Integrand size = 13, antiderivative size = 140 \[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=\frac {1}{2 x \sqrt {1+x^4}}-\frac {3 \sqrt {1+x^4}}{2 x}+\frac {3 x \sqrt {1+x^4}}{2 \left (1+x^2\right )}-\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{2 \sqrt {1+x^4}}+\frac {3 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{4 \sqrt {1+x^4}} \] Output:
1/2/x/(x^4+1)^(1/2)-3/2*(x^4+1)^(1/2)/x+3*x*(x^4+1)^(1/2)/(2*x^2+2)-3/2*(x ^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*EllipticE(sin(2*arctan(x)),1/2*2^(1/2))/(x ^4+1)^(1/2)+3/4*(x^2+1)*((x^4+1)/(x^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan (x),1/2*2^(1/2))/(x^4+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {3}{2},\frac {3}{4},-x^4\right )}{x} \] Input:
Integrate[1/(x^2*(1 + x^4)^(3/2)),x]
Output:
-(Hypergeometric2F1[-1/4, 3/2, 3/4, -x^4]/x)
Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {819, 847, 834, 761, 1510}
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.
\(\displaystyle \int \frac {1}{x^2 \left (x^4+1\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 819 |
\(\displaystyle \frac {3}{2} \int \frac {1}{x^2 \sqrt {x^4+1}}dx+\frac {1}{2 x \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {3}{2} \left (\int \frac {x^2}{\sqrt {x^4+1}}dx-\frac {\sqrt {x^4+1}}{x}\right )+\frac {1}{2 x \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {3}{2} \left (\int \frac {1}{\sqrt {x^4+1}}dx-\int \frac {1-x^2}{\sqrt {x^4+1}}dx-\frac {\sqrt {x^4+1}}{x}\right )+\frac {1}{2 x \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {3}{2} \left (-\int \frac {1-x^2}{\sqrt {x^4+1}}dx+\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\sqrt {x^4+1}}{x}\right )+\frac {1}{2 x \sqrt {x^4+1}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {3}{2} \left (\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (x),\frac {1}{2}\right )}{2 \sqrt {x^4+1}}-\frac {\left (x^2+1\right ) \sqrt {\frac {x^4+1}{\left (x^2+1\right )^2}} E\left (2 \arctan (x)\left |\frac {1}{2}\right .\right )}{\sqrt {x^4+1}}-\frac {\sqrt {x^4+1}}{x}+\frac {\sqrt {x^4+1} x}{x^2+1}\right )+\frac {1}{2 x \sqrt {x^4+1}}\) |
Input:
Int[1/(x^2*(1 + x^4)^(3/2)),x]
Output:
1/(2*x*Sqrt[1 + x^4]) + (3*(-(Sqrt[1 + x^4]/x) + (x*Sqrt[1 + x^4])/(1 + x^ 2) - ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticE[2*ArcTan[x], 1/2])/S qrt[1 + x^4] + ((1 + x^2)*Sqrt[(1 + x^4)/(1 + x^2)^2]*EllipticF[2*ArcTan[x ], 1/2])/(2*Sqrt[1 + x^4])))/2
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-( c*x)^(m + 1))*((a + b*x^n)^(p + 1)/(a*c*n*(p + 1))), x] + Simp[(m + n*(p + 1) + 1)/(a*n*(p + 1)) Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a , b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Time = 0.54 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.12
method | result | size |
meijerg | \(-\frac {\operatorname {hypergeom}\left (\left [-\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {3}{4}\right ], -x^{4}\right )}{x}\) | \(17\) |
risch | \(-\frac {3 x^{4}+2}{2 x \sqrt {x^{4}+1}}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(102\) |
default | \(-\frac {\sqrt {x^{4}+1}}{x}-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
elliptic | \(-\frac {\sqrt {x^{4}+1}}{x}-\frac {x^{3}}{2 \sqrt {x^{4}+1}}+\frac {3 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\operatorname {EllipticE}\left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}\) | \(107\) |
Input:
int(1/x^2/(x^4+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/x*hypergeom([-1/4,3/2],[3/4],-x^4)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=-\frac {3 \, \sqrt {i} {\left (i \, x^{5} + i \, x\right )} E(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + 3 \, \sqrt {i} {\left (-i \, x^{5} - i \, x\right )} F(\arcsin \left (\sqrt {i} x\right )\,|\,-1) + {\left (3 \, x^{4} + 2\right )} \sqrt {x^{4} + 1}}{2 \, {\left (x^{5} + x\right )}} \] Input:
integrate(1/x^2/(x^4+1)^(3/2),x, algorithm="fricas")
Output:
-1/2*(3*sqrt(I)*(I*x^5 + I*x)*elliptic_e(arcsin(sqrt(I)*x), -1) + 3*sqrt(I )*(-I*x^5 - I*x)*elliptic_f(arcsin(sqrt(I)*x), -1) + (3*x^4 + 2)*sqrt(x^4 + 1))/(x^5 + x)
Result contains complex when optimal does not.
Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=\frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{2} \\ \frac {3}{4} \end {matrix}\middle | {x^{4} e^{i \pi }} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate(1/x**2/(x**4+1)**(3/2),x)
Output:
gamma(-1/4)*hyper((-1/4, 3/2), (3/4,), x**4*exp_polar(I*pi))/(4*x*gamma(3/ 4))
\[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(x^4+1)^(3/2),x, algorithm="maxima")
Output:
integrate(1/((x^4 + 1)^(3/2)*x^2), x)
\[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (x^{4} + 1\right )}^{\frac {3}{2}} x^{2}} \,d x } \] Input:
integrate(1/x^2/(x^4+1)^(3/2),x, algorithm="giac")
Output:
integrate(1/((x^4 + 1)^(3/2)*x^2), x)
Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.11 \[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=-\frac {{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {3}{2};\ \frac {3}{4};\ -x^4\right )}{x} \] Input:
int(1/(x^2*(x^4 + 1)^(3/2)),x)
Output:
-hypergeom([-1/4, 3/2], 3/4, -x^4)/x
\[ \int \frac {1}{x^2 \left (1+x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {x^{4}+1}}{x^{10}+2 x^{6}+x^{2}}d x \] Input:
int(1/x^2/(x^4+1)^(3/2),x)
Output:
int(sqrt(x**4 + 1)/(x**10 + 2*x**6 + x**2),x)