Integrand size = 24, antiderivative size = 91 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {2 a x^2}{b^5}+\frac {x^4}{4 b^4}+\frac {a^5}{6 b^6 \left (a+b x^2\right )^3}-\frac {5 a^4}{4 b^6 \left (a+b x^2\right )^2}+\frac {5 a^3}{b^6 \left (a+b x^2\right )}+\frac {5 a^2 \log \left (a+b x^2\right )}{b^6} \] Output:
-2*a*x^2/b^5+1/4*x^4/b^4+1/6*a^5/b^6/(b*x^2+a)^3-5/4*a^4/b^6/(b*x^2+a)^2+5 *a^3/b^6/(b*x^2+a)+5*a^2*ln(b*x^2+a)/b^6
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.86 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {-24 a b x^2+3 b^2 x^4+\frac {2 a^5}{\left (a+b x^2\right )^3}-\frac {15 a^4}{\left (a+b x^2\right )^2}+\frac {60 a^3}{a+b x^2}+60 a^2 \log \left (a+b x^2\right )}{12 b^6} \] Input:
Integrate[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
(-24*a*b*x^2 + 3*b^2*x^4 + (2*a^5)/(a + b*x^2)^3 - (15*a^4)/(a + b*x^2)^2 + (60*a^3)/(a + b*x^2) + 60*a^2*Log[a + b*x^2])/(12*b^6)
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1380, 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.
\(\displaystyle \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^{11}}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^{11}}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {x^{10}}{\left (b x^2+a\right )^4}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {a^5}{b^5 \left (b x^2+a\right )^4}+\frac {5 a^4}{b^5 \left (b x^2+a\right )^3}-\frac {10 a^3}{b^5 \left (b x^2+a\right )^2}+\frac {10 a^2}{b^5 \left (b x^2+a\right )}-\frac {4 a}{b^5}+\frac {x^2}{b^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^5}{3 b^6 \left (a+b x^2\right )^3}-\frac {5 a^4}{2 b^6 \left (a+b x^2\right )^2}+\frac {10 a^3}{b^6 \left (a+b x^2\right )}+\frac {10 a^2 \log \left (a+b x^2\right )}{b^6}-\frac {4 a x^2}{b^5}+\frac {x^4}{2 b^4}\right )\) |
Input:
Int[x^11/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
((-4*a*x^2)/b^5 + x^4/(2*b^4) + a^5/(3*b^6*(a + b*x^2)^3) - (5*a^4)/(2*b^6 *(a + b*x^2)^2) + (10*a^3)/(b^6*(a + b*x^2)) + (10*a^2*Log[a + b*x^2])/b^6 )/2
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[1/c^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]
Time = 0.14 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84
method | result | size |
norman | \(\frac {\frac {x^{10}}{4 b}-\frac {5 a \,x^{8}}{4 b^{2}}+\frac {55 a^{5}}{6 b^{6}}+\frac {15 a^{3} x^{4}}{b^{4}}+\frac {45 a^{4} x^{2}}{2 b^{5}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {5 a^{2} \ln \left (b \,x^{2}+a \right )}{b^{6}}\) | \(76\) |
default | \(\frac {\left (-b \,x^{2}+4 a \right )^{2}}{4 b^{6}}+\frac {a^{2} \left (\frac {10 \ln \left (b \,x^{2}+a \right )}{b}+\frac {a^{3}}{3 b \left (b \,x^{2}+a \right )^{3}}+\frac {10 a}{b \left (b \,x^{2}+a \right )}-\frac {5 a^{2}}{2 b \left (b \,x^{2}+a \right )^{2}}\right )}{2 b^{5}}\) | \(90\) |
risch | \(\frac {x^{4}}{4 b^{4}}-\frac {2 a \,x^{2}}{b^{5}}+\frac {4 a^{2}}{b^{6}}+\frac {5 a^{3} b \,x^{4}+\frac {35 a^{4} x^{2}}{4}+\frac {47 a^{5}}{12 b}}{b^{5} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {5 a^{2} \ln \left (b \,x^{2}+a \right )}{b^{6}}\) | \(102\) |
parallelrisch | \(\frac {3 x^{10} b^{5}-15 a \,x^{8} b^{4}+60 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}+180 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}+180 a^{3} x^{4} b^{2}+180 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b +270 x^{2} a^{4} b +60 \ln \left (b \,x^{2}+a \right ) a^{5}+110 a^{5}}{12 b^{6} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}\) | \(146\) |
Input:
int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(1/4/b*x^10-5/4*a/b^2*x^8+55/6*a^5/b^6+15*a^3/b^4*x^4+45/2*a^4/b^5*x^2)/(b *x^2+a)^3+5*a^2*ln(b*x^2+a)/b^6
