Integrand size = 30, antiderivative size = 297 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 a^5 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d \left (a+b x^2\right )}+\frac {10 a^4 b (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^3 \left (a+b x^2\right )}+\frac {4 a^3 b^2 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{19/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{19 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{23/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{23 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{27/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{27 d^{11} \left (a+b x^2\right )} \]
2/7*a^5*(d*x)^(7/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)+10/11*a^4*b*(d*x)^(11/ 2)*((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)+4/3*a^3*b^2*(d*x)^(15/2)*((b*x^2+a)^2 )^(1/2)/d^5/(b*x^2+a)+20/19*a^2*b^3*(d*x)^(19/2)*((b*x^2+a)^2)^(1/2)/d^7/( b*x^2+a)+10/23*a*b^4*(d*x)^(23/2)*((b*x^2+a)^2)^(1/2)/d^9/(b*x^2+a)+2/27*b ^5*(d*x)^(27/2)*((b*x^2+a)^2)^(1/2)/d^11/(b*x^2+a)
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2 x (d x)^{5/2} \sqrt {\left (a+b x^2\right )^2} \left (129789 a^5+412965 a^4 b x^2+605682 a^3 b^2 x^4+478170 a^2 b^3 x^6+197505 a b^4 x^8+33649 b^5 x^{10}\right )}{908523 \left (a+b x^2\right )} \]
(2*x*(d*x)^(5/2)*Sqrt[(a + b*x^2)^2]*(129789*a^5 + 412965*a^4*b*x^2 + 6056 82*a^3*b^2*x^4 + 478170*a^2*b^3*x^6 + 197505*a*b^4*x^8 + 33649*b^5*x^10))/ (908523*(a + b*x^2))
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.48, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, 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.
\(\displaystyle \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b^5 (d x)^{5/2} \left (b x^2+a\right )^5dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int (d x)^{5/2} \left (b x^2+a\right )^5dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^5 (d x)^{25/2}}{d^{10}}+\frac {5 a b^4 (d x)^{21/2}}{d^8}+\frac {10 a^2 b^3 (d x)^{17/2}}{d^6}+\frac {10 a^3 b^2 (d x)^{13/2}}{d^4}+\frac {5 a^4 b (d x)^{9/2}}{d^2}+a^5 (d x)^{5/2}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {2 a^5 (d x)^{7/2}}{7 d}+\frac {10 a^4 b (d x)^{11/2}}{11 d^3}+\frac {4 a^3 b^2 (d x)^{15/2}}{3 d^5}+\frac {20 a^2 b^3 (d x)^{19/2}}{19 d^7}+\frac {10 a b^4 (d x)^{23/2}}{23 d^9}+\frac {2 b^5 (d x)^{27/2}}{27 d^{11}}\right )}{a+b x^2}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((2*a^5*(d*x)^(7/2))/(7*d) + (10*a^4*b*(d *x)^(11/2))/(11*d^3) + (4*a^3*b^2*(d*x)^(15/2))/(3*d^5) + (20*a^2*b^3*(d*x )^(19/2))/(19*d^7) + (10*a*b^4*(d*x)^(23/2))/(23*d^9) + (2*b^5*(d*x)^(27/2 ))/(27*d^11)))/(a + b*x^2)
3.8.42.3.1 Defintions of rubi rules used
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 = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28
method | result | size |
gosper | \(\frac {2 x \left (33649 x^{10} b^{5}+197505 a \,x^{8} b^{4}+478170 a^{2} x^{6} b^{3}+605682 a^{3} x^{4} b^{2}+412965 x^{2} a^{4} b +129789 a^{5}\right ) \left (d x \right )^{\frac {5}{2}} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{908523 \left (b \,x^{2}+a \right )^{5}}\) | \(83\) |
default | \(\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (d x \right )^{\frac {7}{2}} \left (33649 x^{10} b^{5}+197505 a \,x^{8} b^{4}+478170 a^{2} x^{6} b^{3}+605682 a^{3} x^{4} b^{2}+412965 x^{2} a^{4} b +129789 a^{5}\right )}{908523 d \left (b \,x^{2}+a \right )^{5}}\) | \(85\) |
