Integrand size = 26, antiderivative size = 238 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {5 a^2}{b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^5}{8 b^6 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a^4}{6 b^6 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 a^3}{2 b^6 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
-5*a^2/b^6/((b*x^2+a)^2)^(1/2)+1/8*a^5/b^6/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2) -5/6*a^4/b^6/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)+5/2*a^3/b^6/(b*x^2+a)/((b*x^2 +a)^2)^(1/2)+1/2*x^2*(b*x^2+a)/b^5/((b*x^2+a)^2)^(1/2)-5/2*a*(b*x^2+a)*ln( b*x^2+a)/b^6/((b*x^2+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.43 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-77 a^5-248 a^4 b x^2-252 a^3 b^2 x^4-48 a^2 b^3 x^6+48 a b^4 x^8+12 b^5 x^{10}-60 a \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 b^6 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
(-77*a^5 - 248*a^4*b*x^2 - 252*a^3*b^2*x^4 - 48*a^2*b^3*x^6 + 48*a*b^4*x^8 + 12*b^5*x^10 - 60*a*(a + b*x^2)^4*Log[a + b*x^2])/(24*b^6*(a + b*x^2)^3* Sqrt[(a + b*x^2)^2])
Time = 0.47 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.53, 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.
\(\displaystyle \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {x^{11}}{b^5 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^{11}}{\left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^{10}}{\left (b x^2+a\right )^5}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (-\frac {a^5}{b^5 \left (b x^2+a\right )^5}+\frac {5 a^4}{b^5 \left (b x^2+a\right )^4}-\frac {10 a^3}{b^5 \left (b x^2+a\right )^3}+\frac {10 a^2}{b^5 \left (b x^2+a\right )^2}-\frac {5 a}{b^5 \left (b x^2+a\right )}+\frac {1}{b^5}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {a^5}{4 b^6 \left (a+b x^2\right )^4}-\frac {5 a^4}{3 b^6 \left (a+b x^2\right )^3}+\frac {5 a^3}{b^6 \left (a+b x^2\right )^2}-\frac {10 a^2}{b^6 \left (a+b x^2\right )}-\frac {5 a \log \left (a+b x^2\right )}{b^6}+\frac {x^2}{b^5}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2),x]
Output:
((a + b*x^2)*(x^2/b^5 + a^5/(4*b^6*(a + b*x^2)^4) - (5*a^4)/(3*b^6*(a + b* x^2)^3) + (5*a^3)/(b^6*(a + b*x^2)^2) - (10*a^2)/(b^6*(a + b*x^2)) - (5*a* Log[a + b*x^2])/b^6))/(2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
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 = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.40
method | result | size |
pseudoelliptic | \(-\frac {5 \,\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (a \left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )-\frac {x^{10} b^{5}}{5}-\frac {4 a \,x^{8} b^{4}}{5}+\frac {4 a^{2} x^{6} b^{3}}{5}+\frac {21 a^{3} x^{4} b^{2}}{5}+\frac {62 x^{2} a^{4} b}{15}+\frac {77 a^{5}}{60}\right )}{2 \left (b \,x^{2}+a \right )^{4} b^{6}}\) | \(96\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2}}{2 \left (b \,x^{2}+a \right ) b^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-5 a^{2} b^{2} x^{6}-\frac {25 a^{3} b \,x^{4}}{2}-\frac {65 a^{4} x^{2}}{6}-\frac {77 a^{5}}{24 b}\right )}{\left (b \,x^{2}+a \right )^{5} b^{5}}-\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{6}}\) | \(125\) |
default | \(-\frac {\left (-12 x^{10} b^{5}+60 \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}-48 a \,x^{8} b^{4}+240 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}+48 a^{2} x^{6} b^{3}+360 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+252 a^{3} x^{4} b^{2}+240 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +248 x^{2} a^{4} b +60 \ln \left (b \,x^{2}+a \right ) a^{5}+77 a^{5}\right ) \left (b \,x^{2}+a \right )}{24 b^{6} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(163\) |
