Integrand size = 33, antiderivative size = 223 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 b d-a e}{2 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b d-a e}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {d \left (a+b x^2\right )}{2 a^3 x^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {(3 b d-a e) \left (a+b x^2\right ) \log (x)}{a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {(3 b d-a e) \left (a+b x^2\right ) \log \left (a+b x^2\right )}{2 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
1/2*(a*e-2*b*d)/a^3/((b*x^2+a)^2)^(1/2)+1/4*(a*e-b*d)/a^2/(b*x^2+a)/((b*x^ 2+a)^2)^(1/2)-1/2*d*(b*x^2+a)/a^3/x^2/((b*x^2+a)^2)^(1/2)-(-a*e+3*b*d)*(b* x^2+a)*ln(x)/a^4/((b*x^2+a)^2)^(1/2)+1/2*(-a*e+3*b*d)*(b*x^2+a)*ln(b*x^2+a )/a^4/((b*x^2+a)^2)^(1/2)
Time = 1.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.58 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {a \left (-6 b^2 d x^4+a^2 \left (-2 d+3 e x^2\right )+a b \left (-9 d x^2+2 e x^4\right )\right )+4 (-3 b d+a e) x^2 \left (a+b x^2\right )^2 \log (x)+2 (3 b d-a e) x^2 \left (a+b x^2\right )^2 \log \left (a+b x^2\right )}{4 a^4 x^2 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]
(a*(-6*b^2*d*x^4 + a^2*(-2*d + 3*e*x^2) + a*b*(-9*d*x^2 + 2*e*x^4)) + 4*(- 3*b*d + a*e)*x^2*(a + b*x^2)^2*Log[x] + 2*(3*b*d - a*e)*x^2*(a + b*x^2)^2* Log[a + b*x^2])/(4*a^4*x^2*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.58, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, 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.
\(\displaystyle \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {e x^2+d}{b^3 x^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {e x^2+d}{x^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {e x^2+d}{x^4 \left (b x^2+a\right )^3}dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \left (\frac {d}{a^3 x^4}-\frac {b (a e-3 b d)}{a^4 \left (b x^2+a\right )}+\frac {a e-3 b d}{a^4 x^2}-\frac {b (a e-2 b d)}{a^3 \left (b x^2+a\right )^2}-\frac {b (a e-b d)}{a^2 \left (b x^2+a\right )^3}\right )dx^2}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (-\frac {\log \left (x^2\right ) (3 b d-a e)}{a^4}+\frac {(3 b d-a e) \log \left (a+b x^2\right )}{a^4}-\frac {2 b d-a e}{a^3 \left (a+b x^2\right )}-\frac {d}{a^3 x^2}-\frac {b d-a e}{2 a^2 \left (a+b x^2\right )^2}\right )}{2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
((a + b*x^2)*(-(d/(a^3*x^2)) - (b*d - a*e)/(2*a^2*(a + b*x^2)^2) - (2*b*d - a*e)/(a^3*(a + b*x^2)) - ((3*b*d - a*e)*Log[x^2])/a^4 + ((3*b*d - a*e)*L og[a + b*x^2])/a^4))/(2*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])
3.1.86.3.1 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 = 0.14 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(-\frac {\operatorname {csgn}\left (b \,x^{2}+a \right ) \left (x^{2} \left (b \,x^{2}+a \right )^{2} \left (a e -3 b d \right ) \ln \left (b \,x^{2}+a \right )-x^{2} \left (b \,x^{2}+a \right )^{2} \left (a e -3 b d \right ) \ln \left (x^{2}\right )+a \left (\left (-a b e +3 b^{2} d \right ) x^{4}-\frac {3 a \left (a e -3 b d \right ) x^{2}}{2}+d \,a^{2}\right )\right )}{2 \left (b \,x^{2}+a \right )^{2} x^{2} a^{4}}\) | \(120\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {b \left (a e -3 b d \right ) x^{4}}{2 a^{3}}+\frac {3 \left (a e -3 b d \right ) x^{2}}{4 a^{2}}-\frac {d}{2 a}\right )}{\left (b \,x^{2}+a \right )^{3} x^{2}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -3 b d \right ) \ln \left (x \right )}{\left (b \,x^{2}+a \right ) a^{4}}-\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (a e -3 b d \right ) \ln \left (b \,x^{2}+a \right )}{2 \left (b \,x^{2}+a \right ) a^{4}}\) | \(141\) |
