Integrand size = 22, antiderivative size = 216 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=-\frac {x \left (1+b x^2\right )}{6 \left (-1-b^2 x^4\right )^{3/2}}+\frac {x \left (5+3 b x^2\right )}{12 \sqrt {-1-b^2 x^4}}+\frac {x \sqrt {-1-b^2 x^4}}{4 \left (1+b x^2\right )}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{4 \sqrt {b} \sqrt {-1-b^2 x^4}}+\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{12 \sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:
-1/6*x*(b*x^2+1)/(-b^2*x^4-1)^(3/2)+1/12*x*(3*b*x^2+5)/(-b^2*x^4-1)^(1/2)+ x*(-b^2*x^4-1)^(1/2)/(4*b*x^2+4)+1/4*(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)+1/12*(b*x^2+1)*((b^2*x^4+1)/(b*x^2+1)^2)^(1/2)*InverseJacobiAM(2*arc tan(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 = 10.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.46 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=-\frac {x \left (7+5 b^2 x^4+5 \left (1+b^2 x^4\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+4 b x^2 \left (1+b^2 x^4\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {5}{2},\frac {7}{4},-b^2 x^4\right )\right )}{12 \left (-1-b^2 x^4\right )^{3/2}} \] Input:
Integrate[(1 + b*x^2)/(-1 - b^2*x^4)^(5/2),x]
Output:
-1/12*(x*(7 + 5*b^2*x^4 + 5*(1 + b^2*x^4)^(3/2)*Hypergeometric2F1[1/4, 1/2 , 5/4, -(b^2*x^4)] + 4*b*x^2*(1 + b^2*x^4)^(3/2)*Hypergeometric2F1[3/4, 5/ 2, 7/4, -(b^2*x^4)]))/(-1 - b^2*x^4)^(3/2)
Time = 0.58 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1493, 25, 1493, 25, 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.
| \(\displaystyle \int \frac {b x^2+1}{\left (-b^2 x^4-1\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {1}{6} \int -\frac {3 b x^2+5}{\left (-b^2 x^4-1\right )^{3/2}}dx-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{6} \int \frac {3 b x^2+5}{\left (-b^2 x^4-1\right )^{3/2}}dx-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {1}{6} \left (\frac {x \left (3 b x^2+5\right )}{2 \sqrt {-b^2 x^4-1}}-\frac {1}{2} \int -\frac {5-3 b x^2}{\sqrt {-b^2 x^4-1}}dx\right )-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int \frac {5-3 b x^2}{\sqrt {-b^2 x^4-1}}dx+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {-b^2 x^4-1}}\right )-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (2 \int \frac {1}{\sqrt {-b^2 x^4-1}}dx+3 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\right )+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {-b^2 x^4-1}}\right )-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (3 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx+\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {-b^2 x^4-1}}\right )-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+3 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1}\right )\right )+\frac {x \left (3 b x^2+5\right )}{2 \sqrt {-b^2 x^4-1}}\right )-\frac {x \left (b x^2+1\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\) |
Input:
Int[(1 + b*x^2)/(-1 - b^2*x^4)^(5/2),x]
Output:
-1/6*(x*(1 + b*x^2))/(-1 - b^2*x^4)^(3/2) + ((x*(5 + 3*b*x^2))/(2*Sqrt[-1 - b^2*x^4]) + (3*((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]*Elli pticF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x^4]))/2)/6
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)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x )*(d + e*x^2)*((a + c*x^4)^(p + 1)/(4*a*(p + 1))), x] + Simp[1/(4*a*(p + 1) ) Int[Simp[d*(4*p + 5) + e*(4*p + 7)*x^2, x]*(a + c*x^4)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && Integer Q[2*p]
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 = 1.06 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.42
| method | result | size |
| meijerg | \(\frac {\operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {5}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {5}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{{\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {5}{2}}}+\frac {b \operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {5}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {5}{2}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 {\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {5}{2}}}\) | \(90\) |
