\(\int x^2 (a^2+2 a b x^2+b^2 x^4)^{5/2} \, dx\) [610]

Optimal result

Integrand size = 26, antiderivative size = 252 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )} \]

[Out]

1/3*a^5*x^3*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+a^4*b*x^5*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10/7*a^3*b^2*x^7*((b*x^2+a)^
2)^(1/2)/(b*x^2+a)+10/9*a^2*b^3*x^9*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5/11*a*b^4*x^11*((b*x^2+a)^2)^(1/2)/(b*x^2+a
)+1/13*b^5*x^13*((b*x^2+a)^2)^(1/2)/(b*x^2+a)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1126, 276} \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )} \]

[In]

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

[Out]

(a^5*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*(a + b*x^2)) + (a^4*b*x^5*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b
*x^2) + (10*a^3*b^2*x^7*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(7*(a + b*x^2)) + (10*a^2*b^3*x^9*Sqrt[a^2 + 2*a*b*x^
2 + b^2*x^4])/(9*(a + b*x^2)) + (5*a*b^4*x^11*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(11*(a + b*x^2)) + (b^5*x^13*Sq
rt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*(a + b*x^2))

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 1126

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^2 \left (a b+b^2 x^2\right )^5 \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (a^5 b^5 x^2+5 a^4 b^6 x^4+10 a^3 b^7 x^6+10 a^2 b^8 x^8+5 a b^9 x^{10}+b^{10} x^{12}\right ) \, dx}{b^4 \left (a b+b^2 x^2\right )} \\ & = \frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^3 \sqrt {\left (a+b x^2\right )^2} \left (3003 a^5+9009 a^4 b x^2+12870 a^3 b^2 x^4+10010 a^2 b^3 x^6+4095 a b^4 x^8+693 b^5 x^{10}\right )}{9009 \left (a+b x^2\right )} \]

[In]

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

[Out]

(x^3*Sqrt[(a + b*x^2)^2]*(3003*a^5 + 9009*a^4*b*x^2 + 12870*a^3*b^2*x^4 + 10010*a^2*b^3*x^6 + 4095*a*b^4*x^8 +
693*b^5*x^10))/(9009*(a + b*x^2))

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32

method	result	size
gosper	\(\frac {x^{3} \left (693 x^{10} b^{5}+4095 a \,x^{8} b^{4}+10010 a^{2} x^{6} b^{3}+12870 a^{3} x^{4} b^{2}+9009 x^{2} a^{4} b +3003 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{9009 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
default	\(\frac {x^{3} \left (693 x^{10} b^{5}+4095 a \,x^{8} b^{4}+10010 a^{2} x^{6} b^{3}+12870 a^{3} x^{4} b^{2}+9009 x^{2} a^{4} b +3003 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{9009 \left (b \,x^{2}+a \right )^{5}}\)	\(80\)
risch	\(\frac {a^{5} x^{3} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{3 b \,x^{2}+3 a}+\frac {a^{4} b \,x^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}+\frac {10 a^{3} b^{2} x^{7} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{7 \left (b \,x^{2}+a \right )}+\frac {10 a^{2} b^{3} x^{9} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{9 \left (b \,x^{2}+a \right )}+\frac {5 a \,b^{4} x^{11} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{11 \left (b \,x^{2}+a \right )}+\frac {b^{5} x^{13} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{13 b \,x^{2}+13 a}\)	\(177\)

[In]

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

[Out]

1/9009*x^3*(693*b^5*x^10+4095*a*b^4*x^8+10010*a^2*b^3*x^6+12870*a^3*b^2*x^4+9009*a^4*b*x^2+3003*a^5)*((b*x^2+a
)^2)^(5/2)/(b*x^2+a)^5

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

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

[Out]

1/13*b^5*x^13 + 5/11*a*b^4*x^11 + 10/9*a^2*b^3*x^9 + 10/7*a^3*b^2*x^7 + a^4*b*x^5 + 1/3*a^5*x^3

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \]

[In]

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

[Out]

1/13*b^5*x^13 + 5/11*a*b^4*x^11 + 10/9*a^2*b^3*x^9 + 10/7*a^3*b^2*x^7 + a^4*b*x^5 + 1/3*a^5*x^3

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{11} \, a b^{4} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a^{5} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \]

[In]

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

[Out]

1/13*b^5*x^13*sgn(b*x^2 + a) + 5/11*a*b^4*x^11*sgn(b*x^2 + a) + 10/9*a^2*b^3*x^9*sgn(b*x^2 + a) + 10/7*a^3*b^2
*x^7*sgn(b*x^2 + a) + a^4*b*x^5*sgn(b*x^2 + a) + 1/3*a^5*x^3*sgn(b*x^2 + a)

Mupad [F(-1)]

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

[In]

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

[Out]

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