Integrand size = 22, antiderivative size = 93 \[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=\frac {2}{105} x \left (15+7 b x^2\right ) \sqrt {1-b^2 x^4}+\frac {1}{63} x \left (9+7 b x^2\right ) \left (1-b^2 x^4\right )^{3/2}+\frac {4 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{15 \sqrt {b}}+\frac {32 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{105 \sqrt {b}} \] Output:
2/105*x*(7*b*x^2+15)*(-b^2*x^4+1)^(1/2)+1/63*x*(7*b*x^2+9)*(-b^2*x^4+1)^(3 /2)+4/15*EllipticE(b^(1/2)*x,I)/b^(1/2)+32/105*EllipticF(b^(1/2)*x,I)/b^(1 /2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.48 \[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right )+\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right ) \] Input:
Integrate[(1 + b*x^2)*(1 - b^2*x^4)^(3/2),x]
Output:
x*Hypergeometric2F1[-3/2, 1/4, 5/4, b^2*x^4] + (b*x^3*Hypergeometric2F1[-3 /2, 3/4, 7/4, b^2*x^4])/3
Time = 0.56 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.72, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {1388, 318, 27, 403, 25, 27, 403, 27, 403, 27, 399, 284, 327, 762}
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.
| \(\displaystyle \int \left (b x^2+1\right ) \left (1-b^2 x^4\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle \int \left (1-b x^2\right )^{3/2} \left (b x^2+1\right )^{5/2}dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle -\frac {\int -2 b \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1} \left (8 b x^2+5\right )dx}{9 b}-\frac {1}{9} x \left (b x^2+1\right )^{3/2} \left (1-b x^2\right )^{5/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{9} \int \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1} \left (8 b x^2+5\right )dx-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {2}{9} \left (-\frac {\int -\frac {b \left (1-b x^2\right )^{3/2} \left (59 b x^2+43\right )}{\sqrt {b x^2+1}}dx}{7 b}-\frac {8}{7} x \sqrt {b x^2+1} \left (1-b x^2\right )^{5/2}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {2}{9} \left (\frac {\int \frac {b \left (1-b x^2\right )^{3/2} \left (59 b x^2+43\right )}{\sqrt {b x^2+1}}dx}{7 b}-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \int \frac {\left (1-b x^2\right )^{3/2} \left (59 b x^2+43\right )}{\sqrt {b x^2+1}}dx-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {\int \frac {6 b \sqrt {1-b x^2} \left (33 b x^2+26\right )}{\sqrt {b x^2+1}}dx}{5 b}+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \int \frac {\sqrt {1-b x^2} \left (33 b x^2+26\right )}{\sqrt {b x^2+1}}dx+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (\frac {\int \frac {3 b \left (7 b x^2+15\right )}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{3 b}+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (\int \frac {7 b x^2+15}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 399 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (8 \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx+7 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 284 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (8 \int \frac {1}{\sqrt {1-b^2 x^4}}dx+7 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (8 \int \frac {1}{\sqrt {1-b^2 x^4}}dx+\frac {7 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
| \(\Big \downarrow \) 762 |
| \(\displaystyle \frac {2}{9} \left (\frac {1}{7} \left (\frac {6}{5} \left (\frac {8 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}+\frac {7 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}+11 x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )+\frac {59}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\right )-\frac {8}{7} x \left (1-b x^2\right )^{5/2} \sqrt {b x^2+1}\right )-\frac {1}{9} x \left (1-b x^2\right )^{5/2} \left (b x^2+1\right )^{3/2}\) |
Input:
Int[(1 + b*x^2)*(1 - b^2*x^4)^(3/2),x]
Output:
-1/9*(x*(1 - b*x^2)^(5/2)*(1 + b*x^2)^(3/2)) + (2*((-8*x*(1 - b*x^2)^(5/2) *Sqrt[1 + b*x^2])/7 + ((59*x*(1 - b*x^2)^(3/2)*Sqrt[1 + b*x^2])/5 + (6*(11 *x*Sqrt[1 - b*x^2]*Sqrt[1 + b*x^2] + (7*EllipticE[ArcSin[Sqrt[b]*x], -1])/ Sqrt[b] + (8*EllipticF[ArcSin[Sqrt[b]*x], -1])/Sqrt[b]))/5)/7))/9
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)^(p_.), x_Symbol] :> I nt[(a*c + b*d*x^4)^p, x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0]))
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 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[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) ^2]), x_Symbol] :> Simp[f/b Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/b Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr eeQ[{a, b, c, d, e, f}, x] && !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && 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 = 1.65 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.39
