Integrand size = 35, antiderivative size = 90 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=-\frac {6 \left (b x^2+a x^4\right )^{3/4}}{x \left (b+a x^2\right )}+\frac {2 \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b x^2+a x^4}}\right )}{\sqrt [4]{a}} \]
-6*(a*x^4+b*x^2)^(3/4)/x/(a*x^2+b)+2*arctan(a^(1/4)*x/(a*x^4+b*x^2)^(1/4)) /a^(1/4)+2*arctanh(a^(1/4)*x/(a*x^4+b*x^2)^(1/4))/a^(1/4)
Time = 0.38 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {2 \sqrt {x} \left (-3 \sqrt [4]{a} \sqrt {x}+\sqrt [4]{b+a x^2} \arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )+\sqrt [4]{b+a x^2} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{b+a x^2}}\right )\right )}{\sqrt [4]{a} \sqrt [4]{x^2 \left (b+a x^2\right )}} \]
(2*Sqrt[x]*(-3*a^(1/4)*Sqrt[x] + (b + a*x^2)^(1/4)*ArcTan[(a^(1/4)*Sqrt[x] )/(b + a*x^2)^(1/4)] + (b + a*x^2)^(1/4)*ArcTanh[(a^(1/4)*Sqrt[x])/(b + a* x^2)^(1/4)]))/(a^(1/4)*(x^2*(b + a*x^2))^(1/4))
Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.32, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {2467, 25, 357, 266, 770, 756, 216, 219}
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.
| \(\displaystyle \int \frac {2 a x^2-b}{\left (a x^2+b\right ) \sqrt [4]{a x^4+b x^2}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt {x} \sqrt [4]{a x^2+b} \int -\frac {b-2 a x^2}{\sqrt {x} \left (a x^2+b\right )^{5/4}}dx}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \int \frac {b-2 a x^2}{\sqrt {x} \left (a x^2+b\right )^{5/4}}dx}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 357 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-2 \int \frac {1}{\sqrt {x} \sqrt [4]{a x^2+b}}dx\right )}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-4 \int \frac {1}{\sqrt [4]{a x^2+b}}d\sqrt {x}\right )}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-4 \int \frac {1}{1-a x^2}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-4 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}+\frac {1}{2} \int \frac {1}{\sqrt {a} x+1}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}\right )\right )}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-4 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {a} x}d\frac {\sqrt {x}}{\sqrt [4]{a x^2+b}}+\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [4]{a}}\right )\right )}{\sqrt [4]{a x^4+b x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\sqrt {x} \sqrt [4]{a x^2+b} \left (\frac {6 \sqrt {x}}{\sqrt [4]{a x^2+b}}-4 \left (\frac {\arctan \left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt {x}}{\sqrt [4]{a x^2+b}}\right )}{2 \sqrt [4]{a}}\right )\right )}{\sqrt [4]{a x^4+b x^2}}\) |
-((Sqrt[x]*(b + a*x^2)^(1/4)*((6*Sqrt[x])/(b + a*x^2)^(1/4) - 4*(ArcTan[(a ^(1/4)*Sqrt[x])/(b + a*x^2)^(1/4)]/(2*a^(1/4)) + ArcTanh[(a^(1/4)*Sqrt[x]) /(b + a*x^2)^(1/4)]/(2*a^(1/4)))))/(b*x^2 + a*x^4)^(1/4))
3.13.41.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_ Symbol] :> Simp[(b*c - a*d)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*b*e*(m + 1))), x] + Simp[d/b Int[(e*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Time = 2.41 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.22
| method | result | size |
| pseudoelliptic | \(-\frac {2 \left (3 a^{\frac {1}{4}} x +\frac {\left (2 \arctan \left (\frac {\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )-\ln \left (\frac {a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +\left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\right )\right ) \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}{2}\right )}{a^{\frac {1}{4}} \left (x^{2} \left (a \,x^{2}+b \right )\right )^{\frac {1}{4}}}\) | \(110\) |
-2/a^(1/4)*(3*a^(1/4)*x+1/2*(2*arctan(1/a^(1/4)/x*(x^2*(a*x^2+b))^(1/4))-l n((a^(1/4)*x+(x^2*(a*x^2+b))^(1/4))/(-a^(1/4)*x+(x^2*(a*x^2+b))^(1/4))))*( x^2*(a*x^2+b))^(1/4))/(x^2*(a*x^2+b))^(1/4)
Timed out. \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\text {Timed out} \]
\[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int \frac {2 a x^{2} - b}{\sqrt [4]{x^{2} \left (a x^{2} + b\right )} \left (a x^{2} + b\right )}\, dx \]
\[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int { \frac {2 \, a x^{2} - b}{{\left (a x^{4} + b x^{2}\right )}^{\frac {1}{4}} {\left (a x^{2} + b\right )}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (76) = 152\).
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.17 \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{a} - \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, a} + \frac {\sqrt {2} \left (-a\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a + \frac {b}{x^{2}}}\right )}{2 \, a} - \frac {6}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {1}{4}}} \]
sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x^2)^ (1/4))/(-a)^(1/4))/a + sqrt(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-a )^(1/4) - 2*(a + b/x^2)^(1/4))/(-a)^(1/4))/a - 1/2*sqrt(2)*(-a)^(3/4)*log( sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a + 1/2 *sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^2)^(1/4) + sqrt(-a) + sqrt(a + b/x^2))/a - 6/(a + b/x^2)^(1/4)
Timed out. \[ \int \frac {-b+2 a x^2}{\left (b+a x^2\right ) \sqrt [4]{b x^2+a x^4}} \, dx=\int -\frac {b-2\,a\,x^2}{\left (a\,x^2+b\right )\,{\left (a\,x^4+b\,x^2\right )}^{1/4}} \,d x \]