Integrand size = 26, antiderivative size = 147 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (a+b x^2\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
1/2/a^2/((b*x^2+a)^2)^(1/2)+1/4/a/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+(b*x^2+a)* ln(x)/a^3/((b*x^2+a)^2)^(1/2)-1/2*(b*x^2+a)*ln(b*x^2+a)/a^3/((b*x^2+a)^2)^ (1/2)
Leaf count is larger than twice the leaf count of optimal. \(790\) vs. \(2(147)=294\).
Time = 1.18 (sec) , antiderivative size = 790, normalized size of antiderivative = 5.37 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {4 a^4 b x^2+3 a^3 b^2 x^4-a b^4 x^8-4 \left (a^2\right )^{3/2} b x^2 \sqrt {\left (a+b x^2\right )^2}+a \sqrt {a^2} b^2 x^4 \sqrt {\left (a+b x^2\right )^2}-\sqrt {a^2} b^3 x^6 \sqrt {\left (a+b x^2\right )^2}+2 \left (\left (a^2\right )^{3/2} b^2 x^4+a^4 \left (\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )+a^3 b x^2 \left (2 \sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}\right )\right ) \text {arctanh}\left (\frac {b x^2}{\sqrt {a^2}-\sqrt {\left (a+b x^2\right )^2}}\right )-2 \left (a^5+2 a^4 b x^2-\left (a^2\right )^{3/2} b x^2 \sqrt {\left (a+b x^2\right )^2}+a^3 \left (b^2 x^4-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )\right ) \log \left (x^2\right )+a^5 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+2 a^4 b x^2 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a^3 b^2 x^4 \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-a^3 \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-\left (a^2\right )^{3/2} b x^2 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}-b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a^5 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+2 a^4 b x^2 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )+a^3 b^2 x^4 \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-a^3 \sqrt {a^2} \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )-\left (a^2\right )^{3/2} b x^2 \sqrt {\left (a+b x^2\right )^2} \log \left (\sqrt {a^2}+b x^2-\sqrt {\left (a+b x^2\right )^2}\right )}{2 a^3 \sqrt {a^2} \left (a^2+a b x^2-\sqrt {a^2} \sqrt {\left (a+b x^2\right )^2}\right )^2} \]
(4*a^4*b*x^2 + 3*a^3*b^2*x^4 - a*b^4*x^8 - 4*(a^2)^(3/2)*b*x^2*Sqrt[(a + b *x^2)^2] + a*Sqrt[a^2]*b^2*x^4*Sqrt[(a + b*x^2)^2] - Sqrt[a^2]*b^3*x^6*Sqr t[(a + b*x^2)^2] + 2*((a^2)^(3/2)*b^2*x^4 + a^4*(Sqrt[a^2] - Sqrt[(a + b*x ^2)^2]) + a^3*b*x^2*(2*Sqrt[a^2] - Sqrt[(a + b*x^2)^2]))*ArcTanh[(b*x^2)/( Sqrt[a^2] - Sqrt[(a + b*x^2)^2])] - 2*(a^5 + 2*a^4*b*x^2 - (a^2)^(3/2)*b*x ^2*Sqrt[(a + b*x^2)^2] + a^3*(b^2*x^4 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2]))*Lo g[x^2] + a^5*Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + 2*a^4*b*x^2*Lo g[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + a^3*b^2*x^4*Log[Sqrt[a^2] - b *x^2 - Sqrt[(a + b*x^2)^2]] - a^3*Sqrt[a^2]*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a ^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] - (a^2)^(3/2)*b*x^2*Sqrt[(a + b*x^2)^2] *Log[Sqrt[a^2] - b*x^2 - Sqrt[(a + b*x^2)^2]] + a^5*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] + 2*a^4*b*x^2*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x ^2)^2]] + a^3*b^2*x^4*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] - a^3*S qrt[a^2]*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]] - (a^2)^(3/2)*b*x^2*Sqrt[(a + b*x^2)^2]*Log[Sqrt[a^2] + b*x^2 - Sqrt[(a + b*x^2)^2]])/(2*a^3*Sqrt[a^2]*(a^2 + a*b*x^2 - Sqrt[a^2]*Sqrt[(a + b*x^2)^2 ])^2)
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57, 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.
\(\displaystyle \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {1}{b^3 x \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^2 \left (b x^2+a\right )^3}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (-\frac {b}{a^3 \left (b x^2+a\right )}-\frac {b}{a^2 \left (b x^2+a\right )^2}-\frac {b}{a \left (b x^2+a\right )^3}+\frac {1}{a^3 x^2}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {\log \left (a+b x^2\right )}{a^3}+\frac {\log \left (x^2\right )}{a^3}+\frac {1}{a^2 \left (a+b x^2\right )}+\frac {1}{2 a \left (a+b x^2\right )^2}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
((a + b*x^2)*(1/(2*a*(a + b*x^2)^2) + 1/(a^2*(a + b*x^2)) + Log[x^2]/a^3 - Log[a + b*x^2]/a^3))/(2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
3.7.38.3.1 Defintions of rubi rules used
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 = 0.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.46
method | result | size |
pseudoelliptic | \(\frac {\left (-\left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )+\left (b \,x^{2}+a \right )^{2} \ln \left (x^{2}\right )+a b \,x^{2}+\frac {3 a^{2}}{2}\right ) \operatorname {csgn}\left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right )^{2} a^{3}}\) | \(68\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \,x^{2}}{2 a^{2}}+\frac {3}{4 a}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\) | \(97\) |
default | \(\frac {\left (4 b^{2} \ln \left (x \right ) x^{4}-2 \ln \left (b \,x^{2}+a \right ) x^{4} b^{2}+8 a b \ln \left (x \right ) x^{2}-4 \ln \left (b \,x^{2}+a \right ) x^{2} a b +2 a b \,x^{2}+4 a^{2} \ln \left (x \right )-2 \ln \left (b \,x^{2}+a \right ) a^{2}+3 a^{2}\right ) \left (b \,x^{2}+a \right )}{4 a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(107\) |
1/2*(-(b*x^2+a)^2*ln(b*x^2+a)+(b*x^2+a)^2*ln(x^2)+a*b*x^2+3/2*a^2)*csgn(b* x^2+a)/(b*x^2+a)^2/a^3
Time = 0.27 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, a b x^{2} + 3 \, a^{2} - 2 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{2} x^{4} + 2 \, a^{4} b x^{2} + a^{5}\right )}} \]
1/4*(2*a*b*x^2 + 3*a^2 - 2*(b^2*x^4 + 2*a*b*x^2 + a^2)*log(b*x^2 + a) + 4* (b^2*x^4 + 2*a*b*x^2 + a^2)*log(x))/(a^3*b^2*x^4 + 2*a^4*b*x^2 + a^5)
\[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.39 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, b x^{2} + 3 \, a}{4 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} - \frac {\log \left (b x^{2} + a\right )}{2 \, a^{3}} + \frac {\log \left (x\right )}{a^{3}} \]
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.61 \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\log \left (x^{2}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {\log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b^{2} x^{4} + 8 \, a b x^{2} + 6 \, a^{2}}{4 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]
1/2*log(x^2)/(a^3*sgn(b*x^2 + a)) - 1/2*log(abs(b*x^2 + a))/(a^3*sgn(b*x^2 + a)) + 1/4*(3*b^2*x^4 + 8*a*b*x^2 + 6*a^2)/((b*x^2 + a)^2*a^3*sgn(b*x^2 + a))
Timed out. \[ \int \frac {1}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]