risch | \(\frac {2 d^{3} \sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{4} \left (33649 x^{10} b^{5}+197505 a \,x^{8} b^{4}+478170 a^{2} x^{6} b^{3}+605682 a^{3} x^{4} b^{2}+412965 x^{2} a^{4} b +129789 a^{5}\right )}{908523 \left (b \,x^{2}+a \right ) \sqrt {d x}}\) | \(88\) |
2/908523*x*(33649*b^5*x^10+197505*a*b^4*x^8+478170*a^2*b^3*x^6+605682*a^3* b^2*x^4+412965*a^4*b*x^2+129789*a^5)*(d*x)^(5/2)*((b*x^2+a)^2)^(5/2)/(b*x^ 2+a)^5
Time = 0.24 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.28 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{908523} \, {\left (33649 \, b^{5} d^{2} x^{13} + 197505 \, a b^{4} d^{2} x^{11} + 478170 \, a^{2} b^{3} d^{2} x^{9} + 605682 \, a^{3} b^{2} d^{2} x^{7} + 412965 \, a^{4} b d^{2} x^{5} + 129789 \, a^{5} d^{2} x^{3}\right )} \sqrt {d x} \]
2/908523*(33649*b^5*d^2*x^13 + 197505*a*b^4*d^2*x^11 + 478170*a^2*b^3*d^2* x^9 + 605682*a^3*b^2*d^2*x^7 + 412965*a^4*b*d^2*x^5 + 129789*a^5*d^2*x^3)* sqrt(d*x)
\[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int \left (d x\right )^{\frac {5}{2}} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.49 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{621} \, {\left (23 \, b^{5} d^{\frac {5}{2}} x^{3} + 27 \, a b^{4} d^{\frac {5}{2}} x\right )} x^{\frac {21}{2}} + \frac {8}{437} \, {\left (19 \, a b^{4} d^{\frac {5}{2}} x^{3} + 23 \, a^{2} b^{3} d^{\frac {5}{2}} x\right )} x^{\frac {17}{2}} + \frac {4}{95} \, {\left (15 \, a^{2} b^{3} d^{\frac {5}{2}} x^{3} + 19 \, a^{3} b^{2} d^{\frac {5}{2}} x\right )} x^{\frac {13}{2}} + \frac {8}{165} \, {\left (11 \, a^{3} b^{2} d^{\frac {5}{2}} x^{3} + 15 \, a^{4} b d^{\frac {5}{2}} x\right )} x^{\frac {9}{2}} + \frac {2}{77} \, {\left (7 \, a^{4} b d^{\frac {5}{2}} x^{3} + 11 \, a^{5} d^{\frac {5}{2}} x\right )} x^{\frac {5}{2}} \]
2/621*(23*b^5*d^(5/2)*x^3 + 27*a*b^4*d^(5/2)*x)*x^(21/2) + 8/437*(19*a*b^4 *d^(5/2)*x^3 + 23*a^2*b^3*d^(5/2)*x)*x^(17/2) + 4/95*(15*a^2*b^3*d^(5/2)*x ^3 + 19*a^3*b^2*d^(5/2)*x)*x^(13/2) + 8/165*(11*a^3*b^2*d^(5/2)*x^3 + 15*a ^4*b*d^(5/2)*x)*x^(9/2) + 2/77*(7*a^4*b*d^(5/2)*x^3 + 11*a^5*d^(5/2)*x)*x^ (5/2)
Time = 0.28 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.52 \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\frac {2}{27} \, \sqrt {d x} b^{5} d^{2} x^{13} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{23} \, \sqrt {d x} a b^{4} d^{2} x^{11} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {20}{19} \, \sqrt {d x} a^{2} b^{3} d^{2} x^{9} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {4}{3} \, \sqrt {d x} a^{3} b^{2} d^{2} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {10}{11} \, \sqrt {d x} a^{4} b d^{2} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {2}{7} \, \sqrt {d x} a^{5} d^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) \]
2/27*sqrt(d*x)*b^5*d^2*x^13*sgn(b*x^2 + a) + 10/23*sqrt(d*x)*a*b^4*d^2*x^1 1*sgn(b*x^2 + a) + 20/19*sqrt(d*x)*a^2*b^3*d^2*x^9*sgn(b*x^2 + a) + 4/3*sq rt(d*x)*a^3*b^2*d^2*x^7*sgn(b*x^2 + a) + 10/11*sqrt(d*x)*a^4*b*d^2*x^5*sgn (b*x^2 + a) + 2/7*sqrt(d*x)*a^5*d^2*x^3*sgn(b*x^2 + a)
Timed out. \[ \int (d x)^{5/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2} \, dx=\int {\left (d\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2} \,d x \]