Integrand size = 22, antiderivative size = 36 \[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=-\frac {x \sqrt {1-x^4}}{2 \left (1+x^2\right )}-\frac {1}{2} E(\arcsin (x)|-1)+\operatorname {EllipticF}(\arcsin (x),-1) \] Output:
-1/2*x*(-x^4+1)^(1/2)/(x^2+1)-1/2*EllipticE(x,I)+EllipticF(x,I)
Time = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\frac {1}{2} \left (-\frac {x}{\sqrt {1-x^4}}+\frac {x^3}{\sqrt {1-x^4}}-E(\arcsin (x)|-1)+2 \operatorname {EllipticF}(\arcsin (x),-1)\right ) \] Input:
Integrate[x^2/((1 + x^2)*Sqrt[1 - x^4]),x]
Output:
(-(x/Sqrt[1 - x^4]) + x^3/Sqrt[1 - x^4] - EllipticE[ArcSin[x], -1] + 2*Ell ipticF[ArcSin[x], -1])/2
Time = 0.21 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1388, 373, 326, 284, 327, 762}
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.
\(\displaystyle \int \frac {x^2}{\left (x^2+1\right ) \sqrt {1-x^4}} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \frac {x^2}{\sqrt {1-x^2} \left (x^2+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 373 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {1-x^2}}{\sqrt {x^2+1}}dx-\frac {x \sqrt {1-x^2}}{2 \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 326 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{\sqrt {1-x^2} \sqrt {x^2+1}}dx-\int \frac {\sqrt {x^2+1}}{\sqrt {1-x^2}}dx\right )-\frac {x \sqrt {1-x^2}}{2 \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 284 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{\sqrt {1-x^4}}dx-\int \frac {\sqrt {x^2+1}}{\sqrt {1-x^2}}dx\right )-\frac {x \sqrt {1-x^2}}{2 \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {1}{\sqrt {1-x^4}}dx-E(\arcsin (x)|-1)\right )-\frac {x \sqrt {1-x^2}}{2 \sqrt {x^2+1}}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {1}{2} (2 \operatorname {EllipticF}(\arcsin (x),-1)-E(\arcsin (x)|-1))-\frac {x \sqrt {1-x^2}}{2 \sqrt {x^2+1}}\) |
Input:
Int[x^2/((1 + x^2)*Sqrt[1 - x^4]),x]
Output:
-1/2*(x*Sqrt[1 - x^2])/Sqrt[1 + x^2] + (-EllipticE[ArcSin[x], -1] + 2*Elli pticF[ArcSin[x], -1])/2
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> I nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ b/d Int[Sqrt[c + d*x^2]/Sqrt[a + b*x^2], x], x] - Simp[(b*c - a*d)/d In t[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && PosQ[d/c] && NegQ[b/a]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1)) Int[(e *x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (30 ) = 60\).
Time = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.44
method | result | size |
risch | \(\frac {x \left (x^{2}-1\right )}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (x , i\right )}{2 \sqrt {-x^{4}+1}}\) | \(88\) |
default | \(\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (x , i\right )}{2 \sqrt {-x^{4}+1}}-\frac {\left (-x^{2}+1\right ) x}{2 \sqrt {\left (x^{2}+1\right ) \left (-x^{2}+1\right )}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\) | \(96\) |
elliptic | \(\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \operatorname {EllipticF}\left (x , i\right )}{2 \sqrt {-x^{4}+1}}-\frac {\left (-x^{2}+1\right ) x}{2 \sqrt {\left (x^{2}+1\right ) \left (-x^{2}+1\right )}}+\frac {\sqrt {-x^{2}+1}\, \sqrt {x^{2}+1}\, \left (\operatorname {EllipticF}\left (x , i\right )-\operatorname {EllipticE}\left (x , i\right )\right )}{2 \sqrt {-x^{4}+1}}\) | \(96\) |
Input:
int(x^2/(x^2+1)/(-x^4+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*x*(x^2-1)/(-x^4+1)^(1/2)+1/2*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+1)^(1/ 2)*(EllipticF(x,I)-EllipticE(x,I))+1/2*(-x^2+1)^(1/2)*(x^2+1)^(1/2)/(-x^4+ 1)^(1/2)*EllipticF(x,I)
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.17 \[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=-\frac {{\left (x^{2} + 1\right )} E(\arcsin \left (x\right )\,|\,-1) - 2 \, {\left (x^{2} + 1\right )} F(\arcsin \left (x\right )\,|\,-1) + \sqrt {-x^{4} + 1} x}{2 \, {\left (x^{2} + 1\right )}} \] Input:
integrate(x^2/(x^2+1)/(-x^4+1)^(1/2),x, algorithm="fricas")
Output:
-1/2*((x^2 + 1)*elliptic_e(arcsin(x), -1) - 2*(x^2 + 1)*elliptic_f(arcsin( x), -1) + sqrt(-x^4 + 1)*x)/(x^2 + 1)
\[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\int \frac {x^{2}}{\sqrt {- \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )} \left (x^{2} + 1\right )}\, dx \] Input:
integrate(x**2/(x**2+1)/(-x**4+1)**(1/2),x)
Output:
Integral(x**2/(sqrt(-(x - 1)*(x + 1)*(x**2 + 1))*(x**2 + 1)), x)
\[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {-x^{4} + 1} {\left (x^{2} + 1\right )}} \,d x } \] Input:
integrate(x^2/(x^2+1)/(-x^4+1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^2/(sqrt(-x^4 + 1)*(x^2 + 1)), x)
\[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {-x^{4} + 1} {\left (x^{2} + 1\right )}} \,d x } \] Input:
integrate(x^2/(x^2+1)/(-x^4+1)^(1/2),x, algorithm="giac")
Output:
integrate(x^2/(sqrt(-x^4 + 1)*(x^2 + 1)), x)
Timed out. \[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\int \frac {x^2}{\left (x^2+1\right )\,\sqrt {1-x^4}} \,d x \] Input:
int(x^2/((x^2 + 1)*(1 - x^4)^(1/2)),x)
Output:
int(x^2/((x^2 + 1)*(1 - x^4)^(1/2)), x)
\[ \int \frac {x^2}{\left (1+x^2\right ) \sqrt {1-x^4}} \, dx=\frac {-\sqrt {-x^{4}+1}\, x +\left (\int \frac {\sqrt {-x^{4}+1}}{x^{2}+1}d x \right ) x^{2}+\int \frac {\sqrt {-x^{4}+1}}{x^{2}+1}d x}{2 x^{2}+2} \] Input:
int(x^2/(x^2+1)/(-x^4+1)^(1/2),x)
Output:
( - sqrt( - x**4 + 1)*x + int(sqrt( - x**4 + 1)/(x**2 + 1),x)*x**2 + int(s qrt( - x**4 + 1)/(x**2 + 1),x))/(2*(x**2 + 1))