Time = 0.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.51 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {3 \, b^{5} x^{10} - 15 \, a b^{4} x^{8} - 63 \, a^{2} b^{3} x^{6} - 9 \, a^{3} b^{2} x^{4} + 81 \, a^{4} b x^{2} + 47 \, a^{5} + 60 \, {\left (a^{2} b^{3} x^{6} + 3 \, a^{3} b^{2} x^{4} + 3 \, a^{4} b x^{2} + a^{5}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
1/12*(3*b^5*x^10 - 15*a*b^4*x^8 - 63*a^2*b^3*x^6 - 9*a^3*b^2*x^4 + 81*a^4* b*x^2 + 47*a^5 + 60*(a^2*b^3*x^6 + 3*a^3*b^2*x^4 + 3*a^4*b*x^2 + a^5)*log( b*x^2 + a))/(b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2 + a^3*b^6)
Time = 0.28 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.10 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5 a^{2} \log {\left (a + b x^{2} \right )}}{b^{6}} - \frac {2 a x^{2}}{b^{5}} + \frac {47 a^{5} + 105 a^{4} b x^{2} + 60 a^{3} b^{2} x^{4}}{12 a^{3} b^{6} + 36 a^{2} b^{7} x^{2} + 36 a b^{8} x^{4} + 12 b^{9} x^{6}} + \frac {x^{4}}{4 b^{4}} \] Input:
integrate(x**11/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
5*a**2*log(a + b*x**2)/b**6 - 2*a*x**2/b**5 + (47*a**5 + 105*a**4*b*x**2 + 60*a**3*b**2*x**4)/(12*a**3*b**6 + 36*a**2*b**7*x**2 + 36*a*b**8*x**4 + 1 2*b**9*x**6) + x**4/(4*b**4)
Time = 0.03 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.09 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {60 \, a^{3} b^{2} x^{4} + 105 \, a^{4} b x^{2} + 47 \, a^{5}}{12 \, {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}} + \frac {5 \, a^{2} \log \left (b x^{2} + a\right )}{b^{6}} + \frac {b x^{4} - 8 \, a x^{2}}{4 \, b^{5}} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
1/12*(60*a^3*b^2*x^4 + 105*a^4*b*x^2 + 47*a^5)/(b^9*x^6 + 3*a*b^8*x^4 + 3* a^2*b^7*x^2 + a^3*b^6) + 5*a^2*log(b*x^2 + a)/b^6 + 1/4*(b*x^4 - 8*a*x^2)/ b^5
Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5 \, a^{2} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{6}} + \frac {b^{4} x^{4} - 8 \, a b^{3} x^{2}}{4 \, b^{8}} - \frac {110 \, a^{2} b^{3} x^{6} + 270 \, a^{3} b^{2} x^{4} + 225 \, a^{4} b x^{2} + 63 \, a^{5}}{12 \, {\left (b x^{2} + a\right )}^{3} b^{6}} \] Input:
integrate(x^11/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
5*a^2*log(abs(b*x^2 + a))/b^6 + 1/4*(b^4*x^4 - 8*a*b^3*x^2)/b^8 - 1/12*(11 0*a^2*b^3*x^6 + 270*a^3*b^2*x^4 + 225*a^4*b*x^2 + 63*a^5)/((b*x^2 + a)^3*b ^6)
Time = 17.76 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.08 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {47\,a^5}{12\,b}+\frac {35\,a^4\,x^2}{4}+5\,a^3\,b\,x^4}{a^3\,b^5+3\,a^2\,b^6\,x^2+3\,a\,b^7\,x^4+b^8\,x^6}+\frac {x^4}{4\,b^4}-\frac {2\,a\,x^2}{b^5}+\frac {5\,a^2\,\ln \left (b\,x^2+a\right )}{b^6} \] Input:
int(x^11/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)
Output:
((47*a^5)/(12*b) + (35*a^4*x^2)/4 + 5*a^3*b*x^4)/(a^3*b^5 + b^8*x^6 + 3*a* b^7*x^4 + 3*a^2*b^6*x^2) + x^4/(4*b^4) - (2*a*x^2)/b^5 + (5*a^2*log(a + b* x^2))/b^6
Time = 0.17 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.62 \[ \int \frac {x^{11}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{5}+180 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{4} b \,x^{2}+180 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b^{2} x^{4}+60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{3} x^{6}+50 a^{5}+90 a^{4} b \,x^{2}-60 a^{2} b^{3} x^{6}-15 a \,b^{4} x^{8}+3 b^{5} x^{10}}{12 b^{6} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(x^11/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
(60*log(a + b*x**2)*a**5 + 180*log(a + b*x**2)*a**4*b*x**2 + 180*log(a + b *x**2)*a**3*b**2*x**4 + 60*log(a + b*x**2)*a**2*b**3*x**6 + 50*a**5 + 90*a **4*b*x**2 - 60*a**2*b**3*x**6 - 15*a*b**4*x**8 + 3*b**5*x**10)/(12*b**6*( a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))