Integrand size = 26, antiderivative size = 267 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {2 b}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{8 a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b}{3 a^3 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {3 b}{4 a^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{2 a^5 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 b \left (a+b x^2\right ) \log (x)}{a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 b \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^6 \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
-2*b/a^5/((b*x^2+a)^2)^(1/2)-1/8*b/a^2/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)-1/3 *b/a^3/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)-3/4*b/a^4/(b*x^2+a)/((b*x^2+a)^2)^( 1/2)-1/2*(b*x^2+a)/a^5/x^2/((b*x^2+a)^2)^(1/2)-5*b*(b*x^2+a)*ln(x)/a^6/((b *x^2+a)^2)^(1/2)+5/2*b*(b*x^2+a)*ln(b*x^2+a)/a^6/((b*x^2+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-a \left (12 a^4+125 a^3 b x^2+260 a^2 b^2 x^4+210 a b^3 x^6+60 b^4 x^8\right )-120 b x^2 \left (a+b x^2\right )^4 \log (x)+60 b x^2 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 a^6 x^2 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
Output:
(-(a*(12*a^4 + 125*a^3*b*x^2 + 260*a^2*b^2*x^4 + 210*a*b^3*x^6 + 60*b^4*x^ 8)) - 120*b*x^2*(a + b*x^2)^4*Log[x] + 60*b*x^2*(a + b*x^2)^4*Log[a + b*x^ 2])/(24*a^6*x^2*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])
Time = 0.47 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.49, 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, 54, 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.
| \(\displaystyle \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {1}{b^5 x^3 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^3 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^4 \left (b x^2+a\right )^5}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 54 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \int \left (\frac {5 b^2}{a^6 \left (b x^2+a\right )}+\frac {4 b^2}{a^5 \left (b x^2+a\right )^2}+\frac {3 b^2}{a^4 \left (b x^2+a\right )^3}+\frac {2 b^2}{a^3 \left (b x^2+a\right )^4}+\frac {b^2}{a^2 \left (b x^2+a\right )^5}-\frac {5 b}{a^6 x^2}+\frac {1}{a^5 x^4}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {5 b \log \left (x^2\right )}{a^6}+\frac {5 b \log \left (a+b x^2\right )}{a^6}-\frac {4 b}{a^5 \left (a+b x^2\right )}-\frac {1}{a^5 x^2}-\frac {3 b}{2 a^4 \left (a+b x^2\right )^2}-\frac {2 b}{3 a^3 \left (a+b x^2\right )^3}-\frac {b}{4 a^2 \left (a+b x^2\right )^4}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[1/(x^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
Output:
((a + b*x^2)*(-(1/(a^5*x^2)) - b/(4*a^2*(a + b*x^2)^4) - (2*b)/(3*a^3*(a + b*x^2)^3) - (3*b)/(2*a^4*(a + b*x^2)^2) - (4*b)/(a^5*(a + b*x^2)) - (5*b* Log[x^2])/a^6 + (5*b*Log[a + b*x^2])/a^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[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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 = 0.18 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.42
| method | result | size |
| pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (-5 b \,x^{2} \left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+5 b \,x^{2} \left (b \,x^{2}+a \right )^{4} \ln \left (x^{2}\right )+a \left (5 b^{4} x^{8}+\frac {35}{2} a \,b^{3} x^{6}+\frac {65}{3} a^{2} b^{2} x^{4}+\frac {125}{12} a^{3} b \,x^{2}+a^{4}\right )\right )}{2 \left (b \,x^{2}+a \right )^{4} x^{2} a^{6}}\) | \(112\) |
| risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {5 b^{4} x^{8}}{2 a^{5}}-\frac {35 b^{3} x^{6}}{4 a^{4}}-\frac {65 b^{2} x^{4}}{6 a^{3}}-\frac {125 b \,x^{2}}{24 a^{2}}-\frac {1}{2 a}\right )}{\left (b \,x^{2}+a \right )^{5} x^{2}}-\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{6}}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, b \ln \left (-b \,x^{2}-a \right )}{2 \left (b \,x^{2}+a \right ) a^{6}}\) | \(139\) |
| default | \(\frac {\left (60 \ln \left (b \,x^{2}+a \right ) x^{10} b^{5}-120 \ln \left (x \right ) x^{10} b^{5}+240 \ln \left (b \,x^{2}+a \right ) x^{8} a \,b^{4}-480 \ln \left (x \right ) x^{8} a \,b^{4}-60 a \,x^{8} b^{4}+360 \ln \left (b \,x^{2}+a \right ) x^{6} a^{2} b^{3}-720 \ln \left (x \right ) x^{6} a^{2} b^{3}-210 a^{2} x^{6} b^{3}+240 \ln \left (b \,x^{2}+a \right ) x^{4} a^{3} b^{2}-480 \ln \left (x \right ) x^{4} a^{3} b^{2}-260 a^{3} x^{4} b^{2}+60 \ln \left (b \,x^{2}+a \right ) x^{2} a^{4} b -120 b \,a^{4} \ln \left (x \right ) x^{2}-125 x^{2} a^{4} b -12 a^{5}\right ) \left (b \,x^{2}+a \right )}{24 x^{2} a^{6} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(219\) |
