Integrand size = 26, antiderivative size = 255 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{24 x^{24} \left (a+b x^2\right )}-\frac {5 a^4 b \sqrt {a^2+2 a b x^2+b^2 x^4}}{22 x^{22} \left (a+b x^2\right )}-\frac {a^3 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 x^{20} \left (a+b x^2\right )}-\frac {5 a^2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}{9 x^{18} \left (a+b x^2\right )}-\frac {5 a b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}{16 x^{16} \left (a+b x^2\right )}-\frac {b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}{14 x^{14} \left (a+b x^2\right )} \]
-1/24*a^5*((b*x^2+a)^2)^(1/2)/x^24/(b*x^2+a)-5/22*a^4*b*((b*x^2+a)^2)^(1/2 )/x^22/(b*x^2+a)-1/2*a^3*b^2*((b*x^2+a)^2)^(1/2)/x^20/(b*x^2+a)-5/9*a^2*b^ 3*((b*x^2+a)^2)^(1/2)/x^18/(b*x^2+a)-5/16*a*b^4*((b*x^2+a)^2)^(1/2)/x^16/( b*x^2+a)-1/14*b^5*((b*x^2+a)^2)^(1/2)/x^14/(b*x^2+a)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {\sqrt {\left (a+b x^2\right )^2} \left (462 a^5+2520 a^4 b x^2+5544 a^3 b^2 x^4+6160 a^2 b^3 x^6+3465 a b^4 x^8+792 b^5 x^{10}\right )}{11088 x^{24} \left (a+b x^2\right )} \]
-1/11088*(Sqrt[(a + b*x^2)^2]*(462*a^5 + 2520*a^4*b*x^2 + 5544*a^3*b^2*x^4 + 6160*a^2*b^3*x^6 + 3465*a*b^4*x^8 + 792*b^5*x^10))/(x^24*(a + b*x^2))
Time = 0.24 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.40, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1384, 27, 243, 53, 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.
\(\displaystyle \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {b^5 \left (b x^2+a\right )^5}{x^{25}}dx}{b^5 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{25}}dx}{a+b x^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \frac {\left (b x^2+a\right )^5}{x^{26}}dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (\frac {a^5}{x^{26}}+\frac {5 b a^4}{x^{24}}+\frac {10 b^2 a^3}{x^{22}}+\frac {10 b^3 a^2}{x^{20}}+\frac {5 b^4 a}{x^{18}}+\frac {b^5}{x^{16}}\right )dx^2}{2 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (-\frac {a^5}{12 x^{24}}-\frac {5 a^4 b}{11 x^{22}}-\frac {a^3 b^2}{x^{20}}-\frac {10 a^2 b^3}{9 x^{18}}-\frac {5 a b^4}{8 x^{16}}-\frac {b^5}{7 x^{14}}\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}{2 \left (a+b x^2\right )}\) |
((-1/12*a^5/x^24 - (5*a^4*b)/(11*x^22) - (a^3*b^2)/x^20 - (10*a^2*b^3)/(9* x^18) - (5*a*b^4)/(8*x^16) - b^5/(7*x^14))*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4] )/(2*(a + b*x^2))
3.7.4.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
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)])
Result contains higher order function than in optimal. Order 9 vs. order 2.
