Integrand size = 26, antiderivative size = 252 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {a^4 b x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {5 a b^4 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {b^5 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )} \] Output:
a^5*x^3*((b*x^2+a)^2)^(1/2)/(3*b*x^2+3*a)+a^4*b*x^5*((b*x^2+a)^2)^(1/2)/(b *x^2+a)+10*a^3*b^2*x^7*((b*x^2+a)^2)^(1/2)/(7*b*x^2+7*a)+10*a^2*b^3*x^9*(( b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)+5*a*b^4*x^11*((b*x^2+a)^2)^(1/2)/(11*b*x^2 +11*a)+b^5*x^13*((b*x^2+a)^2)^(1/2)/(13*b*x^2+13*a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^3 \sqrt {\left (a+b x^2\right )^2} \left (3003 a^5+9009 a^4 b x^2+12870 a^3 b^2 x^4+10010 a^2 b^3 x^6+4095 a b^4 x^8+693 b^5 x^{10}\right )}{9009 \left (a+b x^2\right )} \] Input:
Integrate[x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(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))
Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.39, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1384, 27, 244, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^2 \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^2 \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^{12}+5 a b^4 x^{10}+10 a^2 b^3 x^8+10 a^3 b^2 x^6+5 a^4 b x^4+a^5 x^2\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^5 x^3}{3}+a^4 b x^5+\frac {10}{7} a^3 b^2 x^7+\frac {10}{9} a^2 b^3 x^9+\frac {5}{11} a b^4 x^{11}+\frac {b^5 x^{13}}{13}\right )}{a+b x^2}\) |
Input:
Int[x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((a^5*x^3)/3 + a^4*b*x^5 + (10*a^3*b^2*x^ 7)/7 + (10*a^2*b^3*x^9)/9 + (5*a*b^4*x^11)/11 + (b^5*x^13)/13))/(a + b*x^2 )
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 1.05 (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 | \(\frac {x^{3} \left (693 x^{10} b^{5}+4095 a \,x^{8} b^{4}+10010 a^{2} x^{6} b^{3}+12870 a^{3} x^{4} b^{2}+9009 x^{2} a^{4} b +3003 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{9009 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{3} \left (693 x^{10} b^{5}+4095 a \,x^{8} b^{4}+10010 a^{2} x^{6} b^{3}+12870 a^{3} x^{4} b^{2}+9009 x^{2} a^{4} b +3003 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{9009 \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {b^{5} x^{13} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{13 b \,x^{2}+13 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \,b^{4} x^{11}}{11 \left (b \,x^{2}+a \right )}+\frac {10 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{9}}{9 \left (b \,x^{2}+a \right )}+\frac {10 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{7}}{7 \left (b \,x^{2}+a \right )}+\frac {a^{4} b \,x^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}+\frac {a^{5} x^{3} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{3 b \,x^{2}+3 a}\) | \(177\) |
Input:
int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
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
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \] Input:
integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
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
\[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**2*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**2*((a + b*x**2)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.22 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} + \frac {5}{11} \, a b^{4} x^{11} + \frac {10}{9} \, a^{2} b^{3} x^{9} + \frac {10}{7} \, a^{3} b^{2} x^{7} + a^{4} b x^{5} + \frac {1}{3} \, a^{5} x^{3} \] Input:
integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
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
Time = 0.15 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.41 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{13} \, b^{5} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{11} \, a b^{4} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{9} \, a^{2} b^{3} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, a^{3} b^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{4} b x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{3} \, a^{5} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(x^2*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
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)
Timed out. \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{3} \left (693 b^{5} x^{10}+4095 a \,b^{4} x^{8}+10010 a^{2} b^{3} x^{6}+12870 a^{3} b^{2} x^{4}+9009 a^{4} b \,x^{2}+3003 a^{5}\right )}{9009} \] Input:
int(x^2*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x**3*(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