Integrand size = 22, antiderivative size = 156 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=-\frac {x \sqrt {-1-b^2 x^4}}{1+b x^2}-\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-1-b^2 x^4}}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:
-x*(-b^2*x^4-1)^(1/2)/(b*x^2+1)-(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)* EllipticE(sin(2*arctan(b^(1/2)*x)),1/2*2^(1/2))/b^(1/2)/(-b^2*x^4-1)^(1/2) +(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/2 )*x),1/2*2^(1/2))/b^(1/2)/(-b^2*x^4-1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.49 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\frac {\sqrt {1+b^2 x^4} \left (3 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+b x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right )\right )}{3 \sqrt {-1-b^2 x^4}} \] Input:
Integrate[(1 + b*x^2)/Sqrt[-1 - b^2*x^4],x]
Output:
(Sqrt[1 + b^2*x^4]*(3*x*Hypergeometric2F1[1/4, 1/2, 5/4, -(b^2*x^4)] + b*x ^3*Hypergeometric2F1[1/2, 3/4, 7/4, -(b^2*x^4)]))/(3*Sqrt[-1 - b^2*x^4])
Time = 0.42 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1512, 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.
\(\displaystyle \int \frac {b x^2+1}{\sqrt {-b^2 x^4-1}} \, dx\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle 2 \int \frac {1}{\sqrt {-b^2 x^4-1}}dx-\int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-\int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1}\) |
Input:
Int[(1 + b*x^2)/Sqrt[-1 - b^2*x^4],x]
Output:
-((x*Sqrt[-1 - b^2*x^4])/(1 + b*x^2)) - ((1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*EllipticE[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x ^4]) + ((1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*EllipticF[2*ArcTan[S qrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x^4])
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[((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]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Time = 1.03 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.58
method | result | size |
meijerg | \(\frac {b \sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 \sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}}+\frac {\sqrt {\operatorname {signum}\left (b^{2} x^{4}+1\right )}\, x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{\sqrt {-\operatorname {signum}\left (b^{2} x^{4}+1\right )}}\) | \(90\) |
default | \(\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) | \(122\) |
elliptic | \(\frac {\sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{\sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) | \(122\) |
Input:
int((b*x^2+1)/(-b^2*x^4-1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*b/(-signum(b^2*x^4+1))^(1/2)*signum(b^2*x^4+1)^(1/2)*x^3*hypergeom([1/ 2,3/4],[7/4],-b^2*x^4)+1/(-signum(b^2*x^4+1))^(1/2)*signum(b^2*x^4+1)^(1/2 )*x*hypergeom([1/4,1/2],[5/4],-b^2*x^4)
Time = 0.06 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.54 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=-\frac {\sqrt {-b^{2}} {\left (b - 1\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-b^{2}} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-b^{2} x^{4} - 1}}{b x} \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(1/2),x, algorithm="fricas")
Output:
-(sqrt(-b^2)*(b - 1)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^2)^(1/4)/x), -1) + sqrt(-b^2)*x*(-1/b^2)^(3/4)*elliptic_e(arcsin((-1/b^2)^(1/4)/x), -1 ) + sqrt(-b^2*x^4 - 1))/(b*x)
Result contains complex when optimal does not.
Time = 0.85 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.46 \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)/(-b**2*x**4-1)**(1/2),x)
Output:
-I*b*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), b**2*x**4*exp_polar(I*pi))/ (4*gamma(7/4)) - I*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), b**2*x**4*exp_po lar(I*pi))/(4*gamma(5/4))
\[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {-b^{2} x^{4} - 1}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(1/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + 1)/sqrt(-b^2*x^4 - 1), x)
\[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\int { \frac {b x^{2} + 1}{\sqrt {-b^{2} x^{4} - 1}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(1/2),x, algorithm="giac")
Output:
integrate((b*x^2 + 1)/sqrt(-b^2*x^4 - 1), x)
Timed out. \[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=\int \frac {b\,x^2+1}{\sqrt {-b^2\,x^4-1}} \,d x \] Input:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(1/2),x)
Output:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(1/2), x)
\[ \int \frac {1+b x^2}{\sqrt {-1-b^2 x^4}} \, dx=-i \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{2} x^{4}+1}d x +\left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}+1}d x \right ) b \right ) \] Input:
int((b*x^2+1)/(-b^2*x^4-1)^(1/2),x)
Output:
- i*(int(sqrt(b**2*x**4 + 1)/(b**2*x**4 + 1),x) + int((sqrt(b**2*x**4 + 1 )*x**2)/(b**2*x**4 + 1),x)*b)