Integrand size = 24, antiderivative size = 71 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {a^3}{6 b^4 \left (a+b x^2\right )^3}-\frac {3 a^2}{4 b^4 \left (a+b x^2\right )^2}+\frac {3 a}{2 b^4 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{2 b^4} \] Output:
1/6*a^3/b^4/(b*x^2+a)^3-3/4*a^2/b^4/(b*x^2+a)^2+3/2*a/b^4/(b*x^2+a)+1/2*ln (b*x^2+a)/b^4
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.70 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {a \left (11 a^2+27 a b x^2+18 b^2 x^4\right )}{\left (a+b x^2\right )^3}+6 \log \left (a+b x^2\right )}{12 b^4} \] Input:
Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
((a*(11*a^2 + 27*a*b*x^2 + 18*b^2*x^4))/(a + b*x^2)^3 + 6*Log[a + b*x^2])/ (12*b^4)
Time = 0.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99, 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.
\(\displaystyle \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^7}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^7}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {1}{2} \int \frac {x^6}{\left (b x^2+a\right )^4}dx^2\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \int \left (-\frac {a^3}{b^3 \left (b x^2+a\right )^4}+\frac {3 a^2}{b^3 \left (b x^2+a\right )^3}-\frac {3 a}{b^3 \left (b x^2+a\right )^2}+\frac {1}{b^3 \left (b x^2+a\right )}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {a^3}{3 b^4 \left (a+b x^2\right )^3}-\frac {3 a^2}{2 b^4 \left (a+b x^2\right )^2}+\frac {3 a}{b^4 \left (a+b x^2\right )}+\frac {\log \left (a+b x^2\right )}{b^4}\right )\) |
Input:
Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
(a^3/(3*b^4*(a + b*x^2)^3) - (3*a^2)/(2*b^4*(a + b*x^2)^2) + (3*a)/(b^4*(a + b*x^2)) + Log[a + b*x^2]/b^4)/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 = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76
method | result | size |
norman | \(\frac {\frac {11 a^{3}}{12 b^{4}}+\frac {3 a \,x^{4}}{2 b^{2}}+\frac {9 a^{2} x^{2}}{4 b^{3}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(54\) |
default | \(\frac {a^{3}}{6 b^{4} \left (b \,x^{2}+a \right )^{3}}-\frac {3 a^{2}}{4 b^{4} \left (b \,x^{2}+a \right )^{2}}+\frac {3 a}{2 b^{4} \left (b \,x^{2}+a \right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(64\) |
risch | \(\frac {\frac {11 a^{3}}{12 b^{4}}+\frac {3 a \,x^{4}}{2 b^{2}}+\frac {9 a^{2} x^{2}}{4 b^{3}}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {\ln \left (b \,x^{2}+a \right )}{2 b^{4}}\) | \(74\) |
parallelrisch | \(\frac {6 \ln \left (b \,x^{2}+a \right ) x^{6} b^{3}+18 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}+18 b^{2} x^{4} a +18 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b +27 a^{2} b \,x^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3}+11 a^{3}}{12 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}\) | \(122\) |
Input:
int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(11/12*a^3/b^4+3/2*a/b^2*x^4+9/4*a^2/b^3*x^2)/(b*x^2+a)^3+1/2*ln(b*x^2+a)/ b^4
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.44 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {18 \, a b^{2} x^{4} + 27 \, a^{2} b x^{2} + 11 \, a^{3} + 6 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{12 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
1/12*(18*a*b^2*x^4 + 27*a^2*b*x^2 + 11*a^3 + 6*(b^3*x^6 + 3*a*b^2*x^4 + 3* a^2*b*x^2 + a^3)*log(b*x^2 + a))/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2*b^5*x^2 + a^3*b^4)
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.07 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {11 a^{3} + 27 a^{2} b x^{2} + 18 a b^{2} x^{4}}{12 a^{3} b^{4} + 36 a^{2} b^{5} x^{2} + 36 a b^{6} x^{4} + 12 b^{7} x^{6}} + \frac {\log {\left (a + b x^{2} \right )}}{2 b^{4}} \] Input:
integrate(x**7/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
(11*a**3 + 27*a**2*b*x**2 + 18*a*b**2*x**4)/(12*a**3*b**4 + 36*a**2*b**5*x **2 + 36*a*b**6*x**4 + 12*b**7*x**6) + log(a + b*x**2)/(2*b**4)
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {18 \, a b^{2} x^{4} + 27 \, a^{2} b x^{2} + 11 \, a^{3}}{12 \, {\left (b^{7} x^{6} + 3 \, a b^{6} x^{4} + 3 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}} + \frac {\log \left (b x^{2} + a\right )}{2 \, b^{4}} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
1/12*(18*a*b^2*x^4 + 27*a^2*b*x^2 + 11*a^3)/(b^7*x^6 + 3*a*b^6*x^4 + 3*a^2 *b^5*x^2 + a^3*b^4) + 1/2*log(b*x^2 + a)/b^4
Time = 0.12 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.75 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4}} - \frac {11 \, b^{2} x^{6} + 15 \, a b x^{4} + 6 \, a^{2} x^{2}}{12 \, {\left (b x^{2} + a\right )}^{3} b^{3}} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
1/2*log(abs(b*x^2 + a))/b^4 - 1/12*(11*b^2*x^6 + 15*a*b*x^4 + 6*a^2*x^2)/( (b*x^2 + a)^3*b^3)
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {11\,a^3}{12\,b^4}+\frac {3\,a\,x^4}{2\,b^2}+\frac {9\,a^2\,x^2}{4\,b^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {\ln \left (b\,x^2+a\right )}{2\,b^4} \] Input:
int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)
Output:
((11*a^3)/(12*b^4) + (3*a*x^4)/(2*b^2) + (9*a^2*x^2)/(4*b^3))/(a^3 + b^3*x ^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) + log(a + b*x^2)/(2*b^4)
Time = 0.17 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.72 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3}+18 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b \,x^{2}+18 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} x^{4}+6 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{3} x^{6}+5 a^{3}+9 a^{2} b \,x^{2}-6 b^{3} x^{6}}{12 b^{4} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
(6*log(a + b*x**2)*a**3 + 18*log(a + b*x**2)*a**2*b*x**2 + 18*log(a + b*x* *2)*a*b**2*x**4 + 6*log(a + b*x**2)*b**3*x**6 + 5*a**3 + 9*a**2*b*x**2 - 6 *b**3*x**6)/(12*b**4*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))