Integrand size = 30, antiderivative size = 295 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{5 d (d x)^{5/2} \left (a+b x^2\right )}-\frac {10 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{d^3 \sqrt {d x} \left (a+b x^2\right )}+\frac {20 a^3 b^2 (d x)^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac {20 a^2 b^3 (d x)^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{7 d^7 \left (a+b x^2\right )}+\frac {10 a b^4 (d x)^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{11 d^9 \left (a+b x^2\right )}+\frac {2 b^5 (d x)^{15/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{15 d^{11} \left (a+b x^2\right )} \]
-2/5*a^5*((b*x^2+a)^2)^(1/2)/d/(d*x)^(5/2)/(b*x^2+a)+20/3*a^3*b^2*(d*x)^(3 /2)*((b*x^2+a)^2)^(1/2)/d^5/(b*x^2+a)+20/7*a^2*b^3*(d*x)^(7/2)*((b*x^2+a)^ 2)^(1/2)/d^7/(b*x^2+a)+10/11*a*b^4*(d*x)^(11/2)*((b*x^2+a)^2)^(1/2)/d^9/(b *x^2+a)+2/15*b^5*(d*x)^(15/2)*((b*x^2+a)^2)^(1/2)/d^11/(b*x^2+a)-10*a^4*b* ((b*x^2+a)^2)^(1/2)/d^3/(b*x^2+a)/(d*x)^(1/2)
Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.30 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 x \left (\left (a+b x^2\right )^2\right )^{5/2} \left (231 a^5+5775 a^4 b x^2-3850 a^3 b^2 x^4-1650 a^2 b^3 x^6-525 a b^4 x^8-77 b^5 x^{10}\right )}{1155 (d x)^{7/2} \left (a+b x^2\right )^5} \]
(-2*x*((a + b*x^2)^2)^(5/2)*(231*a^5 + 5775*a^4*b*x^2 - 3850*a^3*b^2*x^4 - 1650*a^2*b^3*x^6 - 525*a*b^4*x^8 - 77*b^5*x^10))/(1155*(d*x)^(7/2)*(a + b *x^2)^5)
Time = 0.27 (sec) , antiderivative size = 141, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{(d x)^{7/2}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{(d x)^{7/2}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {b^5 (d x)^{13/2}}{d^{10}}+\frac {5 a b^4 (d x)^{9/2}}{d^8}+\frac {10 a^2 b^3 (d x)^{5/2}}{d^6}+\frac {10 a^3 b^2 \sqrt {d x}}{d^4}+\frac {5 a^4 b}{d^2 (d x)^{3/2}}+\frac {a^5}{(d x)^{7/2}}\right )dx}{a+b x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (-\frac {2 a^5}{5 d (d x)^{5/2}}-\frac {10 a^4 b}{d^3 \sqrt {d x}}+\frac {20 a^3 b^2 (d x)^{3/2}}{3 d^5}+\frac {20 a^2 b^3 (d x)^{7/2}}{7 d^7}+\frac {10 a b^4 (d x)^{11/2}}{11 d^9}+\frac {2 b^5 (d x)^{15/2}}{15 d^{11}}\right )}{a+b x^2}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((-2*a^5)/(5*d*(d*x)^(5/2)) - (10*a^4*b)/ (d^3*Sqrt[d*x]) + (20*a^3*b^2*(d*x)^(3/2))/(3*d^5) + (20*a^2*b^3*(d*x)^(7/ 2))/(7*d^7) + (10*a*b^4*(d*x)^(11/2))/(11*d^9) + (2*b^5*(d*x)^(15/2))/(15* d^11)))/(a + b*x^2)
3.8.48.3.1 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 = 0.04 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.28
method | result | size |
gosper | \(-\frac {2 x \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{1155 \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {7}{2}}}\) | \(83\) |
default | \(-\frac {2 {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}} \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right )}{1155 d \left (b \,x^{2}+a \right )^{5} \left (d x \right )^{\frac {5}{2}}}\) | \(85\) |
