Integrand size = 26, antiderivative size = 255 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )}+\frac {5 a^4 b x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {10 a^3 b^2 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {10 a^2 b^3 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )}+\frac {5 a b^4 x^{13} \sqrt {a^2+2 a b x^2+b^2 x^4}}{13 \left (a+b x^2\right )}+\frac {b^5 x^{15} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 \left (a+b x^2\right )} \] Output:
a^5*x^5*((b*x^2+a)^2)^(1/2)/(5*b*x^2+5*a)+5*a^4*b*x^7*((b*x^2+a)^2)^(1/2)/ (7*b*x^2+7*a)+10*a^3*b^2*x^9*((b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)+10*a^2*b^3* x^11*((b*x^2+a)^2)^(1/2)/(11*b*x^2+11*a)+5*a*b^4*x^13*((b*x^2+a)^2)^(1/2)/ (13*b*x^2+13*a)+b^5*x^15*((b*x^2+a)^2)^(1/2)/(15*b*x^2+15*a)
Time = 1.01 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^5 \sqrt {\left (a+b x^2\right )^2} \left (9009 a^5+32175 a^4 b x^2+50050 a^3 b^2 x^4+40950 a^2 b^3 x^6+17325 a b^4 x^8+3003 b^5 x^{10}\right )}{45045 \left (a+b x^2\right )} \] Input:
Integrate[x^4*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(x^5*Sqrt[(a + b*x^2)^2]*(9009*a^5 + 32175*a^4*b*x^2 + 50050*a^3*b^2*x^4 + 40950*a^2*b^3*x^6 + 17325*a*b^4*x^8 + 3003*b^5*x^10))/(45045*(a + b*x^2))
Time = 0.39 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.40, 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.
\(\displaystyle \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 x^4 \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int x^4 \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^{14}+5 a b^4 x^{12}+10 a^2 b^3 x^{10}+10 a^3 b^2 x^8+5 a^4 b x^6+a^5 x^4\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {a^5 x^5}{5}+\frac {5}{7} a^4 b x^7+\frac {10}{9} a^3 b^2 x^9+\frac {10}{11} a^2 b^3 x^{11}+\frac {5}{13} a b^4 x^{13}+\frac {b^5 x^{15}}{15}\right )}{a+b x^2}\) |
Input:
Int[x^4*(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^5)/5 + (5*a^4*b*x^7)/7 + (10*a^3* b^2*x^9)/9 + (10*a^2*b^3*x^11)/11 + (5*a*b^4*x^13)/13 + (b^5*x^15)/15))/(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 = 1.52 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x^{5} \left (3003 x^{10} b^{5}+17325 a \,x^{8} b^{4}+40950 a^{2} x^{6} b^{3}+50050 a^{3} x^{4} b^{2}+32175 x^{2} a^{4} b +9009 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{45045 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(\frac {x^{5} \left (3003 x^{10} b^{5}+17325 a \,x^{8} b^{4}+40950 a^{2} x^{6} b^{3}+50050 a^{3} x^{4} b^{2}+32175 x^{2} a^{4} b +9009 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{45045 \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(\frac {x^{5} \left (3003 x^{10} b^{5}+17325 a \,x^{8} b^{4}+40950 a^{2} x^{6} b^{3}+50050 a^{3} x^{4} b^{2}+32175 x^{2} a^{4} b +9009 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{45045 \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {a^{5} x^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{5 b \,x^{2}+5 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \,a^{4} x^{7}}{7 \left (b \,x^{2}+a \right )}+\frac {10 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{3} b^{2} x^{9}}{9 \left (b \,x^{2}+a \right )}+\frac {10 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{11}}{11 \left (b \,x^{2}+a \right )}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{4} a \,x^{13}}{13 \left (b \,x^{2}+a \right )}+\frac {b^{5} x^{15} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{15 b \,x^{2}+15 a}\) | \(178\) |
Input:
int(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/45045*x^5*(3003*b^5*x^10+17325*a*b^4*x^8+40950*a^2*b^3*x^6+50050*a^3*b^2 *x^4+32175*a^4*b*x^2+9009*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{13} \, a b^{4} x^{13} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {10}{9} \, a^{3} b^{2} x^{9} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{5} \, a^{5} x^{5} \] Input:
integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/15*b^5*x^15 + 5/13*a*b^4*x^13 + 10/11*a^2*b^3*x^11 + 10/9*a^3*b^2*x^9 + 5/7*a^4*b*x^7 + 1/5*a^5*x^5
\[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \] Input:
integrate(x**4*(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**4*((a + b*x**2)**2)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{15} \, b^{5} x^{15} + \frac {5}{13} \, a b^{4} x^{13} + \frac {10}{11} \, a^{2} b^{3} x^{11} + \frac {10}{9} \, a^{3} b^{2} x^{9} + \frac {5}{7} \, a^{4} b x^{7} + \frac {1}{5} \, a^{5} x^{5} \] Input:
integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/15*b^5*x^15 + 5/13*a*b^4*x^13 + 10/11*a^2*b^3*x^11 + 10/9*a^3*b^2*x^9 + 5/7*a^4*b*x^7 + 1/5*a^5*x^5
Time = 0.15 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.41 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{15} \, b^{5} x^{15} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{13} \, a b^{4} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{11} \, a^{2} b^{3} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{9} \, a^{3} b^{2} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a^{4} b x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {1}{5} \, a^{5} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/15*b^5*x^15*sgn(b*x^2 + a) + 5/13*a*b^4*x^13*sgn(b*x^2 + a) + 10/11*a^2* b^3*x^11*sgn(b*x^2 + a) + 10/9*a^3*b^2*x^9*sgn(b*x^2 + a) + 5/7*a^4*b*x^7* sgn(b*x^2 + a) + 1/5*a^5*x^5*sgn(b*x^2 + a)
Timed out. \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int x^4\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int(x^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^4*(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 \[ \int x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x^{5} \left (3003 b^{5} x^{10}+17325 a \,b^{4} x^{8}+40950 a^{2} b^{3} x^{6}+50050 a^{3} b^{2} x^{4}+32175 a^{4} b \,x^{2}+9009 a^{5}\right )}{45045} \] Input:
int(x^4*(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x**5*(9009*a**5 + 32175*a**4*b*x**2 + 50050*a**3*b**2*x**4 + 40950*a**2*b **3*x**6 + 17325*a*b**4*x**8 + 3003*b**5*x**10))/45045