Integrand size = 22, antiderivative size = 248 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {a^5 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a^4 b x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {2 a^3 b^2 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^2 b^3 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {5 a b^4 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )}+\frac {b^5 x^{11} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 \left (a+b x^2\right )} \] Output:
a^5*x*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+5*a^4*b*x^3*((b*x^2+a)^2)^(1/2)/(3*b*x ^2+3*a)+2*a^3*b^2*x^5*((b*x^2+a)^2)^(1/2)/(b*x^2+a)+10*a^2*b^3*x^7*((b*x^2 +a)^2)^(1/2)/(7*b*x^2+7*a)+5*a*b^4*x^9*((b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)+b ^5*x^11*((b*x^2+a)^2)^(1/2)/(11*b*x^2+11*a)
Time = 1.01 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.33 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (693 a^5 x+1155 a^4 b x^3+1386 a^3 b^2 x^5+990 a^2 b^3 x^7+385 a b^4 x^9+63 b^5 x^{11}\right )}{693 \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(Sqrt[(a + b*x^2)^2]*(693*a^5*x + 1155*a^4*b*x^3 + 1386*a^3*b^2*x^5 + 990* a^2*b^3*x^7 + 385*a*b^4*x^9 + 63*b^5*x^11))/(693*(a + b*x^2))
Time = 0.40 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.41, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1384, 210, 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.
\(\displaystyle \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^2 x^2+a b\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^{10} x^{10}+5 a b^9 x^8+10 a^2 b^8 x^6+10 a^3 b^7 x^4+5 a^4 b^6 x^2+a^5 b^5\right )dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (a^5 b^5 x+\frac {5}{3} a^4 b^6 x^3+2 a^3 b^7 x^5+\frac {10}{7} a^2 b^8 x^7+\frac {5}{9} a b^9 x^9+\frac {b^{10} x^{11}}{11}\right )}{b^5 \left (a+b x^2\right )}\) |
Input:
Int[(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*b^5*x + (5*a^4*b^6*x^3)/3 + 2*a^3*b^ 7*x^5 + (10*a^2*b^8*x^7)/7 + (5*a*b^9*x^9)/9 + (b^10*x^11)/11))/(b^5*(a + b*x^2))
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, 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 = 0.64 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.31
method | result | size |
gosper | \(\frac {x \left (63 x^{10} b^{5}+385 a \,x^{8} b^{4}+990 a^{2} x^{6} b^{3}+1386 a^{3} x^{4} b^{2}+1155 x^{2} a^{4} b +693 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{693 \left (b \,x^{2}+a \right )^{5}}\) | \(78\) |
default | \(\frac {x \left (63 x^{10} b^{5}+385 a \,x^{8} b^{4}+990 a^{2} x^{6} b^{3}+1386 a^{3} x^{4} b^{2}+1155 x^{2} a^{4} b +693 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{693 \left (b \,x^{2}+a \right )^{5}}\) | \(78\) |
orering | \(\frac {x \left (63 x^{10} b^{5}+385 a \,x^{8} b^{4}+990 a^{2} x^{6} b^{3}+1386 a^{3} x^{4} b^{2}+1155 x^{2} a^{4} b +693 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{693 \left (b \,x^{2}+a \right )^{5}}\) | \(87\) |
risch | \(\frac {b^{5} x^{11} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{11 b \,x^{2}+11 a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \,b^{4} x^{9}}{9 \left (b \,x^{2}+a \right )}+\frac {10 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{2} b^{3} x^{7}}{7 \left (b \,x^{2}+a \right )}+\frac {2 a^{3} b^{2} x^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a^{4} b \,x^{3}}{3 \left (b \,x^{2}+a \right )}+\frac {a^{5} x \sqrt {\left (b \,x^{2}+a \right )^{2}}}{b \,x^{2}+a}\) | \(175\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/693*x*(63*b^5*x^10+385*a*b^4*x^8+990*a^2*b^3*x^6+1386*a^3*b^2*x^4+1155*a ^4*b*x^2+693*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.22 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{11} \, b^{5} x^{11} + \frac {5}{9} \, a b^{4} x^{9} + \frac {10}{7} \, a^{2} b^{3} x^{7} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} x \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/11*b^5*x^11 + 5/9*a*b^4*x^9 + 10/7*a^2*b^3*x^7 + 2*a^3*b^2*x^5 + 5/3*a^4 *b*x^3 + a^5*x
\[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int \left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {5}{2}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(5/2), x)
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.22 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{11} \, b^{5} x^{11} + \frac {5}{9} \, a b^{4} x^{9} + \frac {10}{7} \, a^{2} b^{3} x^{7} + 2 \, a^{3} b^{2} x^{5} + \frac {5}{3} \, a^{4} b x^{3} + a^{5} x \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/11*b^5*x^11 + 5/9*a*b^4*x^9 + 10/7*a^2*b^3*x^7 + 2*a^3*b^2*x^5 + 5/3*a^4 *b*x^3 + a^5*x
Time = 0.13 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.41 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {1}{11} \, b^{5} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{9} \, a b^{4} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{7} \, a^{2} b^{3} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{3} b^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{3} \, a^{4} b x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} x \mathrm {sgn}\left (b x^{2} + a\right ) \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/11*b^5*x^11*sgn(b*x^2 + a) + 5/9*a*b^4*x^9*sgn(b*x^2 + a) + 10/7*a^2*b^3 *x^7*sgn(b*x^2 + a) + 2*a^3*b^2*x^5*sgn(b*x^2 + a) + 5/3*a^4*b*x^3*sgn(b*x ^2 + a) + a^5*x*sgn(b*x^2 + a)
Timed out. \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int {\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.23 \[ \int \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {x \left (63 b^{5} x^{10}+385 a \,b^{4} x^{8}+990 a^{2} b^{3} x^{6}+1386 a^{3} b^{2} x^{4}+1155 a^{4} b \,x^{2}+693 a^{5}\right )}{693} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(x*(693*a**5 + 1155*a**4*b*x**2 + 1386*a**3*b**2*x**4 + 990*a**2*b**3*x**6 + 385*a*b**4*x**8 + 63*b**5*x**10))/693