Integrand size = 28, antiderivative size = 89 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=-\frac {2 a^4}{3 d (d x)^{3/2}}+\frac {8 a^3 b \sqrt {d x}}{d^3}+\frac {12 a^2 b^2 (d x)^{5/2}}{5 d^5}+\frac {8 a b^3 (d x)^{9/2}}{9 d^7}+\frac {2 b^4 (d x)^{13/2}}{13 d^9} \] Output:
-2/3*a^4/d/(d*x)^(3/2)+8*a^3*b*(d*x)^(1/2)/d^3+12/5*a^2*b^2*(d*x)^(5/2)/d^ 5+8/9*a*b^3*(d*x)^(9/2)/d^7+2/13*b^4*(d*x)^(13/2)/d^9
Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.62 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=-\frac {2 x \left (195 a^4-2340 a^3 b x^2-702 a^2 b^2 x^4-260 a b^3 x^6-45 b^4 x^8\right )}{585 (d x)^{5/2}} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/(d*x)^(5/2),x]
Output:
(-2*x*(195*a^4 - 2340*a^3*b*x^2 - 702*a^2*b^2*x^4 - 260*a*b^3*x^6 - 45*b^4 *x^8))/(585*(d*x)^(5/2))
Time = 0.37 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1380, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \frac {\int \frac {b^4 \left (b x^2+a\right )^4}{(d x)^{5/2}}dx}{b^4}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {\left (a+b x^2\right )^4}{(d x)^{5/2}}dx\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \int \left (\frac {a^4}{(d x)^{5/2}}+\frac {4 a^3 b}{d^2 \sqrt {d x}}+\frac {6 a^2 b^2 (d x)^{3/2}}{d^4}+\frac {4 a b^3 (d x)^{7/2}}{d^6}+\frac {b^4 (d x)^{11/2}}{d^8}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 a^4}{3 d (d x)^{3/2}}+\frac {8 a^3 b \sqrt {d x}}{d^3}+\frac {12 a^2 b^2 (d x)^{5/2}}{5 d^5}+\frac {8 a b^3 (d x)^{9/2}}{9 d^7}+\frac {2 b^4 (d x)^{13/2}}{13 d^9}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^2/(d*x)^(5/2),x]
Output:
(-2*a^4)/(3*d*(d*x)^(3/2)) + (8*a^3*b*Sqrt[d*x])/d^3 + (12*a^2*b^2*(d*x)^( 5/2))/(5*d^5) + (8*a*b^3*(d*x)^(9/2))/(9*d^7) + (2*b^4*(d*x)^(13/2))/(13*d ^9)
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[1/c^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]
Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.58
method | result | size |
gosper | \(-\frac {2 \left (-45 b^{4} x^{8}-260 a \,b^{3} x^{6}-702 a^{2} b^{2} x^{4}-2340 a^{3} b \,x^{2}+195 a^{4}\right ) x}{585 \left (d x \right )^{\frac {5}{2}}}\) | \(52\) |
pseudoelliptic | \(-\frac {2 \left (-\frac {3}{13} b^{4} x^{8}-\frac {4}{3} a \,b^{3} x^{6}-\frac {18}{5} a^{2} b^{2} x^{4}-12 a^{3} b \,x^{2}+a^{4}\right )}{3 \sqrt {d x}\, d^{2} x}\) | \(55\) |
trager | \(-\frac {2 \left (-45 b^{4} x^{8}-260 a \,b^{3} x^{6}-702 a^{2} b^{2} x^{4}-2340 a^{3} b \,x^{2}+195 a^{4}\right ) \sqrt {d x}}{585 d^{3} x^{2}}\) | \(57\) |
risch | \(-\frac {2 \left (-45 b^{4} x^{8}-260 a \,b^{3} x^{6}-702 a^{2} b^{2} x^{4}-2340 a^{3} b \,x^{2}+195 a^{4}\right )}{585 d^{2} x \sqrt {d x}}\) | \(57\) |
derivativedivides | \(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {13}{2}}}{13}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {9}{2}}}{9}+\frac {12 a^{2} b^{2} d^{4} \left (d x \right )^{\frac {5}{2}}}{5}+8 \sqrt {d x}\, a^{3} b \,d^{6}-\frac {2 a^{4} d^{8}}{3 \left (d x \right )^{\frac {3}{2}}}}{d^{9}}\) | \(74\) |
default | \(\frac {\frac {2 b^{4} \left (d x \right )^{\frac {13}{2}}}{13}+\frac {8 a \,b^{3} d^{2} \left (d x \right )^{\frac {9}{2}}}{9}+\frac {12 a^{2} b^{2} d^{4} \left (d x \right )^{\frac {5}{2}}}{5}+8 \sqrt {d x}\, a^{3} b \,d^{6}-\frac {2 a^{4} d^{8}}{3 \left (d x \right )^{\frac {3}{2}}}}{d^{9}}\) | \(74\) |
orering | \(-\frac {2 \left (-45 b^{4} x^{8}-260 a \,b^{3} x^{6}-702 a^{2} b^{2} x^{4}-2340 a^{3} b \,x^{2}+195 a^{4}\right ) x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}{585 \left (b \,x^{2}+a \right )^{4} \left (d x \right )^{\frac {5}{2}}}\) | \(81\) |
