Integrand size = 22, antiderivative size = 66 \[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {1}{15} x \left (5+3 b x^2\right ) \sqrt {1-b^2 x^4}+\frac {2 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{5 \sqrt {b}}+\frac {4 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{15 \sqrt {b}} \] Output:
1/15*x*(3*b*x^2+5)*(-b^2*x^4+1)^(1/2)+2/5*EllipticE(b^(1/2)*x,I)/b^(1/2)+4 /15*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 = 6.62 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.68 \[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},b^2 x^4\right )+\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},b^2 x^4\right ) \] Input:
Integrate[(1 + b*x^2)*Sqrt[1 - b^2*x^4],x]
Output:
x*Hypergeometric2F1[-1/2, 1/4, 5/4, b^2*x^4] + (b*x^3*Hypergeometric2F1[-1 /2, 3/4, 7/4, b^2*x^4])/3
Time = 0.43 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.53, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1388, 318, 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.
\(\displaystyle \int \left (b x^2+1\right ) \sqrt {1-b^2 x^4} \, dx\) |
\(\Big \downarrow \) 1388 |
\(\displaystyle \int \sqrt {1-b x^2} \left (b x^2+1\right )^{3/2}dx\) |
\(\Big \downarrow \) 318 |
\(\displaystyle -\frac {\int -\frac {2 b \sqrt {1-b x^2} \left (4 b x^2+3\right )}{\sqrt {b x^2+1}}dx}{5 b}-\frac {1}{5} x \sqrt {b x^2+1} \left (1-b x^2\right )^{3/2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int \frac {\sqrt {1-b x^2} \left (4 b x^2+3\right )}{\sqrt {b x^2+1}}dx-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 403 |
\(\displaystyle \frac {2}{5} \left (\frac {\int \frac {b \left (3 b x^2+5\right )}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx}{3 b}+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \int \frac {3 b x^2+5}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (2 \int \frac {1}{\sqrt {1-b x^2} \sqrt {b x^2+1}}dx+3 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 284 |
\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (2 \int \frac {1}{\sqrt {1-b^2 x^4}}dx+3 \int \frac {\sqrt {b x^2+1}}{\sqrt {1-b x^2}}dx\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (2 \int \frac {1}{\sqrt {1-b^2 x^4}}dx+\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {2}{5} \left (\frac {1}{3} \left (\frac {2 \operatorname {EllipticF}\left (\arcsin \left (\sqrt {b} x\right ),-1\right )}{\sqrt {b}}+\frac {3 E\left (\left .\arcsin \left (\sqrt {b} x\right )\right |-1\right )}{\sqrt {b}}\right )+\frac {4}{3} x \sqrt {1-b x^2} \sqrt {b x^2+1}\right )-\frac {1}{5} x \left (1-b x^2\right )^{3/2} \sqrt {b x^2+1}\) |
Input:
Int[(1 + b*x^2)*Sqrt[1 - b^2*x^4],x]
Output:
-1/5*(x*(1 - b*x^2)^(3/2)*Sqrt[1 + b*x^2]) + (2*((4*x*Sqrt[1 - b*x^2]*Sqrt [1 + b*x^2])/3 + ((3*EllipticE[ArcSin[Sqrt[b]*x], -1])/Sqrt[b] + (2*Ellipt icF[ArcSin[Sqrt[b]*x], -1])/Sqrt[b])/3))/5
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 = 1.59 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.55
method | result | size |
meijerg | \(x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], b^{2} x^{4}\right )+\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], b^{2} x^{4}\right )}{3}\) | \(36\) |
