Integrand size = 23, antiderivative size = 47 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=\frac {x \left (1-b x^2\right )}{2 \sqrt {1-b^2 x^4}}+\frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{2 \sqrt {b}} \] Output:
1/2*x*(-b*x^2+1)/(-b^2*x^4+1)^(1/2)+1/2*EllipticE(b^(1/2)*x,I)/b^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.36 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=\frac {1}{6} x \left (\frac {3}{\sqrt {1-b^2 x^4}}+3 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},b^2 x^4\right )-2 b x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},b^2 x^4\right )\right ) \] Input:
Integrate[(1 - b*x^2)/(1 - b^2*x^4)^(3/2),x]
Output:
(x*(3/Sqrt[1 - b^2*x^4] + 3*Hypergeometric2F1[1/4, 1/2, 5/4, b^2*x^4] - 2* b*x^2*Hypergeometric2F1[3/4, 3/2, 7/4, b^2*x^4]))/6
Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1388, 316, 25, 27, 327}
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.
\(\displaystyle \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \frac {1}{\sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}}dx\) |
\(\Big \downarrow \) 316 |
\(\displaystyle \frac {x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}-\frac {\int -\frac {b \sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx}{2 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {b \sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx}{2 b}+\frac {x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx+\frac {x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{2 \sqrt {b}}+\frac {x \sqrt {1-b x^2}}{2 \sqrt {b x^2+1}}\) |
Input:
Int[(1 - b*x^2)/(1 - b^2*x^4)^(3/2),x]
Output:
(x*Sqrt[1 - b*x^2])/(2*Sqrt[1 + b*x^2]) + EllipticE[ArcSin[Sqrt[b]*x], -1] /(2*Sqrt[b])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
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[(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]))
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 1.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77
method | result | size |
meijerg | \(x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {3}{2}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )-\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{2}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\) | \(36\) |
elliptic | \(\frac {\left (-b^{2} x^{2}+b \right ) x}{2 b \sqrt {\left (x^{2}+\frac {1}{b}\right ) \left (-b^{2} x^{2}+b \right )}}+\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{2 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{2 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(137\) |
default | \(-b \left (\frac {x^{3}}{2 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )+\frac {x}{2 \sqrt {-\left (x^{4}-\frac {1}{b^{2}}\right ) b^{2}}}+\frac {\sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{2 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(145\) |
Input:
int((-b*x^2+1)/(-b^2*x^4+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
x*hypergeom([1/4,3/2],[5/4],b^2*x^4)-1/3*b*x^3*hypergeom([3/4,3/2],[7/4],b ^2*x^4)
Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (35) = 70\).
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.64 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {-b^{2} x^{4} + 1} b x + {\left (b^{2} x^{2} + b\right )} \sqrt {b} E(\arcsin \left (\sqrt {b} x\right )\,|\,-1) - {\left ({\left (b^{2} - b\right )} x^{2} + b - 1\right )} \sqrt {b} F(\arcsin \left (\sqrt {b} x\right )\,|\,-1)}{2 \, {\left (b^{2} x^{2} + b\right )}} \] Input:
integrate((-b*x^2+1)/(-b^2*x^4+1)^(3/2),x, algorithm="fricas")
Output:
1/2*(sqrt(-b^2*x^4 + 1)*b*x + (b^2*x^2 + b)*sqrt(b)*elliptic_e(arcsin(sqrt (b)*x), -1) - ((b^2 - b)*x^2 + b - 1)*sqrt(b)*elliptic_f(arcsin(sqrt(b)*x) , -1))/(b^2*x^2 + b)
Time = 2.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.49 \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=- \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{2} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((-b*x**2+1)/(-b**2*x**4+1)**(3/2),x)
Output:
-b*x**3*gamma(3/4)*hyper((3/4, 3/2), (7/4,), b**2*x**4*exp_polar(2*I*pi))/ (4*gamma(7/4)) + x*gamma(1/4)*hyper((1/4, 3/2), (5/4,), b**2*x**4*exp_pola r(2*I*pi))/(4*gamma(5/4))
\[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=\int { -\frac {b x^{2} - 1}{{\left (-b^{2} x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x^2+1)/(-b^2*x^4+1)^(3/2),x, algorithm="maxima")
Output:
-integrate((b*x^2 - 1)/(-b^2*x^4 + 1)^(3/2), x)
\[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=\int { -\frac {b x^{2} - 1}{{\left (-b^{2} x^{4} + 1\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((-b*x^2+1)/(-b^2*x^4+1)^(3/2),x, algorithm="giac")
Output:
integrate(-(b*x^2 - 1)/(-b^2*x^4 + 1)^(3/2), x)
Timed out. \[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=-\int \frac {b\,x^2-1}{{\left (1-b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(-(b*x^2 - 1)/(1 - b^2*x^4)^(3/2),x)
Output:
-int((b*x^2 - 1)/(1 - b^2*x^4)^(3/2), x)
\[ \int \frac {1-b x^2}{\left (1-b^2 x^4\right )^{3/2}} \, dx=-\left (\int \frac {\sqrt {-b^{2} x^{4}+1}}{b^{3} x^{6}+b^{2} x^{4}-b \,x^{2}-1}d x \right ) \] Input:
int((-b*x^2+1)/(-b^2*x^4+1)^(3/2),x)
Output:
- int(sqrt( - b**2*x**4 + 1)/(b**3*x**6 + b**2*x**4 - b*x**2 - 1),x)