Integrand size = 26, antiderivative size = 223 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {1}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{6 a^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log (x)}{a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
1/2/a^4/((b*x^2+a)^2)^(1/2)+1/8/a/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)+1/6/a^2/ (b*x^2+a)^2/((b*x^2+a)^2)^(1/2)+1/4/a^3/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+(b*x ^2+a)*ln(x)/a^5/((b*x^2+a)^2)^(1/2)-1/2*(b*x^2+a)*ln(b*x^2+a)/a^5/((b*x^2+ a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {a \left (25 a^3+52 a^2 b x^2+42 a b^2 x^4+12 b^3 x^6\right )+24 \left (a+b x^2\right )^4 \log (x)-12 \left (a+b x^2\right )^4 \log \left (a+b x^2\right )}{24 a^5 \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
Output:
(a*(25*a^3 + 52*a^2*b*x^2 + 42*a*b^2*x^4 + 12*b^3*x^6) + 24*(a + b*x^2)^4* Log[x] - 12*(a + b*x^2)^4*Log[a + b*x^2])/(24*a^5*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])
Time = 0.45 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.52, 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.
\(\displaystyle \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {1}{b^5 x \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^2 \left (b x^2+a\right )^5}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (-\frac {b}{a^5 \left (b x^2+a\right )}-\frac {b}{a^4 \left (b x^2+a\right )^2}-\frac {b}{a^3 \left (b x^2+a\right )^3}-\frac {b}{a^2 \left (b x^2+a\right )^4}-\frac {b}{a \left (b x^2+a\right )^5}+\frac {1}{a^5 x^2}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {\log \left (a+b x^2\right )}{a^5}+\frac {\log \left (x^2\right )}{a^5}+\frac {1}{a^4 \left (a+b x^2\right )}+\frac {1}{2 a^3 \left (a+b x^2\right )^2}+\frac {1}{3 a^2 \left (a+b x^2\right )^3}+\frac {1}{4 a \left (a+b x^2\right )^4}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[1/(x*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
Output:
((a + b*x^2)*(1/(4*a*(a + b*x^2)^4) + 1/(3*a^2*(a + b*x^2)^3) + 1/(2*a^3*( a + b*x^2)^2) + 1/(a^4*(a + b*x^2)) + Log[x^2]/a^5 - Log[a + b*x^2]/a^5))/ (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 = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.40
method | result | size |
pseudoelliptic | \(\frac {\left (-\left (b \,x^{2}+a \right )^{4} \ln \left (b \,x^{2}+a \right )+\left (b \,x^{2}+a \right )^{4} \ln \left (x^{2}\right )+a \,b^{3} x^{6}+\frac {7 a^{2} b^{2} x^{4}}{2}+\frac {13 a^{3} b \,x^{2}}{3}+\frac {25 a^{4}}{12}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right )^{4} a^{5}}\) | \(90\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b^{3} x^{6}}{2 a^{4}}+\frac {7 b^{2} x^{4}}{4 a^{3}}+\frac {13 b \,x^{2}}{6 a^{2}}+\frac {25}{24 a}\right )}{\left (b \,x^{2}+a \right )^{5}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{5}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{5}}\) | \(119\) |
default | \(-\frac {\left (12 \ln \left (b \,x^{2}+a \right ) b^{4} x^{8}-24 \ln \left (x \right ) b^{4} x^{8}+48 \ln \left (b \,x^{2}+a \right ) a \,b^{3} x^{6}-96 \ln \left (x \right ) a \,b^{3} x^{6}-12 a \,b^{3} x^{6}+72 \ln \left (b \,x^{2}+a \right ) a^{2} b^{2} x^{4}-144 \ln \left (x \right ) a^{2} b^{2} x^{4}-42 a^{2} b^{2} x^{4}+48 \ln \left (b \,x^{2}+a \right ) x^{2} a^{3} b -96 \ln \left (x \right ) a^{3} b \,x^{2}-52 a^{3} b \,x^{2}+12 \ln \left (b \,x^{2}+a \right ) a^{4}-24 a^{4} \ln \left (x \right )-25 a^{4}\right ) \left (b \,x^{2}+a \right )}{24 a^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(193\) |
Input:
