Integrand size = 33, antiderivative size = 161 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {d}{2 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {b d-a e}{4 a b \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {d \left (a+b x^2\right ) \log (x)}{a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
1/2*d/a^2/((b*x^2+a)^2)^(1/2)+1/4*(-a*e+b*d)/a/b/(b*x^2+a)/((b*x^2+a)^2)^( 1/2)+d*(b*x^2+a)*ln(x)/a^3/((b*x^2+a)^2)^(1/2)-1/2*d*(b*x^2+a)*ln(b*x^2+a) /a^3/((b*x^2+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.57 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a \left (3 a b d-a^2 e+2 b^2 d x^2\right )+4 b d \left (a+b x^2\right )^2 \log (x)-2 b d \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^3 b \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]
(a*(3*a*b*d - a^2*e + 2*b^2*d*x^2) + 4*b*d*(a + b*x^2)^2*Log[x] - 2*b*d*(a + b*x^2)^2*Log[a + b*x^2])/(4*a^3*b*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.61, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {1384, 27, 354, 86, 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.
\(\displaystyle \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {e x^2+d}{b^3 x \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {e x^2+d}{x \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {e x^2+d}{x^2 \left (b x^2+a\right )^3}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (-\frac {b d}{a^3 \left (b x^2+a\right )}+\frac {d}{a^3 x^2}-\frac {b d}{a^2 \left (b x^2+a\right )^2}+\frac {a e-b d}{a \left (b x^2+a\right )^3}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {d \log \left (a+b x^2\right )}{a^3}+\frac {d \log \left (x^2\right )}{a^3}+\frac {d}{a^2 \left (a+b x^2\right )}+\frac {b d-a e}{2 a b \left (a+b x^2\right )^2}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
((a + b*x^2)*((b*d - a*e)/(2*a*b*(a + b*x^2)^2) + d/(a^2*(a + b*x^2)) + (d *Log[x^2])/a^3 - (d*Log[a + b*x^2])/a^3))/(2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^ 4])
3.1.84.3.1 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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(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 = 87, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (2 d b \left (b \,x^{2}+a \right )^{2} \ln \left (b \,x^{2}+a \right )-2 d b \left (b \,x^{2}+a \right )^{2} \ln \left (x^{2}\right )+a \left (-2 b^{2} d \,x^{2}+e \,a^{2}-3 d a b \right )\right )}{4 a^{3} b \left (b \,x^{2}+a \right )^{2}}\) | \(87\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b d \,x^{2}}{2 a^{2}}-\frac {a e -3 b d}{4 a b}\right )}{\left (b \,x^{2}+a \right )^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{3}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, d \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{3}}\) | \(111\) |
default | \(\frac {\left (4 \ln \left (x \right ) b^{3} d \,x^{4}-2 \ln \left (b \,x^{2}+a \right ) b^{3} d \,x^{4}+8 \ln \left (x \right ) a \,b^{2} d \,x^{2}-4 \ln \left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{2}+2 b^{2} d \,x^{2} a +4 \ln \left (x \right ) a^{2} b d -2 \ln \left (b \,x^{2}+a \right ) a^{2} b d -a^{3} e +3 a^{2} b d \right ) \left (b \,x^{2}+a \right )}{4 b \,a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(133\) |
-1/4*csgn(b*x^2+a)*(2*d*b*(b*x^2+a)^2*ln(b*x^2+a)-2*d*b*(b*x^2+a)^2*ln(x^2 )+a*(-2*b^2*d*x^2+a^2*e-3*a*b*d))/a^3/b/(b*x^2+a)^2
Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {2 \, a b^{2} d x^{2} + 3 \, a^{2} b d - a^{3} e - 2 \, {\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left (b^{3} d x^{4} + 2 \, a b^{2} d x^{2} + a^{2} b d\right )} \log \left (x\right )}{4 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} \]
1/4*(2*a*b^2*d*x^2 + 3*a^2*b*d - a^3*e - 2*(b^3*d*x^4 + 2*a*b^2*d*x^2 + a^ 2*b*d)*log(b*x^2 + a) + 4*(b^3*d*x^4 + 2*a*b^2*d*x^2 + a^2*b*d)*log(x))/(a ^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)
\[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {d + e x^{2}}{x \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{4} \, d {\left (\frac {2 \, b x^{2} + 3 \, a}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} - \frac {2 \, \log \left (b x^{2} + a\right )}{a^{3}} + \frac {4 \, \log \left (x\right )}{a^{3}}\right )} - \frac {e}{4 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )}} \]
1/4*d*((2*b*x^2 + 3*a)/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) - 2*log(b*x^2 + a )/a^3 + 4*log(x)/a^3) - 1/4*e/(b^3*x^4 + 2*a*b^2*x^2 + a^2*b)
Time = 0.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {d \log \left (x^{2}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {d \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b^{3} d x^{4} + 8 \, a b^{2} d x^{2} + 6 \, a^{2} b d - a^{3} e}{4 \, {\left (b x^{2} + a\right )}^{2} a^{3} b \mathrm {sgn}\left (b x^{2} + a\right )} \]
1/2*d*log(x^2)/(a^3*sgn(b*x^2 + a)) - 1/2*d*log(abs(b*x^2 + a))/(a^3*sgn(b *x^2 + a)) + 1/4*(3*b^3*d*x^4 + 8*a*b^2*d*x^2 + 6*a^2*b*d - a^3*e)/((b*x^2 + a)^2*a^3*b*sgn(b*x^2 + a))
Timed out. \[ \int \frac {d+e x^2}{x \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {e\,x^2+d}{x\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]