Integrand size = 26, antiderivative size = 196 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {2 a}{b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^4}{8 b^5 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {2 a^3}{3 b^5 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a^2}{2 b^5 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
2*a/b^5/((b*x^2+a)^2)^(1/2)-1/8*a^4/b^5/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)+2/ 3*a^3/b^5/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)-3/2*a^2/b^5/(b*x^2+a)/((b*x^2+a) ^2)^(1/2)+1/2*(b*x^2+a)*ln(b*x^2+a)/b^5/((b*x^2+a)^2)^(1/2)
Time = 0.74 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.33 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\frac {b x^2 \left (-a \sqrt {\left (a+b x^2\right )^2} \left (12 a^6+30 a^5 b x^2+22 a^4 b^2 x^4+3 a^3 b^3 x^6-3 a^2 b^4 x^8+3 a b^5 x^{10}-3 b^6 x^{12}\right )+\sqrt {a^2} \left (12 a^7+42 a^6 b x^2+52 a^5 b^2 x^4+25 a^4 b^3 x^6+3 b^7 x^{14}\right )\right )}{a^4 \left (a+b x^2\right )^3 \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )}+12 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-12 \log \left (b^5 \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )\right )}{24 b^5} \]
((b*x^2*(-(a*Sqrt[(a + b*x^2)^2]*(12*a^6 + 30*a^5*b*x^2 + 22*a^4*b^2*x^4 + 3*a^3*b^3*x^6 - 3*a^2*b^4*x^8 + 3*a*b^5*x^10 - 3*b^6*x^12)) + Sqrt[a^2]*( 12*a^7 + 42*a^6*b*x^2 + 52*a^5*b^2*x^4 + 25*a^4*b^3*x^6 + 3*b^7*x^14)))/(a ^4*(a + b*x^2)^3*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2])) + 12*Log [Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - 12*Log[b^5*(Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2])])/(24*b^5)
Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.59, 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, 49, 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.
\(\displaystyle \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {x^9}{b^5 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^9}{\left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^8}{\left (b x^2+a\right )^5}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (\frac {a^4}{b^4 \left (b x^2+a\right )^5}-\frac {4 a^3}{b^4 \left (b x^2+a\right )^4}+\frac {6 a^2}{b^4 \left (b x^2+a\right )^3}-\frac {4 a}{b^4 \left (b x^2+a\right )^2}+\frac {1}{b^4 \left (b x^2+a\right )}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {a^4}{4 b^5 \left (a+b x^2\right )^4}+\frac {4 a^3}{3 b^5 \left (a+b x^2\right )^3}-\frac {3 a^2}{b^5 \left (a+b x^2\right )^2}+\frac {4 a}{b^5 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{b^5}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
((a + b*x^2)*(-1/4*a^4/(b^5*(a + b*x^2)^4) + (4*a^3)/(3*b^5*(a + b*x^2)^3) - (3*a^2)/(b^5*(a + b*x^2)^2) + (4*a)/(b^5*(a + b*x^2)) + Log[a + b*x^2]/ b^5))/(2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
3.7.46.3.1 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}, x] && IGtQ[m, 0] && IGtQ[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 3.
Time = 0.09 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.39
method | result | size |
pseudoelliptic | \(\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (\left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+4 a \,b^{3} x^{6}+9 a^{2} b^{2} x^{4}+\frac {22 a^{3} b \,x^{2}}{3}+\frac {25 a^{4}}{12}\right )}{2 \left (b \,x^{2}+a \right )^{4} b^{5}}\) | \(76\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {2 a \,x^{6}}{b^{2}}+\frac {9 a^{2} x^{4}}{2 b^{3}}+\frac {11 a^{3} x^{2}}{3 b^{4}}+\frac {25 a^{4}}{24 b^{5}}\right )}{\left (b \,x^{2}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{5}}\) | \(96\) |
default | \(\frac {\left (12 \ln \left (b \,x^{2}+a \right ) x^{8} b^{4}+48 \ln \left (b \,x^{2}+a \right ) x^{6} a \,b^{3}+48 a \,b^{3} x^{6}+72 \ln \left (b \,x^{2}+a \right ) x^{4} a^{2} b^{2}+108 a^{2} b^{2} x^{4}+48 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b +88 a^{3} b \,x^{2}+12 \ln \left (b \,x^{2}+a \right ) a^{4}+25 a^{4}\right ) \left (b \,x^{2}+a \right )}{24 b^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(141\) |
1/2*csgn(b*x^2+a)*((b*x^2+a)^4*ln(b*x^2+a)+4*a*b^3*x^6+9*a^2*b^2*x^4+22/3* a^3*b*x^2+25/12*a^4)/(b*x^2+a)^4/b^5
Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.69 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4} + 12 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} \]
1/24*(48*a*b^3*x^6 + 108*a^2*b^2*x^4 + 88*a^3*b*x^2 + 25*a^4 + 12*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*log(b*x^2 + a))/(b^9*x ^8 + 4*a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5)
\[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{9}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.51 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {48 \, a b^{3} x^{6} + 108 \, a^{2} b^{2} x^{4} + 88 \, a^{3} b x^{2} + 25 \, a^{4}}{24 \, {\left (b^{9} x^{8} + 4 \, a b^{8} x^{6} + 6 \, a^{2} b^{7} x^{4} + 4 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{5}} \]
1/24*(48*a*b^3*x^6 + 108*a^2*b^2*x^4 + 88*a^3*b*x^2 + 25*a^4)/(b^9*x^8 + 4 *a*b^8*x^6 + 6*a^2*b^7*x^4 + 4*a^3*b^6*x^2 + a^4*b^5) + 1/2*log(b*x^2 + a) /b^5
Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.43 \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {25 \, b^{3} x^{8} + 52 \, a b^{2} x^{6} + 42 \, a^{2} b x^{4} + 12 \, a^{3} x^{2}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \]
1/2*log(abs(b*x^2 + a))/(b^5*sgn(b*x^2 + a)) - 1/24*(25*b^3*x^8 + 52*a*b^2 *x^6 + 42*a^2*b*x^4 + 12*a^3*x^2)/((b*x^2 + a)^4*b^4*sgn(b*x^2 + a))
Timed out. \[ \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^9}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]