Integrand size = 26, antiderivative size = 247 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{x \left (a+b x^2\right )}+\frac {5 a^4 b x \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {10 a^3 b^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 \left (a+b x^2\right )}+\frac {2 a^2 b^3 x^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{a+b x^2}+\frac {5 a b^4 x^7 \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 \left (a+b x^2\right )}+\frac {b^5 x^9 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 \left (a+b x^2\right )} \] Output:
-a^5*((b*x^2+a)^2)^(1/2)/x/(b*x^2+a)+5*a^4*b*x*((b*x^2+a)^2)^(1/2)/(b*x^2+ a)+10*a^3*b^2*x^3*((b*x^2+a)^2)^(1/2)/(3*b*x^2+3*a)+2*a^2*b^3*x^5*((b*x^2+ a)^2)^(1/2)/(b*x^2+a)+5*a*b^4*x^7*((b*x^2+a)^2)^(1/2)/(7*b*x^2+7*a)+b^5*x^ 9*((b*x^2+a)^2)^(1/2)/(9*b*x^2+9*a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.34 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (-63 a^5+315 a^4 b x^2+210 a^3 b^2 x^4+126 a^2 b^3 x^6+45 a b^4 x^8+7 b^5 x^{10}\right )}{63 x \left (a+b x^2\right )} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^2,x]
Output:
(Sqrt[(a + b*x^2)^2]*(-63*a^5 + 315*a^4*b*x^2 + 210*a^3*b^2*x^4 + 126*a^2* b^3*x^6 + 45*a*b^4*x^8 + 7*b^5*x^10))/(63*x*(a + b*x^2))
Time = 0.40 (sec) , antiderivative size = 93, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^2}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^2}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b^5 x^8+5 a b^4 x^6+10 a^2 b^3 x^4+10 a^3 b^2 x^2+5 a^4 b+\frac {a^5}{x^2}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {a^5}{x}+5 a^4 b x+\frac {10}{3} a^3 b^2 x^3+2 a^2 b^3 x^5+\frac {5}{7} a b^4 x^7+\frac {b^5 x^9}{9}\right )}{a+b x^2}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)/x^2,x]
Output:
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*(-(a^5/x) + 5*a^4*b*x + (10*a^3*b^2*x^3)/ 3 + 2*a^2*b^3*x^5 + (5*a*b^4*x^7)/7 + (b^5*x^9)/9))/(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 = 1.00 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.32
method | result | size |
gosper | \(-\frac {\left (-7 x^{10} b^{5}-45 a \,x^{8} b^{4}-126 a^{2} x^{6} b^{3}-210 a^{3} x^{4} b^{2}-315 x^{2} a^{4} b +63 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{63 x \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (-7 x^{10} b^{5}-45 a \,x^{8} b^{4}-126 a^{2} x^{6} b^{3}-210 a^{3} x^{4} b^{2}-315 x^{2} a^{4} b +63 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{63 x \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
orering | \(-\frac {\left (-7 x^{10} b^{5}-45 a \,x^{8} b^{4}-126 a^{2} x^{6} b^{3}-210 a^{3} x^{4} b^{2}-315 x^{2} a^{4} b +63 a^{5}\right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{\frac {5}{2}}}{63 x \left (b \,x^{2}+a \right )^{5}}\) | \(89\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \left (\frac {1}{9} b^{4} x^{9}+\frac {5}{7} a \,b^{3} x^{7}+2 a^{2} b^{2} x^{5}+\frac {10}{3} a^{3} b \,x^{3}+5 a^{4} x \right )}{b \,x^{2}+a}-\frac {a^{5} \sqrt {\left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}\) | \(96\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^2,x,method=_RETURNVERBOSE)
Output:
-1/63*(-7*b^5*x^10-45*a*b^4*x^8-126*a^2*b^3*x^6-210*a^3*b^2*x^4-315*a^4*b* x^2+63*a^5)*((b*x^2+a)^2)^(5/2)/x/(b*x^2+a)^5
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {7 \, b^{5} x^{10} + 45 \, a b^{4} x^{8} + 126 \, a^{2} b^{3} x^{6} + 210 \, a^{3} b^{2} x^{4} + 315 \, a^{4} b x^{2} - 63 \, a^{5}}{63 \, x} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^2,x, algorithm="fricas")
Output:
1/63*(7*b^5*x^10 + 45*a*b^4*x^8 + 126*a^2*b^3*x^6 + 210*a^3*b^2*x^4 + 315* a^4*b*x^2 - 63*a^5)/x
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{2}}\, dx \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**(5/2)/x**2,x)
Output:
Integral(((a + b*x**2)**2)**(5/2)/x**2, x)
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {1}{9} \, b^{5} x^{9} + \frac {5}{7} \, a b^{4} x^{7} + 2 \, a^{2} b^{3} x^{5} + \frac {10}{3} \, a^{3} b^{2} x^{3} + 5 \, a^{4} b x - \frac {a^{5}}{x} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^2,x, algorithm="maxima")
Output:
1/9*b^5*x^9 + 5/7*a*b^4*x^7 + 2*a^2*b^3*x^5 + 10/3*a^3*b^2*x^3 + 5*a^4*b*x - a^5/x
Time = 0.12 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {1}{9} \, b^{5} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {5}{7} \, a b^{4} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 2 \, a^{2} b^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{3} \, a^{3} b^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a^{4} b x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{x} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^2,x, algorithm="giac")
Output:
1/9*b^5*x^9*sgn(b*x^2 + a) + 5/7*a*b^4*x^7*sgn(b*x^2 + a) + 2*a^2*b^3*x^5* sgn(b*x^2 + a) + 10/3*a^3*b^2*x^3*sgn(b*x^2 + a) + 5*a^4*b*x*sgn(b*x^2 + a ) - a^5*sgn(b*x^2 + a)/x
Timed out. \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\int \frac {{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}}{x^2} \,d x \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^2,x)
Output:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)/x^2, x)
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.24 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^2} \, dx=\frac {7 b^{5} x^{10}+45 a \,b^{4} x^{8}+126 a^{2} b^{3} x^{6}+210 a^{3} b^{2} x^{4}+315 a^{4} b \,x^{2}-63 a^{5}}{63 x} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^(5/2)/x^2,x)
Output:
( - 63*a**5 + 315*a**4*b*x**2 + 210*a**3*b**2*x**4 + 126*a**2*b**3*x**6 + 45*a*b**4*x**8 + 7*b**5*x**10)/(63*x)