Input:
int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-5/2*csgn(b*x^2+a)*(a*(b*x^2+a)^4*ln(b*x^2+a)-1/5*x^10*b^5-4/5*a*x^8*b^4+4 /5*a^2*x^6*b^3+21/5*a^3*x^4*b^2+62/15*x^2*a^4*b+77/60*a^5)/(b*x^2+a)^4/b^6
Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.66 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, b^{5} x^{10} + 48 \, a b^{4} x^{8} - 48 \, a^{2} b^{3} x^{6} - 252 \, a^{3} b^{2} x^{4} - 248 \, a^{4} b x^{2} - 77 \, a^{5} - 60 \, {\left (a b^{4} x^{8} + 4 \, a^{2} b^{3} x^{6} + 6 \, a^{3} b^{2} x^{4} + 4 \, a^{4} b x^{2} + a^{5}\right )} \log \left (b x^{2} + a\right )}{24 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/24*(12*b^5*x^10 + 48*a*b^4*x^8 - 48*a^2*b^3*x^6 - 252*a^3*b^2*x^4 - 248* a^4*b*x^2 - 77*a^5 - 60*(a*b^4*x^8 + 4*a^2*b^3*x^6 + 6*a^3*b^2*x^4 + 4*a^4 *b*x^2 + a^5)*log(b*x^2 + a))/(b^10*x^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4* a^3*b^7*x^2 + a^4*b^6)
\[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{11}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(x**11/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(x**11/((a + b*x**2)**2)**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.46 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {120 \, a^{2} b^{3} x^{6} + 300 \, a^{3} b^{2} x^{4} + 260 \, a^{4} b x^{2} + 77 \, a^{5}}{24 \, {\left (b^{10} x^{8} + 4 \, a b^{9} x^{6} + 6 \, a^{2} b^{8} x^{4} + 4 \, a^{3} b^{7} x^{2} + a^{4} b^{6}\right )}} + \frac {x^{2}}{2 \, b^{5}} - \frac {5 \, a \log \left (b x^{2} + a\right )}{2 \, b^{6}} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/24*(120*a^2*b^3*x^6 + 300*a^3*b^2*x^4 + 260*a^4*b*x^2 + 77*a^5)/(b^10*x ^8 + 4*a*b^9*x^6 + 6*a^2*b^8*x^4 + 4*a^3*b^7*x^2 + a^4*b^6) + 1/2*x^2/b^5 - 5/2*a*log(b*x^2 + a)/b^6
Time = 0.12 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.48 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {x^{2}}{2 \, b^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {5 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {125 \, a b^{4} x^{8} + 380 \, a^{2} b^{3} x^{6} + 450 \, a^{3} b^{2} x^{4} + 240 \, a^{4} b x^{2} + 48 \, a^{5}}{24 \, {\left (b x^{2} + a\right )}^{4} b^{6} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/2*x^2/(b^5*sgn(b*x^2 + a)) - 5/2*a*log(abs(b*x^2 + a))/(b^6*sgn(b*x^2 + a)) + 1/24*(125*a*b^4*x^8 + 380*a^2*b^3*x^6 + 450*a^3*b^2*x^4 + 240*a^4*b* x^2 + 48*a^5)/((b*x^2 + a)^4*b^6*sgn(b*x^2 + a))
Timed out. \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{11}}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:
int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)
Output:
int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)
Time = 0.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.74 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{5}-240 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{4} b \,x^{2}-360 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b^{2} x^{4}-240 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{3} x^{6}-60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{4} x^{8}-65 a^{5}-200 a^{4} b \,x^{2}-180 a^{3} b^{2} x^{4}+60 a \,b^{4} x^{8}+12 b^{5} x^{10}}{24 b^{6} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
( - 60*log(a + b*x**2)*a**5 - 240*log(a + b*x**2)*a**4*b*x**2 - 360*log(a + b*x**2)*a**3*b**2*x**4 - 240*log(a + b*x**2)*a**2*b**3*x**6 - 60*log(a + b*x**2)*a*b**4*x**8 - 65*a**5 - 200*a**4*b*x**2 - 180*a**3*b**2*x**4 + 60 *a*b**4*x**8 + 12*b**5*x**10)/(24*b**6*(a**4 + 4*a**3*b*x**2 + 6*a**2*b**2 *x**4 + 4*a*b**3*x**6 + b**4*x**8))