default | \(\frac {\left (4 \ln \left (x \right ) a \,b^{2} e \,x^{6}-12 \ln \left (x \right ) b^{3} d \,x^{6}-2 \ln \left (b \,x^{2}+a \right ) a \,b^{2} e \,x^{6}+6 \ln \left (b \,x^{2}+a \right ) b^{3} d \,x^{6}+8 \ln \left (x \right ) a^{2} b e \,x^{4}-24 \ln \left (x \right ) a \,b^{2} d \,x^{4}-4 \ln \left (b \,x^{2}+a \right ) a^{2} b e \,x^{4}+12 \ln \left (b \,x^{2}+a \right ) a \,b^{2} d \,x^{4}+2 a^{2} b e \,x^{4}-6 a \,b^{2} d \,x^{4}+4 \ln \left (x \right ) a^{3} e \,x^{2}-12 \ln \left (x \right ) a^{2} b d \,x^{2}-2 \ln \left (b \,x^{2}+a \right ) a^{3} e \,x^{2}+6 \ln \left (b \,x^{2}+a \right ) a^{2} b d \,x^{2}+3 a^{3} e \,x^{2}-9 a^{2} b d \,x^{2}-2 a^{3} d \right ) \left (b \,x^{2}+a \right )}{4 x^{2} a^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(249\) |
-1/2*csgn(b*x^2+a)*(x^2*(b*x^2+a)^2*(a*e-3*b*d)*ln(b*x^2+a)-x^2*(b*x^2+a)^ 2*(a*e-3*b*d)*ln(x^2)+a*((-a*b*e+3*b^2*d)*x^4-3/2*a*(a*e-3*b*d)*x^2+d*a^2) )/(b*x^2+a)^2/x^2/a^4
Time = 0.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.92 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + 2 \, a^{3} d + 3 \, {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2} - 2 \, {\left ({\left (3 \, b^{3} d - a b^{2} e\right )} x^{6} + 2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2}\right )} \log \left (b x^{2} + a\right ) + 4 \, {\left ({\left (3 \, b^{3} d - a b^{2} e\right )} x^{6} + 2 \, {\left (3 \, a b^{2} d - a^{2} b e\right )} x^{4} + {\left (3 \, a^{2} b d - a^{3} e\right )} x^{2}\right )} \log \left (x\right )}{4 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}} \]
-1/4*(2*(3*a*b^2*d - a^2*b*e)*x^4 + 2*a^3*d + 3*(3*a^2*b*d - a^3*e)*x^2 - 2*((3*b^3*d - a*b^2*e)*x^6 + 2*(3*a*b^2*d - a^2*b*e)*x^4 + (3*a^2*b*d - a^ 3*e)*x^2)*log(b*x^2 + a) + 4*((3*b^3*d - a*b^2*e)*x^6 + 2*(3*a*b^2*d - a^2 *b*e)*x^4 + (3*a^2*b*d - a^3*e)*x^2)*log(x))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2)
\[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {d + e x^{2}}{x^{3} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.62 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {1}{4} \, d {\left (\frac {6 \, b^{2} x^{4} + 9 \, a b x^{2} + 2 \, a^{2}}{a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{4} + a^{5} x^{2}} - \frac {6 \, b \log \left (b x^{2} + a\right )}{a^{4}} + \frac {12 \, b \log \left (x\right )}{a^{4}}\right )} + \frac {1}{4} \, e {\left (\frac {2 \, b x^{2} + 3 \, a}{a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}} - \frac {2 \, \log \left (b x^{2} + a\right )}{a^{3}} + \frac {4 \, \log \left (x\right )}{a^{3}}\right )} \]
-1/4*d*((6*b^2*x^4 + 9*a*b*x^2 + 2*a^2)/(a^3*b^2*x^6 + 2*a^4*b*x^4 + a^5*x ^2) - 6*b*log(b*x^2 + a)/a^4 + 12*b*log(x)/a^4) + 1/4*e*((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)
Time = 0.27 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.82 \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {{\left (3 \, b d - a e\right )} \log \left (x^{2}\right )}{2 \, a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {{\left (3 \, b^{2} d - a b e\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, a^{4} b \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {9 \, b^{3} d x^{4} - 3 \, a b^{2} e x^{4} + 22 \, a b^{2} d x^{2} - 8 \, a^{2} b e x^{2} + 14 \, a^{2} b d - 6 \, a^{3} e}{4 \, {\left (b x^{2} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b d x^{2} - a e x^{2} - a d}{2 \, a^{4} x^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \]
-1/2*(3*b*d - a*e)*log(x^2)/(a^4*sgn(b*x^2 + a)) + 1/2*(3*b^2*d - a*b*e)*l og(abs(b*x^2 + a))/(a^4*b*sgn(b*x^2 + a)) - 1/4*(9*b^3*d*x^4 - 3*a*b^2*e*x ^4 + 22*a*b^2*d*x^2 - 8*a^2*b*e*x^2 + 14*a^2*b*d - 6*a^3*e)/((b*x^2 + a)^2 *a^4*sgn(b*x^2 + a)) + 1/2*(3*b*d*x^2 - a*e*x^2 - a*d)/(a^4*x^2*sgn(b*x^2 + a))
Timed out. \[ \int \frac {d+e x^2}{x^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {e\,x^2+d}{x^3\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]