risch | \(-\frac {x \left (3 b \,x^{2}+5\right ) \left (b^{2} x^{4}-1\right )}{15 \sqrt {-b^{2} x^{4}+1}}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(133\) |
elliptic | \(\frac {b \,x^{3} \sqrt {-b^{2} x^{4}+1}}{5}+\frac {x \sqrt {-b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}-\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}\) | \(134\) |
default | \(\frac {x \sqrt {-b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )}{3 \sqrt {b}\, \sqrt {-b^{2} x^{4}+1}}+b \left (\frac {x^{3} \sqrt {-b^{2} x^{4}+1}}{5}-\frac {2 \sqrt {-b \,x^{2}+1}\, \sqrt {b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (\sqrt {b}\, x , i\right )-\operatorname {EllipticE}\left (\sqrt {b}\, x , i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {-b^{2} x^{4}+1}}\right )\) | \(136\) |
Input:
int((b*x^2+1)*(-b^2*x^4+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
x*hypergeom([-1/2,1/4],[5/4],b^2*x^4)+1/3*b*x^3*hypergeom([-1/2,3/4],[7/4] ,b^2*x^4)
Time = 0.08 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.41 \[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {\frac {2 \, \sqrt {-b^{2}} {\left (5 \, b + 3\right )} x F(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} - \frac {6 \, \sqrt {-b^{2}} x E(\arcsin \left (\frac {1}{\sqrt {b} x}\right )\,|\,-1)}{\sqrt {b}} + {\left (3 \, b^{3} x^{4} + 5 \, b^{2} x^{2} - 6 \, b\right )} \sqrt {-b^{2} x^{4} + 1}}{15 \, b^{2} x} \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(1/2),x, algorithm="fricas")
Output:
1/15*(2*sqrt(-b^2)*(5*b + 3)*x*elliptic_f(arcsin(1/(sqrt(b)*x)), -1)/sqrt( b) - 6*sqrt(-b^2)*x*elliptic_e(arcsin(1/(sqrt(b)*x)), -1)/sqrt(b) + (3*b^3 *x^4 + 5*b^2*x^2 - 6*b)*sqrt(-b^2*x^4 + 1))/(b^2*x)
Time = 0.97 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{2 i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((b*x**2+1)*(-b**2*x**4+1)**(1/2),x)
Output:
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)*hyper((-1/2, 1/4), (5/4,), b**2*x**4*exp_pol ar(2*I*pi))/(4*gamma(5/4))
\[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\int { \sqrt {-b^{2} x^{4} + 1} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(-b^2*x^4 + 1)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\int { \sqrt {-b^{2} x^{4} + 1} {\left (b x^{2} + 1\right )} \,d x } \] Input:
integrate((b*x^2+1)*(-b^2*x^4+1)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(-b^2*x^4 + 1)*(b*x^2 + 1), x)
Timed out. \[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\int \sqrt {1-b^2\,x^4}\,\left (b\,x^2+1\right ) \,d x \] Input:
int((1 - b^2*x^4)^(1/2)*(b*x^2 + 1),x)
Output:
int((1 - b^2*x^4)^(1/2)*(b*x^2 + 1), x)
\[ \int \left (1+b x^2\right ) \sqrt {1-b^2 x^4} \, dx=\frac {\sqrt {-b^{2} x^{4}+1}\, b \,x^{3}}{5}+\frac {\sqrt {-b^{2} x^{4}+1}\, x}{3}-\frac {2 \left (\int \frac {\sqrt {-b^{2} x^{4}+1}}{b^{2} x^{4}-1}d x \right )}{3}-\frac {2 \left (\int \frac {\sqrt {-b^{2} x^{4}+1}\, x^{2}}{b^{2} x^{4}-1}d x \right ) b}{5} \] Input:
int((b*x^2+1)*(-b^2*x^4+1)^(1/2),x)
Output:
(3*sqrt( - b**2*x**4 + 1)*b*x**3 + 5*sqrt( - b**2*x**4 + 1)*x - 10*int(sqr t( - b**2*x**4 + 1)/(b**2*x**4 - 1),x) - 6*int((sqrt( - b**2*x**4 + 1)*x** 2)/(b**2*x**4 - 1),x)*b)/15