Integrand size = 26, antiderivative size = 158 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {3 a^2}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {a^3}{4 b^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x^2 \left (a+b x^2\right )}{2 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 a \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 b^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
-3/2*a^2/b^4/((b*x^2+a)^2)^(1/2)+1/4*a^3/b^4/(b*x^2+a)/((b*x^2+a)^2)^(1/2) +1/2*x^2*(b*x^2+a)/b^3/((b*x^2+a)^2)^(1/2)-3/2*a*(b*x^2+a)*ln(b*x^2+a)/b^4 /((b*x^2+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.51 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-5 a^3-4 a^2 b x^2+4 a b^2 x^4+2 b^3 x^6-6 a \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 b^4 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
(-5*a^3 - 4*a^2*b*x^2 + 4*a*b^2*x^4 + 2*b^3*x^6 - 6*a*(a + b*x^2)^2*Log[a + b*x^2])/(4*b^4*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.41 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58, 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.
\(\displaystyle \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {x^7}{b^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^7}{\left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^6}{\left (b x^2+a\right )^3}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (-\frac {a^3}{b^3 \left (b x^2+a\right )^3}+\frac {3 a^2}{b^3 \left (b x^2+a\right )^2}-\frac {3 a}{b^3 \left (b x^2+a\right )}+\frac {1}{b^3}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {a^3}{2 b^4 \left (a+b x^2\right )^2}-\frac {3 a^2}{b^4 \left (a+b x^2\right )}-\frac {3 a \log \left (a+b x^2\right )}{b^4}+\frac {x^2}{b^3}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[x^7/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
((a + b*x^2)*(x^2/b^3 + a^3/(2*b^4*(a + b*x^2)^2) - (3*a^2)/(b^4*(a + b*x^ 2)) - (3*a*Log[a + b*x^2])/b^4))/(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 = 0.15 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.47
method | result | size |
pseudoelliptic | \(-\frac {3 \,\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (a \left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )-\frac {b^{3} x^{6}}{3}-\frac {2 b^{2} x^{4} a}{3}+\frac {2 a^{2} b \,x^{2}}{3}+\frac {5 a^{3}}{6}\right )}{2 \left (b \,x^{2}+a \right )^{2} b^{4}}\) | \(74\) |
default | \(-\frac {\left (-2 b^{3} x^{6}+6 \ln \left (b \,x^{2}+a \right ) x^{4} a \,b^{2}-4 b^{2} x^{4} a +12 \ln \left (b \,x^{2}+a \right ) x^{2} a^{2} b +4 a^{2} b \,x^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{3}+5 a^{3}\right ) \left (b \,x^{2}+a \right )}{4 b^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(103\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x^{2}}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {3 a^{2} x^{2}}{2}-\frac {5 a^{3}}{4 b}\right )}{\left (b \,x^{2}+a \right )^{3} b^{3}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, a \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) b^{4}}\) | \(105\) |
Input:
int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-3/2*csgn(b*x^2+a)*(a*(b*x^2+a)^2*ln(b*x^2+a)-1/3*b^3*x^6-2/3*b^2*x^4*a+2/ 3*a^2*b*x^2+5/6*a^3)/(b*x^2+a)^2/b^4
Time = 0.06 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.58 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, b^{3} x^{6} + 4 \, a b^{2} x^{4} - 4 \, a^{2} b x^{2} - 5 \, a^{3} - 6 \, {\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \log \left (b x^{2} + a\right )}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
1/4*(2*b^3*x^6 + 4*a*b^2*x^4 - 4*a^2*b*x^2 - 5*a^3 - 6*(a*b^2*x^4 + 2*a^2* b*x^2 + a^3)*log(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)
\[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**7/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(x**7/((a + b*x**2)**2)**(3/2), x)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.42 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {6 \, a^{2} b x^{2} + 5 \, a^{3}}{4 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}} + \frac {x^{2}}{2 \, b^{3}} - \frac {3 \, a \log \left (b x^{2} + a\right )}{2 \, b^{4}} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
-1/4*(6*a^2*b*x^2 + 5*a^3)/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4) + 1/2*x^2/b^3 - 3/2*a*log(b*x^2 + a)/b^4
Time = 0.11 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.58 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {x^{2}}{2 \, b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {3 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {9 \, a b^{2} x^{4} + 12 \, a^{2} b x^{2} + 4 \, a^{3}}{4 \, {\left (b x^{2} + a\right )}^{2} b^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
1/2*x^2/(b^3*sgn(b*x^2 + a)) - 3/2*a*log(abs(b*x^2 + a))/(b^4*sgn(b*x^2 + a)) + 1/4*(9*a*b^2*x^4 + 12*a^2*b*x^2 + 4*a^3)/((b*x^2 + a)^2*b^4*sgn(b*x^ 2 + a))
Timed out. \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int(x^7/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.60 \[ \int \frac {x^7}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3}-12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b \,x^{2}-6 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{2} x^{4}-3 a^{3}+6 a \,b^{2} x^{4}+2 b^{3} x^{6}}{4 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^7/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
( - 6*log(a + b*x**2)*a**3 - 12*log(a + b*x**2)*a**2*b*x**2 - 6*log(a + b* x**2)*a*b**2*x**4 - 3*a**3 + 6*a*b**2*x**4 + 2*b**3*x**6)/(4*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4))