Time = 4.19 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.26
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\frac {12}{7} x^{10} b^{5}+\frac {15}{2} a \,x^{8} b^{4}+\frac {40}{3} a^{2} x^{6} b^{3}+12 a^{3} x^{4} b^{2}+\frac {60}{11} x^{2} a^{4} b +a^{5}\right )}{24 x^{24}}\) | \(66\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {1}{24} a^{5}-\frac {5}{22} x^{2} a^{4} b -\frac {1}{2} a^{3} x^{4} b^{2}-\frac {5}{9} a^{2} x^{6} b^{3}-\frac {5}{16} a \,x^{8} b^{4}-\frac {1}{14} x^{10} b^{5}\right )}{\left (b \,x^{2}+a \right ) x^{24}}\) | \(79\) |
gosper | \(-\frac {\left (792 x^{10} b^{5}+3465 a \,x^{8} b^{4}+6160 a^{2} x^{6} b^{3}+5544 a^{3} x^{4} b^{2}+2520 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{11088 x^{24} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
default | \(-\frac {\left (792 x^{10} b^{5}+3465 a \,x^{8} b^{4}+6160 a^{2} x^{6} b^{3}+5544 a^{3} x^{4} b^{2}+2520 x^{2} a^{4} b +462 a^{5}\right ) {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}{11088 x^{24} \left (b \,x^{2}+a \right )^{5}}\) | \(80\) |
-1/24*csgn(b*x^2+a)*(12/7*x^10*b^5+15/2*a*x^8*b^4+40/3*a^2*x^6*b^3+12*a^3* x^4*b^2+60/11*x^2*a^4*b+a^5)/x^24
Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.23 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {792 \, b^{5} x^{10} + 3465 \, a b^{4} x^{8} + 6160 \, a^{2} b^{3} x^{6} + 5544 \, a^{3} b^{2} x^{4} + 2520 \, a^{4} b x^{2} + 462 \, a^{5}}{11088 \, x^{24}} \]
-1/11088*(792*b^5*x^10 + 3465*a*b^4*x^8 + 6160*a^2*b^3*x^6 + 5544*a^3*b^2* x^4 + 2520*a^4*b*x^2 + 462*a^5)/x^24
\[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=\int \frac {\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}{x^{25}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.22 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {b^{5}}{14 \, x^{14}} - \frac {5 \, a b^{4}}{16 \, x^{16}} - \frac {5 \, a^{2} b^{3}}{9 \, x^{18}} - \frac {a^{3} b^{2}}{2 \, x^{20}} - \frac {5 \, a^{4} b}{22 \, x^{22}} - \frac {a^{5}}{24 \, x^{24}} \]
-1/14*b^5/x^14 - 5/16*a*b^4/x^16 - 5/9*a^2*b^3/x^18 - 1/2*a^3*b^2/x^20 - 5 /22*a^4*b/x^22 - 1/24*a^5/x^24
Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {792 \, b^{5} x^{10} \mathrm {sgn}\left (b x^{2} + a\right ) + 3465 \, a b^{4} x^{8} \mathrm {sgn}\left (b x^{2} + a\right ) + 6160 \, a^{2} b^{3} x^{6} \mathrm {sgn}\left (b x^{2} + a\right ) + 5544 \, a^{3} b^{2} x^{4} \mathrm {sgn}\left (b x^{2} + a\right ) + 2520 \, a^{4} b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + 462 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )}{11088 \, x^{24}} \]
-1/11088*(792*b^5*x^10*sgn(b*x^2 + a) + 3465*a*b^4*x^8*sgn(b*x^2 + a) + 61 60*a^2*b^3*x^6*sgn(b*x^2 + a) + 5544*a^3*b^2*x^4*sgn(b*x^2 + a) + 2520*a^4 *b*x^2*sgn(b*x^2 + a) + 462*a^5*sgn(b*x^2 + a))/x^24
Time = 13.23 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}{x^{25}} \, dx=-\frac {a^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{24\,x^{24}\,\left (b\,x^2+a\right )}-\frac {b^5\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{14\,x^{14}\,\left (b\,x^2+a\right )}-\frac {5\,a\,b^4\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{16\,x^{16}\,\left (b\,x^2+a\right )}-\frac {5\,a^4\,b\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{22\,x^{22}\,\left (b\,x^2+a\right )}-\frac {5\,a^2\,b^3\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{9\,x^{18}\,\left (b\,x^2+a\right )}-\frac {a^3\,b^2\,\sqrt {a^2+2\,a\,b\,x^2+b^2\,x^4}}{2\,x^{20}\,\left (b\,x^2+a\right )} \]
- (a^5*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(24*x^24*(a + b*x^2)) - (b^5*(a^ 2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(14*x^14*(a + b*x^2)) - (5*a*b^4*(a^2 + b^ 2*x^4 + 2*a*b*x^2)^(1/2))/(16*x^16*(a + b*x^2)) - (5*a^4*b*(a^2 + b^2*x^4 + 2*a*b*x^2)^(1/2))/(22*x^22*(a + b*x^2)) - (5*a^2*b^3*(a^2 + b^2*x^4 + 2* a*b*x^2)^(1/2))/(9*x^18*(a + b*x^2)) - (a^3*b^2*(a^2 + b^2*x^4 + 2*a*b*x^2 )^(1/2))/(2*x^20*(a + b*x^2))