Input:
int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/2*csgn(b*x^2+a)*(-5*b*x^2*(b*x^2+a)^4*ln(b*x^2+a)+5*b*x^2*(b*x^2+a)^4*l n(x^2)+a*(5*b^4*x^8+35/2*a*b^3*x^6+65/3*a^2*b^2*x^4+125/12*a^3*b*x^2+a^4)) /(b*x^2+a)^4/x^2/a^6
Time = 0.07 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {60 \, a b^{4} x^{8} + 210 \, a^{2} b^{3} x^{6} + 260 \, a^{3} b^{2} x^{4} + 125 \, a^{4} b x^{2} + 12 \, a^{5} - 60 \, {\left (b^{5} x^{10} + 4 \, a b^{4} x^{8} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )} \log \left (b x^{2} + a\right ) + 120 \, {\left (b^{5} x^{10} + 4 \, a b^{4} x^{8} + 6 \, a^{2} b^{3} x^{6} + 4 \, a^{3} b^{2} x^{4} + a^{4} b x^{2}\right )} \log \left (x\right )}{24 \, {\left (a^{6} b^{4} x^{10} + 4 \, a^{7} b^{3} x^{8} + 6 \, a^{8} b^{2} x^{6} + 4 \, a^{9} b x^{4} + a^{10} x^{2}\right )}} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
-1/24*(60*a*b^4*x^8 + 210*a^2*b^3*x^6 + 260*a^3*b^2*x^4 + 125*a^4*b*x^2 + 12*a^5 - 60*(b^5*x^10 + 4*a*b^4*x^8 + 6*a^2*b^3*x^6 + 4*a^3*b^2*x^4 + a^4* b*x^2)*log(b*x^2 + a) + 120*(b^5*x^10 + 4*a*b^4*x^8 + 6*a^2*b^3*x^6 + 4*a^ 3*b^2*x^4 + a^4*b*x^2)*log(x))/(a^6*b^4*x^10 + 4*a^7*b^3*x^8 + 6*a^8*b^2*x ^6 + 4*a^9*b*x^4 + a^10*x^2)
\[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x**3/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(1/(x**3*((a + b*x**2)**2)**(5/2)), x)
Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {60 \, b^{4} x^{8} + 210 \, a b^{3} x^{6} + 260 \, a^{2} b^{2} x^{4} + 125 \, a^{3} b x^{2} + 12 \, a^{4}}{24 \, {\left (a^{5} b^{4} x^{10} + 4 \, a^{6} b^{3} x^{8} + 6 \, a^{7} b^{2} x^{6} + 4 \, a^{8} b x^{4} + a^{9} x^{2}\right )}} + \frac {5 \, b \log \left (b x^{2} + a\right )}{2 \, a^{6}} - \frac {5 \, b \log \left (x\right )}{a^{6}} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
-1/24*(60*b^4*x^8 + 210*a*b^3*x^6 + 260*a^2*b^2*x^4 + 125*a^3*b*x^2 + 12*a ^4)/(a^5*b^4*x^10 + 4*a^6*b^3*x^8 + 6*a^7*b^2*x^6 + 4*a^8*b*x^4 + a^9*x^2) + 5/2*b*log(b*x^2 + a)/a^6 - 5*b*log(x)/a^6
Time = 0.14 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.54 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {5 \, b \log \left (x^{2}\right )}{2 \, a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {5 \, b \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {5 \, b x^{2} - a}{2 \, a^{6} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {125 \, b^{5} x^{8} + 548 \, a b^{4} x^{6} + 912 \, a^{2} b^{3} x^{4} + 688 \, a^{3} b^{2} x^{2} + 202 \, a^{4} b}{24 \, {\left (b x^{2} + a\right )}^{4} a^{6} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
-5/2*b*log(x^2)/(a^6*sgn(b*x^2 + a)) + 5/2*b*log(abs(b*x^2 + a))/(a^6*sgn( b*x^2 + a)) + 1/2*(5*b*x^2 - a)/(a^6*x^2*sgn(b*x^2 + a)) - 1/24*(125*b^5*x ^8 + 548*a*b^4*x^6 + 912*a^2*b^3*x^4 + 688*a^3*b^2*x^2 + 202*a^4*b)/((b*x^ 2 + a)^4*a^6*sgn(b*x^2 + a))
Timed out. \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:
int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)),x)
Output:
int(1/(x^3*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)), x)
Time = 0.23 (sec) , antiderivative size = 241, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {60 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{4} b \,x^{2}+240 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b^{2} x^{4}+360 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{3} x^{6}+240 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{4} x^{8}+60 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{5} x^{10}-120 \,\mathrm {log}\left (x \right ) a^{4} b \,x^{2}-480 \,\mathrm {log}\left (x \right ) a^{3} b^{2} x^{4}-720 \,\mathrm {log}\left (x \right ) a^{2} b^{3} x^{6}-480 \,\mathrm {log}\left (x \right ) a \,b^{4} x^{8}-120 \,\mathrm {log}\left (x \right ) b^{5} x^{10}-12 a^{5}-110 a^{4} b \,x^{2}-200 a^{3} b^{2} x^{4}-120 a^{2} b^{3} x^{6}+15 b^{5} x^{10}}{24 a^{6} x^{2} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(1/x^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
(60*log(a + b*x**2)*a**4*b*x**2 + 240*log(a + b*x**2)*a**3*b**2*x**4 + 360 *log(a + b*x**2)*a**2*b**3*x**6 + 240*log(a + b*x**2)*a*b**4*x**8 + 60*log (a + b*x**2)*b**5*x**10 - 120*log(x)*a**4*b*x**2 - 480*log(x)*a**3*b**2*x* *4 - 720*log(x)*a**2*b**3*x**6 - 480*log(x)*a*b**4*x**8 - 120*log(x)*b**5* x**10 - 12*a**5 - 110*a**4*b*x**2 - 200*a**3*b**2*x**4 - 120*a**2*b**3*x** 6 + 15*b**5*x**10)/(24*a**6*x**2*(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))