*x^16*sgn(b*x^2 + a) + 5/14*a*b^4*x^14*sgn(b*x^2 + a) + 5/6*a^2*b^3*x^12*sgn(b*x^2 + a) + a^3*b^2*x^10
*sgn(b*x^2 + a) + 5/8*a^4*b*x^8*sgn(b*x^2 + a) + 1/6*a^5*x^6*sgn(b*x^2 + a)

Mupad [F(-1)]

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

[In]

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

[Out]

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