| elliptic | \(\frac {\left (-\frac {x^{3}}{6 b^{3}}-\frac {x}{6 b^{4}}\right ) \sqrt {-b^{2} x^{4}-1}}{\left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {2 b^{2} \left (\frac {x^{3}}{8 b}+\frac {5 x}{24 b^{2}}\right )}{\sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{12 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}+\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{4 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) | \(194\) |
| default | \(-\frac {x \sqrt {-b^{2} x^{4}-1}}{6 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {5 x}{12 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {5 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{12 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}+b \left (-\frac {x^{3} \sqrt {-b^{2} x^{4}-1}}{6 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {x^{3}}{4 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}+\frac {i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{4 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}\, b}\right )\) | \(221\) |
Input:
int((b*x^2+1)/(-b^2*x^4-1)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/(-signum(b^2*x^4+1))^(5/2)*signum(b^2*x^4+1)^(5/2)*x*hypergeom([1/4,5/2] ,[5/4],-b^2*x^4)+1/3*b/(-signum(b^2*x^4+1))^(5/2)*signum(b^2*x^4+1)^(5/2)* x^3*hypergeom([3/4,5/2],[7/4],-b^2*x^4)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.77 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=-\frac {3 \, {\left (i \, b^{5} x^{8} + 2 i \, b^{3} x^{4} + i \, b\right )} \left (-b^{2}\right )^{\frac {3}{4}} E(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - {\left (i \, {\left (3 \, b^{5} + 5 \, b^{4}\right )} x^{8} + 2 i \, {\left (3 \, b^{3} + 5 \, b^{2}\right )} x^{4} + 3 i \, b + 5 i\right )} \left (-b^{2}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + {\left (3 \, b^{5} x^{7} + 5 \, b^{4} x^{5} + 5 \, b^{3} x^{3} + 7 \, b^{2} x\right )} \sqrt {-b^{2} x^{4} - 1}}{12 \, {\left (b^{6} x^{8} + 2 \, b^{4} x^{4} + b^{2}\right )}} \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(5/2),x, algorithm="fricas")
Output:
-1/12*(3*(I*b^5*x^8 + 2*I*b^3*x^4 + I*b)*(-b^2)^(3/4)*elliptic_e(arcsin((- b^2)^(1/4)*x), -1) - (I*(3*b^5 + 5*b^4)*x^8 + 2*I*(3*b^3 + 5*b^2)*x^4 + 3* I*b + 5*I)*(-b^2)^(3/4)*elliptic_f(arcsin((-b^2)^(1/4)*x), -1) + (3*b^5*x^ 7 + 5*b^4*x^5 + 5*b^3*x^3 + 7*b^2*x)*sqrt(-b^2*x^4 - 1))/(b^6*x^8 + 2*b^4* x^4 + b^2)
Result contains complex when optimal does not.
Time = 9.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.33 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=- \frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)/(-b**2*x**4-1)**(5/2),x)
Output:
-I*b*x**3*gamma(3/4)*hyper((3/4, 5/2), (7/4,), b**2*x**4*exp_polar(I*pi))/ (4*gamma(7/4)) - I*x*gamma(1/4)*hyper((1/4, 5/2), (5/4,), b**2*x**4*exp_po lar(I*pi))/(4*gamma(5/4))
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (-b^{2} x^{4} - 1\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(5/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + 1)/(-b^2*x^4 - 1)^(5/2), x)
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (-b^{2} x^{4} - 1\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(5/2),x, algorithm="giac")
Output:
integrate((b*x^2 + 1)/(-b^2*x^4 - 1)^(5/2), x)
Timed out. \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=\int \frac {b\,x^2+1}{{\left (-b^2\,x^4-1\right )}^{5/2}} \,d x \] Input:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(5/2),x)
Output:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(5/2), x)
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{5/2}} \, dx=-i \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{6} x^{12}+3 b^{4} x^{8}+3 b^{2} x^{4}+1}d x +\left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{6} x^{12}+3 b^{4} x^{8}+3 b^{2} x^{4}+1}d x \right ) b \right ) \] Input:
int((b*x^2+1)/(-b^2*x^4-1)^(5/2),x)
Output:
- i*(int(sqrt(b**2*x**4 + 1)/(b**6*x**12 + 3*b**4*x**8 + 3*b**2*x**4 + 1) ,x) + int((sqrt(b**2*x**4 + 1)*x**2)/(b**6*x**12 + 3*b**4*x**8 + 3*b**2*x* *4 + 1),x)*b)