Integrand size = 22, antiderivative size = 243 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=-\frac {x \left (1+b x^2\right )}{10 \left (-1-b^2 x^4\right )^{5/2}}+\frac {x \left (9+7 b x^2\right )}{60 \left (-1-b^2 x^4\right )^{3/2}}-\frac {x \left (15+7 b x^2\right )}{40 \sqrt {-1-b^2 x^4}}-\frac {7 x \sqrt {-1-b^2 x^4}}{40 \left (1+b x^2\right )}-\frac {7 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{40 \sqrt {b} \sqrt {-1-b^2 x^4}}-\frac {\left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{10 \sqrt {b} \sqrt {-1-b^2 x^4}} \] Output:
-1/10*x*(b*x^2+1)/(-b^2*x^4-1)^(5/2)+1/60*x*(7*b*x^2+9)/(-b^2*x^4-1)^(3/2) -1/40*x*(7*b*x^2+15)/(-b^2*x^4-1)^(1/2)-7*x*(-b^2*x^4-1)^(1/2)/(40*b*x^2+4 0)-7/40*(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)-1/10*(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 = 10.05 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.44 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=-\frac {x \left (75+108 b^2 x^4+45 b^4 x^8+45 \left (1+b^2 x^4\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-b^2 x^4\right )+40 b x^2 \left (1+b^2 x^4\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {7}{2},\frac {7}{4},-b^2 x^4\right )\right )}{120 \left (-1-b^2 x^4\right )^{5/2}} \] Input:
Integrate[(1 + b*x^2)/(-1 - b^2*x^4)^(7/2),x]
Output:
-1/120*(x*(75 + 108*b^2*x^4 + 45*b^4*x^8 + 45*(1 + b^2*x^4)^(5/2)*Hypergeo metric2F1[1/4, 1/2, 5/4, -(b^2*x^4)] + 40*b*x^2*(1 + b^2*x^4)^(5/2)*Hyperg eometric2F1[3/4, 7/2, 7/4, -(b^2*x^4)]))/(-1 - b^2*x^4)^(5/2)
Time = 0.66 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1493, 25, 1493, 27, 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.
| \(\displaystyle \int \frac {b x^2+1}{\left (-b^2 x^4-1\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {1}{10} \int -\frac {7 b x^2+9}{\left (-b^2 x^4-1\right )^{5/2}}dx-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {1}{10} \int \frac {7 b x^2+9}{\left (-b^2 x^4-1\right )^{5/2}}dx-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {1}{10} \left (\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}-\frac {1}{6} \int -\frac {3 \left (7 b x^2+15\right )}{\left (-b^2 x^4-1\right )^{3/2}}dx\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \int \frac {7 b x^2+15}{\left (-b^2 x^4-1\right )^{3/2}}dx+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 1493 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \left (\frac {1}{2} \int -\frac {15-7 b x^2}{\sqrt {-b^2 x^4-1}}dx-\frac {x \left (7 b x^2+15\right )}{2 \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \left (-\frac {1}{2} \int \frac {15-7 b x^2}{\sqrt {-b^2 x^4-1}}dx-\frac {x \left (7 b x^2+15\right )}{2 \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \left (\frac {1}{2} \left (-8 \int \frac {1}{\sqrt {-b^2 x^4-1}}dx-7 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx\right )-\frac {x \left (7 b x^2+15\right )}{2 \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \left (\frac {1}{2} \left (-7 \int \frac {1-b x^2}{\sqrt {-b^2 x^4-1}}dx-\frac {4 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}\right )-\frac {x \left (7 b x^2+15\right )}{2 \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {1}{10} \left (\frac {1}{2} \left (\frac {1}{2} \left (-\frac {4 \left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}-7 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {-b^2 x^4-1}}+\frac {x \sqrt {-b^2 x^4-1}}{b x^2+1}\right )\right )-\frac {x \left (7 b x^2+15\right )}{2 \sqrt {-b^2 x^4-1}}\right )+\frac {x \left (7 b x^2+9\right )}{6 \left (-b^2 x^4-1\right )^{3/2}}\right )-\frac {x \left (b x^2+1\right )}{10 \left (-b^2 x^4-1\right )^{5/2}}\) |