| method | result | size |
| meijerg | \(x \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )+\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\) | \(36\) |
| risch | \(\frac {x \left (35 b^{3} x^{6}+45 b^{2} x^{4}-77 b \,x^{2}-135\right ) \left (b^{2} x^{4}-1\right )}{315 \sqrt {-b^{2} x^{4}+1}}+\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{7 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{15 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(149\) |
| elliptic | \(-\frac {b^{3} x^{7} \sqrt {-b^{2} x^{4}+1}}{9}-\frac {b^{2} x^{5} \sqrt {-b^{2} x^{4}+1}}{7}+\frac {11 b \,x^{3} \sqrt {-b^{2} x^{4}+1}}{45}+\frac {3 x \sqrt {-b^{2} x^{4}+1}}{7}+\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{7 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{15 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(174\) |
| default | \(-\frac {b^{2} x^{5} \sqrt {-b^{2} x^{4}+1}}{7}+\frac {3 x \sqrt {-b^{2} x^{4}+1}}{7}+\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{7 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+b \left (-\frac {b^{2} x^{7} \sqrt {-b^{2} x^{4}+1}}{9}+\frac {11 x^{3} \sqrt {-b^{2} x^{4}+1}}{45}-\frac {4 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{15 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )\) | \(176\) |
Input:
int((b*x^2+1)*(-b^2*x^4+1)^(3/2),x,method=_RETURNVERBOSE)
Output:
x*hypergeom([-3/2,1/4],[5/4],b^2*x^4)+1/3*b*x^3*hypergeom([-3/2,3/4],[7/4] ,b^2*x^4)
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.18 \[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=\frac {\frac {12 \, \sqrt {-b^{2}} {\left (15 \, b + 7\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \frac {84 \, \sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - {\left (35 \, b^{5} x^{8} + 45 \, b^{4} x^{6} - 77 \, b^{3} x^{4} - 135 \, b^{2} x^{2} + 84 \, b\right )} \sqrt {-b^{2} x^{4} + 1}}{315 \, b^{2} x} \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(3/2),x, algorithm="fricas")
Output:
1/315*(12*sqrt(-b^2)*(15*b + 7)*x*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sq rt(b) - 84*sqrt(-b^2)*x*elliptic_e(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) - (3 5*b^5*x^8 + 45*b^4*x^6 - 77*b^3*x^4 - 135*b^2*x^2 + 84*b)*sqrt(-b^2*x^4 + 1))/(b^2*x)
Time = 1.66 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.67 \[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=- \frac {b^{3} x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {11}{4}\right )} - \frac {b^{2} x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {9}{4}\right )} + \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)*(-b**2*x**4+1)**(3/2),x)
Output:
-b**3*x**7*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b**2*x**4*exp_polar(2*I* pi))/(4*gamma(11/4)) - b**2*x**5*gamma(5/4)*hyper((-1/2, 5/4), (9/4,), b** 2*x**4*exp_polar(2*I*pi))/(4*gamma(9/4)) + b*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), b**2*x**4*exp_polar(2*I*pi))/(4*gamma(7/4)) + x*gamma(1/4)*h yper((-1/2, 1/4), (5/4,), b**2*x**4*exp_polar(2*I*pi))/(4*gamma(5/4))
\[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=\int { {\left (-b^{2} x^{4} + 1\right )}^{\frac {3}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(3/2),x, algorithm="maxima")
Output:
integrate((-b^2*x^4 + 1)^(3/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=\int { {\left (-b^{2} x^{4} + 1\right )}^{\frac {3}{2}} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(3/2),x, algorithm="giac")
Output:
integrate((-b^2*x^4 + 1)^(3/2)*(b*x^2 + 1), x)
Timed out. \[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=\int {\left (1-b^2\,x^4\right )}^{3/2}\,\left (b\,x^2+1\right ) \,d x \] Input:
int((1 - b^2*x^4)^(3/2)*(b*x^2 + 1),x)
Output:
int((1 - b^2*x^4)^(3/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \left (1-b^2 x^4\right )^{3/2} \, dx=-\frac {\sqrt {-b^{2} x^{4}+1}\, b^{3} x^{7}}{9}-\frac {\sqrt {-b^{2} x^{4}+1}\, b^{2} x^{5}}{7}+\frac {11 \sqrt {-b^{2} x^{4}+1}\, b \,x^{3}}{45}+\frac {3 \sqrt {-b^{2} x^{4}+1}\, x}{7}-\frac {4 \left (\int \frac {\sqrt {-b^{2} x^{4}+1}}{b^{2} x^{4}-1}d x \right )}{7}-\frac {4 \left (\int \frac {\sqrt {-b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}-1}d x \right ) b}{15} \] Input:
int((b*x^2+1)*(-b^2*x^4+1)^(3/2),x)
Output:
( - 35*sqrt( - b**2*x**4 + 1)*b**3*x**7 - 45*sqrt( - b**2*x**4 + 1)*b**2*x **5 + 77*sqrt( - b**2*x**4 + 1)*b*x**3 + 135*sqrt( - b**2*x**4 + 1)*x - 18 0*int(sqrt( - b**2*x**4 + 1)/(b**2*x**4 - 1),x) - 84*int((sqrt( - b**2*x** 4 + 1)*x**2)/(b**2*x**4 - 1),x)*b)/315