risch | \(-\frac {2 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-77 x^{10} b^{5}-525 a \,x^{8} b^{4}-1650 a^{2} x^{6} b^{3}-3850 a^{3} x^{4} b^{2}+5775 x^{2} a^{4} b +231 a^{5}\right )}{1155 d^{3} \left (b \,x^{2}+a \right ) x^{2} \sqrt {d x}}\) | \(88\) |
-2/1155*x*(-77*b^5*x^10-525*a*b^4*x^8-1650*a^2*b^3*x^6-3850*a^3*b^2*x^4+57 75*a^4*b*x^2+231*a^5)*((b*x^2+a)^2)^(5/2)/(b*x^2+a)^5/(d*x)^(7/2)
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (77 \, b^{5} x^{10} + 525 \, a b^{4} x^{8} + 1650 \, a^{2} b^{3} x^{6} + 3850 \, a^{3} b^{2} x^{4} - 5775 \, a^{4} b x^{2} - 231 \, a^{5}\right )} \sqrt {d x}}{1155 \, d^{4} x^{3}} \]
2/1155*(77*b^5*x^10 + 525*a*b^4*x^8 + 1650*a^2*b^3*x^6 + 3850*a^3*b^2*x^4 - 5775*a^4*b*x^2 - 231*a^5)*sqrt(d*x)/(d^4*x^3)
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{\left (d x\right )^{\frac {7}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {2 \, {\left (7 \, {\left (11 \, b^{5} \sqrt {d} x^{3} + 15 \, a b^{4} \sqrt {d} x\right )} x^{\frac {9}{2}} + 60 \, {\left (7 \, a b^{4} \sqrt {d} x^{3} + 11 \, a^{2} b^{3} \sqrt {d} x\right )} x^{\frac {5}{2}} + 330 \, {\left (3 \, a^{2} b^{3} \sqrt {d} x^{3} + 7 \, a^{3} b^{2} \sqrt {d} x\right )} \sqrt {x} + \frac {1540 \, {\left (a^{3} b^{2} \sqrt {d} x^{3} - 3 \, a^{4} b \sqrt {d} x\right )}}{x^{\frac {3}{2}}} - \frac {231 \, {\left (5 \, a^{4} b \sqrt {d} x^{3} + a^{5} \sqrt {d} x\right )}}{x^{\frac {7}{2}}}\right )}}{1155 \, d^{4}} \]
2/1155*(7*(11*b^5*sqrt(d)*x^3 + 15*a*b^4*sqrt(d)*x)*x^(9/2) + 60*(7*a*b^4* sqrt(d)*x^3 + 11*a^2*b^3*sqrt(d)*x)*x^(5/2) + 330*(3*a^2*b^3*sqrt(d)*x^3 + 7*a^3*b^2*sqrt(d)*x)*sqrt(x) + 1540*(a^3*b^2*sqrt(d)*x^3 - 3*a^4*b*sqrt(d )*x)/x^(3/2) - 231*(5*a^4*b*sqrt(d)*x^3 + a^5*sqrt(d)*x)/x^(7/2))/d^4
Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.55 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=-\frac {2 \, {\left (\frac {231 \, {\left (25 \, a^{4} b d^{3} x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + a^{5} d^{3} \mathrm {sgn}\left (b x^{2} + a\right )\right )}}{\sqrt {d x} d^{2} x^{2}} - \frac {77 \, \sqrt {d x} b^{5} d^{105} x^{7} \mathrm {sgn}\left (b x^{2} + a\right ) + 525 \, \sqrt {d x} a b^{4} d^{105} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 1650 \, \sqrt {d x} a^{2} b^{3} d^{105} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 3850 \, \sqrt {d x} a^{3} b^{2} d^{105} x \mathrm {sgn}\left (b x^{2} + a\right )}{d^{105}}\right )}}{1155 \, d^{4}} \]
-2/1155*(231*(25*a^4*b*d^3*x^2*sgn(b*x^2 + a) + a^5*d^3*sgn(b*x^2 + a))/(s qrt(d*x)*d^2*x^2) - (77*sqrt(d*x)*b^5*d^105*x^7*sgn(b*x^2 + a) + 525*sqrt( d*x)*a*b^4*d^105*x^5*sgn(b*x^2 + a) + 1650*sqrt(d*x)*a^2*b^3*d^105*x^3*sgn (b*x^2 + a) + 3850*sqrt(d*x)*a^3*b^2*d^105*x*sgn(b*x^2 + a))/d^105)/d^4
Time = 13.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.40 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{(d x)^{7/2}} \, dx=\frac {\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}\,\left (\frac {2\,b^4\,x^{10}}{15\,d^3}-\frac {10\,a^4\,x^2}{d^3}-\frac {2\,a^5}{5\,b\,d^3}+\frac {20\,a^3\,b\,x^4}{3\,d^3}+\frac {10\,a\,b^3\,x^8}{11\,d^3}+\frac {20\,a^2\,b^2\,x^6}{7\,d^3}\right )}{x^4\,\sqrt {d\,x}+\frac {a\,x^2\,\sqrt {d\,x}}{b}} \]