Input:
Int[(1 + b*x^2)/(-1 - b^2*x^4)^(7/2),x]
Output:
-1/10*(x*(1 + b*x^2))/(-1 - b^2*x^4)^(5/2) + ((x*(9 + 7*b*x^2))/(6*(-1 - b ^2*x^4)^(3/2)) + (-1/2*(x*(15 + 7*b*x^2))/Sqrt[-1 - b^2*x^4] + (-7*((x*Sqr t[-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])) - ( 4*(1 + b*x^2)*Sqrt[(1 + b^2*x^4)/(1 + b*x^2)^2]*EllipticF[2*ArcTan[Sqrt[b] *x], 1/2])/(Sqrt[b]*Sqrt[-1 - b^2*x^4]))/2)/2)/10
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 1.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.37
| method | result | size |
| meijerg | \(\frac {\operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {7}{2}} x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {7}{2}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )}{{\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {7}{2}}}+\frac {b \operatorname {signum}\left (b^{2} x^{4}+1\right )^{\frac {7}{2}} x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {7}{2}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3 {\left (-\operatorname {signum}\left (b^{2} x^{4}+1\right )\right )}^{\frac {7}{2}}}\) | \(90\) |
| elliptic | \(\frac {\left (\frac {x^{3}}{10 b^{5}}+\frac {x}{10 b^{6}}\right ) \sqrt {-b^{2} x^{4}-1}}{\left (x^{4}+\frac {1}{b^{2}}\right )^{3}}+\frac {\left (\frac {7 x^{3}}{60 b^{3}}+\frac {3 x}{20 b^{4}}\right ) \sqrt {-b^{2} x^{4}-1}}{\left (x^{4}+\frac {1}{b^{2}}\right )^{2}}+\frac {2 b^{2} \left (-\frac {7 x^{3}}{80 b}-\frac {3 x}{16 b^{2}}\right )}{\sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {3 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{8 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}-\frac {7 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{40 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}\) | \(231\) |
| default | \(\frac {x \sqrt {-b^{2} x^{4}-1}}{10 b^{6} \left (x^{4}+\frac {1}{b^{2}}\right )^{3}}+\frac {3 x \sqrt {-b^{2} x^{4}-1}}{20 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}-\frac {3 x}{8 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {3 \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )}{8 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}}+b \left (\frac {x^{3} \sqrt {-b^{2} x^{4}-1}}{10 b^{6} \left (x^{4}+\frac {1}{b^{2}}\right )^{3}}+\frac {7 x^{3} \sqrt {-b^{2} x^{4}-1}}{60 b^{4} \left (x^{4}+\frac {1}{b^{2}}\right )^{2}}-\frac {7 x^{3}}{40 \sqrt {-\left (x^{4}+\frac {1}{b^{2}}\right ) b^{2}}}-\frac {7 i \sqrt {i b \,x^{2}+1}\, \sqrt {-i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-i b}, i\right )\right )}{40 \sqrt {-i b}\, \sqrt {-b^{2} x^{4}-1}\, b}\right )\) | \(277\) |
Input:
int((b*x^2+1)/(-b^2*x^4-1)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/(-signum(b^2*x^4+1))^(7/2)*signum(b^2*x^4+1)^(7/2)*x*hypergeom([1/4,7/2] ,[5/4],-b^2*x^4)+1/3*b/(-signum(b^2*x^4+1))^(7/2)*signum(b^2*x^4+1)^(7/2)* x^3*hypergeom([3/4,7/2],[7/4],-b^2*x^4)