Input:
int((b^2*x^4+2*a*b*x^2+a^2)^2/(d*x)^(5/2),x,method=_RETURNVERBOSE)
Output:
-2/585*(-45*b^4*x^8-260*a*b^3*x^6-702*a^2*b^2*x^4-2340*a^3*b*x^2+195*a^4)* x/(d*x)^(5/2)
Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=\frac {2 \, {\left (45 \, b^{4} x^{8} + 260 \, a b^{3} x^{6} + 702 \, a^{2} b^{2} x^{4} + 2340 \, a^{3} b x^{2} - 195 \, a^{4}\right )} \sqrt {d x}}{585 \, d^{3} x^{2}} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^2/(d*x)^(5/2),x, algorithm="fricas")
Output:
2/585*(45*b^4*x^8 + 260*a*b^3*x^6 + 702*a^2*b^2*x^4 + 2340*a^3*b*x^2 - 195 *a^4)*sqrt(d*x)/(d^3*x^2)
Time = 0.32 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.98 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=- \frac {2 a^{4} x}{3 \left (d x\right )^{\frac {5}{2}}} + \frac {8 a^{3} b x^{3}}{\left (d x\right )^{\frac {5}{2}}} + \frac {12 a^{2} b^{2} x^{5}}{5 \left (d x\right )^{\frac {5}{2}}} + \frac {8 a b^{3} x^{7}}{9 \left (d x\right )^{\frac {5}{2}}} + \frac {2 b^{4} x^{9}}{13 \left (d x\right )^{\frac {5}{2}}} \] Input:
integrate((b**2*x**4+2*a*b*x**2+a**2)**2/(d*x)**(5/2),x)
Output:
-2*a**4*x/(3*(d*x)**(5/2)) + 8*a**3*b*x**3/(d*x)**(5/2) + 12*a**2*b**2*x** 5/(5*(d*x)**(5/2)) + 8*a*b**3*x**7/(9*(d*x)**(5/2)) + 2*b**4*x**9/(13*(d*x )**(5/2))
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {195 \, a^{4}}{\left (d x\right )^{\frac {3}{2}}} - \frac {45 \, \left (d x\right )^{\frac {13}{2}} b^{4} + 260 \, \left (d x\right )^{\frac {9}{2}} a b^{3} d^{2} + 702 \, \left (d x\right )^{\frac {5}{2}} a^{2} b^{2} d^{4} + 2340 \, \sqrt {d x} a^{3} b d^{6}}{d^{8}}\right )}}{585 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^2/(d*x)^(5/2),x, algorithm="maxima")
Output:
-2/585*(195*a^4/(d*x)^(3/2) - (45*(d*x)^(13/2)*b^4 + 260*(d*x)^(9/2)*a*b^3 *d^2 + 702*(d*x)^(5/2)*a^2*b^2*d^4 + 2340*sqrt(d*x)*a^3*b*d^6)/d^8)/d
Time = 0.15 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=-\frac {2 \, {\left (\frac {195 \, a^{4}}{\sqrt {d x} d x} - \frac {45 \, \sqrt {d x} b^{4} d^{102} x^{6} + 260 \, \sqrt {d x} a b^{3} d^{102} x^{4} + 702 \, \sqrt {d x} a^{2} b^{2} d^{102} x^{2} + 2340 \, \sqrt {d x} a^{3} b d^{102}}{d^{104}}\right )}}{585 \, d} \] Input:
integrate((b^2*x^4+2*a*b*x^2+a^2)^2/(d*x)^(5/2),x, algorithm="giac")
Output:
-2/585*(195*a^4/(sqrt(d*x)*d*x) - (45*sqrt(d*x)*b^4*d^102*x^6 + 260*sqrt(d *x)*a*b^3*d^102*x^4 + 702*sqrt(d*x)*a^2*b^2*d^102*x^2 + 2340*sqrt(d*x)*a^3 *b*d^102)/d^104)/d
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=\frac {2\,b^4\,{\left (d\,x\right )}^{13/2}}{13\,d^9}-\frac {2\,a^4}{3\,d\,{\left (d\,x\right )}^{3/2}}+\frac {12\,a^2\,b^2\,{\left (d\,x\right )}^{5/2}}{5\,d^5}+\frac {8\,a^3\,b\,\sqrt {d\,x}}{d^3}+\frac {8\,a\,b^3\,{\left (d\,x\right )}^{9/2}}{9\,d^7} \] Input:
int((a^2 + b^2*x^4 + 2*a*b*x^2)^2/(d*x)^(5/2),x)
Output:
(2*b^4*(d*x)^(13/2))/(13*d^9) - (2*a^4)/(3*d*(d*x)^(3/2)) + (12*a^2*b^2*(d *x)^(5/2))/(5*d^5) + (8*a^3*b*(d*x)^(1/2))/d^3 + (8*a*b^3*(d*x)^(9/2))/(9* d^7)
Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.64 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^2}{(d x)^{5/2}} \, dx=\frac {2 \sqrt {d}\, \left (45 b^{4} x^{8}+260 a \,b^{3} x^{6}+702 a^{2} b^{2} x^{4}+2340 a^{3} b \,x^{2}-195 a^{4}\right )}{585 \sqrt {x}\, d^{3} x} \] Input:
int((b^2*x^4+2*a*b*x^2+a^2)^2/(d*x)^(5/2),x)
Output:
(2*sqrt(d)*( - 195*a**4 + 2340*a**3*b*x**2 + 702*a**2*b**2*x**4 + 260*a*b* *3*x**6 + 45*b**4*x**8))/(585*sqrt(x)*d**3*x)