Integrand size = 26, antiderivative size = 249 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 x^5 \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 x^3 \left (a+b x^2\right )}-\frac {10 a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {10 a^2 b^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {b^5 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 \left (a+b x^2\right )} \] Output:
-1/5*a^5*((b*x^2+a)^2)^(1/2)/x^5/(b*x^2+a)-5/3*a^4*b*((b*x^2+a)^2)^(1/2)/x ^3/(b*x^2+a)-10*a^3*b^2*((b*x^2+a)^2)^(1/2)/x/(b*x^2+a)+10*a^2*b^3*x*((b*x ^2+a)^2)^(1/2)/(b*x^2+a)+5*a*b^4*x^3*((b*x^2+a)^2)^(1/2)/(3*b*x^2+3*a)+b^5 *x^5*((b*x^2+a)^2)^(1/2)/(5*b*x^2+5*a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (-3 a^5-25 a^4 b x^2-150 a^3 b^2 x^4+150 a^2 b^3 x^6+25 a b^4 x^8+3 b^5 x^{10}\right )}{15 x^5 \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^6,x]
Output:
(Sqrt[(a + b*x^2)^2]*(-3*a^5 - 25*a^4*b*x^2 - 150*a^3*b^2*x^4 + 150*a^2*b^ 3*x^6 + 25*a*b^4*x^8 + 3*b^5*x^10))/(15*x^5*(a + b*x^2))
Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.38, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^6}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^6}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^6}+\frac {5 b a^4}{x^4}+\frac {10 b^2 a^3}{x^2}+10 b^3 a^2+5 b^4 x^2 a+b^5 x^4\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{5 x^5}-\frac {5 a^4 b}{3 x^3}-\frac {10 a^3 b^2}{x}+10 a^2 b^3 x+\frac {5}{3} a b^4 x^3+\frac {b^5 x^5}{5}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^6,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-1/5*a^5/x^5 - (5*a^4*b)/(3*x^3) - (10*a ^3*b^2)/x + 10*a^2*b^3*x + (5*a*b^4*x^3)/3 + (b^5*x^5)/5))/(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 = 2.15 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {\left (-3 x^{10} b^{5}-25 a \,x^{8} b^{4}-150 a^{2} x^{6} b^{3}+150 a^{3} x^{4} b^{2}+25 x^{2} a^{4} b +3 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{15 x^{5} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (-3 x^{10} b^{5}-25 a \,x^{8} b^{4}-150 a^{2} x^{6} b^{3}+150 a^{3} x^{4} b^{2}+25 x^{2} a^{4} b +3 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{15 x^{5} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (-3 x^{10} b^{5}-25 a \,x^{8} b^{4}-150 a^{2} x^{6} b^{3}+150 a^{3} x^{4} b^{2}+25 x^{2} a^{4} b +3 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{15 x^{5} \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b^{3} \left (\frac {1}{5} x^{5} b^{2}+\frac {5}{3} a b \,x^{3}+10 a^{2} x \right )}{b \,x^{2}+a}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-10 a^{3} x^{4} b^{2}-\frac {5}{3} x^{2} a^{4} b -\frac {1}{5} a^{5}\right )}{\left (b \,x^{2}+a \right ) x^{5}}\) | \(98\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x,method=_RETURNVERBOSE)
Output:
-1/15*(-3*b^5*x^10-25*a*b^4*x^8-150*a^2*b^3*x^6+150*a^3*b^2*x^4+25*a^4*b*x ^2+3*a^5)*((b*x^2+a)^2)^(5/2)/x^5/(b*x^2+a)^5
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\frac {3 \, b^{5} x^{10} + 25 \, a b^{4} x^{8} + 150 \, a^{2} b^{3} x^{6} - 150 \, a^{3} b^{2} x^{4} - 25 \, a^{4} b x^{2} - 3 \, a^{5}}{15 \, x^{5}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="fricas")
Output:
1/15*(3*b^5*x^10 + 25*a*b^4*x^8 + 150*a^2*b^3*x^6 - 150*a^3*b^2*x^4 - 25*a ^4*b*x^2 - 3*a^5)/x^5
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{6}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**6,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**6, x)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\frac {1}{5} \, b^{5} x^{5} + \frac {5}{3} \, a b^{4} x^{3} + 10 \, a^{2} b^{3} x - \frac {10 \, a^{3} b^{2}}{x} - \frac {5 \, a^{4} b}{3 \, x^{3}} - \frac {a^{5}}{5 \, x^{5}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="maxima")
Output:
1/5*b^5*x^5 + 5/3*a*b^4*x^3 + 10*a^2*b^3*x - 10*a^3*b^2/x - 5/3*a^4*b/x^3 - 1/5*a^5/x^5
Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\frac {1}{5} \, b^{5} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{3} \, a b^{4} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 10 \, a^{2} b^{3} x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {150 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 25 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 3 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{15 \, x^{5}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x, algorithm="giac")
Output:
1/5*b^5*x^5*sgn(b*x^2 + a) + 5/3*a*b^4*x^3*sgn(b*x^2 + a) + 10*a^2*b^3*x*s gn(b*x^2 + a) - 1/15*(150*a^3*b^2*x^4*sgn(b*x^2 + a) + 25*a^4*b*x^2*sgn(b* x^2 + a) + 3*a^5*sgn(b*x^2 + a))/x^5
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^6} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^6,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^6, x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^6} \, dx=\frac {3 b^{5} x^{10}+25 a \,b^{4} x^{8}+150 a^{2} b^{3} x^{6}-150 a^{3} b^{2} x^{4}-25 a^{4} b \,x^{2}-3 a^{5}}{15 x^{5}} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^6,x)
Output:
( - 3*a**5 - 25*a**4*b*x**2 - 150*a**3*b**2*x**4 + 150*a**2*b**3*x**6 + 25 *a*b**4*x**8 + 3*b**5*x**10)/(15*x**5)