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.89 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=-\frac {21 \, {\left (-i \, b^{7} x^{12} - 3 i \, b^{5} x^{8} - 3 i \, b^{3} x^{4} - i \, b\right )} \left (-b^{2}\right )^{\frac {3}{4}} E(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) + 3 \, {\left (i \, {\left (7 \, b^{7} + 15 \, b^{6}\right )} x^{12} + 3 i \, {\left (7 \, b^{5} + 15 \, b^{4}\right )} x^{8} + 3 i \, {\left (7 \, b^{3} + 15 \, b^{2}\right )} x^{4} + 7 i \, b + 15 i\right )} \left (-b^{2}\right )^{\frac {3}{4}} F(\arcsin \left (\left (-b^{2}\right )^{\frac {1}{4}} x\right )\,|\,-1) - {\left (21 \, b^{7} x^{11} + 45 \, b^{6} x^{9} + 56 \, b^{5} x^{7} + 108 \, b^{4} x^{5} + 47 \, b^{3} x^{3} + 75 \, b^{2} x\right )} \sqrt {-b^{2} x^{4} - 1}}{120 \, {\left (b^{8} x^{12} + 3 \, b^{6} x^{8} + 3 \, b^{4} x^{4} + b^{2}\right )}} \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(7/2),x, algorithm="fricas")
Output:
-1/120*(21*(-I*b^7*x^12 - 3*I*b^5*x^8 - 3*I*b^3*x^4 - I*b)*(-b^2)^(3/4)*el liptic_e(arcsin((-b^2)^(1/4)*x), -1) + 3*(I*(7*b^7 + 15*b^6)*x^12 + 3*I*(7 *b^5 + 15*b^4)*x^8 + 3*I*(7*b^3 + 15*b^2)*x^4 + 7*I*b + 15*I)*(-b^2)^(3/4) *elliptic_f(arcsin((-b^2)^(1/4)*x), -1) - (21*b^7*x^11 + 45*b^6*x^9 + 56*b ^5*x^7 + 108*b^4*x^5 + 47*b^3*x^3 + 75*b^2*x)*sqrt(-b^2*x^4 - 1))/(b^8*x^1 2 + 3*b^6*x^8 + 3*b^4*x^4 + b^2)
Result contains complex when optimal does not.
Time = 33.35 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.29 \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=\frac {i b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {i x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)/(-b**2*x**4-1)**(7/2),x)
Output:
I*b*x**3*gamma(3/4)*hyper((3/4, 7/2), (7/4,), b**2*x**4*exp_polar(I*pi))/( 4*gamma(7/4)) + I*x*gamma(1/4)*hyper((1/4, 7/2), (5/4,), b**2*x**4*exp_pol ar(I*pi))/(4*gamma(5/4))
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (-b^{2} x^{4} - 1\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(7/2),x, algorithm="maxima")
Output:
integrate((b*x^2 + 1)/(-b^2*x^4 - 1)^(7/2), x)
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=\int { \frac {b x^{2} + 1}{{\left (-b^{2} x^{4} - 1\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate((b*x^2+1)/(-b^2*x^4-1)^(7/2),x, algorithm="giac")
Output:
integrate((b*x^2 + 1)/(-b^2*x^4 - 1)^(7/2), x)
Timed out. \[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=\int \frac {b\,x^2+1}{{\left (-b^2\,x^4-1\right )}^{7/2}} \,d x \] Input:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(7/2),x)
Output:
int((b*x^2 + 1)/(- b^2*x^4 - 1)^(7/2), x)
\[ \int \frac {1+b x^2}{\left (-1-b^2 x^4\right )^{7/2}} \, dx=i \left (\int \frac {\sqrt {b^{2} x^{4}+1}}{b^{8} x^{16}+4 b^{6} x^{12}+6 b^{4} x^{8}+4 b^{2} x^{4}+1}d x +\left (\int \frac {\sqrt {b^{2} x^{4}+1}\, x^{2}}{b^{8} x^{16}+4 b^{6} x^{12}+6 b^{4} x^{8}+4 b^{2} x^{4}+1}d x \right ) b \right ) \] Input:
int((b*x^2+1)/(-b^2*x^4-1)^(7/2),x)
Output:
i*(int(sqrt(b**2*x**4 + 1)/(b**8*x**16 + 4*b**6*x**12 + 6*b**4*x**8 + 4*b* *2*x**4 + 1),x) + int((sqrt(b**2*x**4 + 1)*x**2)/(b**8*x**16 + 4*b**6*x**1 2 + 6*b**4*x**8 + 4*b**2*x**4 + 1),x)*b)