int(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/2*(-(b*x^2+a)^4*ln(b*x^2+a)+(b*x^2+a)^4*ln(x^2)+a*b^3*x^6+7/2*a^2*b^2*x^ 4+13/3*a^3*b*x^2+25/12*a^4)*csgn(b*x^2+a)/(b*x^2+a)^4/a^5
Time = 0.08 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.80 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, a b^{3} x^{6} + 42 \, a^{2} b^{2} x^{4} + 52 \, a^{3} b x^{2} + 25 \, a^{4} - 12 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right ) + 24 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \log \left (x\right )}{24 \, {\left (a^{5} b^{4} x^{8} + 4 \, a^{6} b^{3} x^{6} + 6 \, a^{7} b^{2} x^{4} + 4 \, a^{8} b x^{2} + a^{9}\right )}} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
Output:
1/24*(12*a*b^3*x^6 + 42*a^2*b^2*x^4 + 52*a^3*b*x^2 + 25*a^4 - 12*(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)*log(b*x^2 + a) + 24*(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)*log(x))/(a^5*b^4* x^8 + 4*a^6*b^3*x^6 + 6*a^7*b^2*x^4 + 4*a^8*b*x^2 + a^9)
\[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/x/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
Output:
Integral(1/(x*((a + b*x**2)**2)**(5/2)), x)
Time = 0.04 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {12 \, b^{3} x^{6} + 42 \, a b^{2} x^{4} + 52 \, a^{2} b x^{2} + 25 \, a^{3}}{24 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{5}} + \frac {\log \left (x\right )}{a^{5}} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
Output:
1/24*(12*b^3*x^6 + 42*a*b^2*x^4 + 52*a^2*b*x^2 + 25*a^3)/(a^4*b^4*x^8 + 4* a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 + a^8) - 1/2*log(b*x^2 + a)/a^5 + log(x)/a^5
Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\log \left (x^{2}\right )}{2 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {25 \, b^{4} x^{8} + 112 \, a b^{3} x^{6} + 192 \, a^{2} b^{2} x^{4} + 152 \, a^{3} b x^{2} + 50 \, a^{4}}{24 \, {\left (b x^{2} + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
Output:
1/2*log(x^2)/(a^5*sgn(b*x^2 + a)) - 1/2*log(abs(b*x^2 + a))/(a^5*sgn(b*x^2 + a)) + 1/24*(25*b^4*x^8 + 112*a*b^3*x^6 + 192*a^2*b^2*x^4 + 152*a^3*b*x^ 2 + 50*a^4)/((b*x^2 + a)^4*a^5*sgn(b*x^2 + a))
Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:
int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)),x)
Output:
int(1/(x*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)), x)
Time = 0.23 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-12 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{4}-48 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b \,x^{2}-72 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{2} x^{4}-48 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{3} x^{6}-12 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{4} x^{8}+24 \,\mathrm {log}\left (x \right ) a^{4}+96 \,\mathrm {log}\left (x \right ) a^{3} b \,x^{2}+144 \,\mathrm {log}\left (x \right ) a^{2} b^{2} x^{4}+96 \,\mathrm {log}\left (x \right ) a \,b^{3} x^{6}+24 \,\mathrm {log}\left (x \right ) b^{4} x^{8}+22 a^{4}+40 a^{3} b \,x^{2}+24 a^{2} b^{2} x^{4}-3 b^{4} x^{8}}{24 a^{5} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:
int(1/x/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
Output:
( - 12*log(a + b*x**2)*a**4 - 48*log(a + b*x**2)*a**3*b*x**2 - 72*log(a + b*x**2)*a**2*b**2*x**4 - 48*log(a + b*x**2)*a*b**3*x**6 - 12*log(a + b*x** 2)*b**4*x**8 + 24*log(x)*a**4 + 96*log(x)*a**3*b*x**2 + 144*log(x)*a**2*b* *2*x**4 + 96*log(x)*a*b**3*x**6 + 24*log(x)*b**4*x**8 + 22*a**4 + 40*a**3* b*x**2 + 24*a**2*b**2*x**4 - 3*b**4*x**8)/(24*a**5*(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))