Integrand size = 22, antiderivative size = 186 \[ \int \left (1-b x^2\right ) \sqrt {1+b^2 x^4} \, dx=\frac {1}{15} x \left (5-3 b x^2\right ) \sqrt {1+b^2 x^4}-\frac {2 x \sqrt {1+b^2 x^4}}{5 \left (1+b x^2\right )}+\frac {2 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{5 \sqrt {b} \sqrt {1+b^2 x^4}}+\frac {2 \left (1+b x^2\right ) \sqrt {\frac {1+b^2 x^4}{\left (1+b x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{15 \sqrt {b} \sqrt {1+b^2 x^4}} \] Output:
1/15*x*(-3*b*x^2+5)*(b^2*x^4+1)^(1/2)-2*x*(b^2*x^4+1)^(1/2)/(5*b*x^2+5)+2/ 5*(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)+2/15*(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 = 5.92 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.25 \[ \int \left (1-b x^2\right ) \sqrt {1+b^2 x^4} \, dx=x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},-b^2 x^4\right )-\frac {1}{3} b x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},-b^2 x^4\right ) \] 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 = 0.48 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1491, 27, 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.
\(\displaystyle \int \left (1-b x^2\right ) \sqrt {b^2 x^4+1} \, dx\) |
\(\Big \downarrow \) 1491 |
\(\displaystyle \frac {1}{15} \int \frac {2 \left (5-3 b x^2\right )}{\sqrt {b^2 x^4+1}}dx+\frac {1}{15} x \sqrt {b^2 x^4+1} \left (5-3 b x^2\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{15} \int \frac {5-3 b x^2}{\sqrt {b^2 x^4+1}}dx+\frac {1}{15} x \sqrt {b^2 x^4+1} \left (5-3 b x^2\right )\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle \frac {2}{15} \left (2 \int \frac {1}{\sqrt {b^2 x^4+1}}dx+3 \int \frac {1-b x^2}{\sqrt {b^2 x^4+1}}dx\right )+\frac {1}{15} x \sqrt {b^2 x^4+1} \left (5-3 b x^2\right )\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2}{15} \left (3 \int \frac {1-b x^2}{\sqrt {b^2 x^4+1}}dx+\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}\right )+\frac {1}{15} x \sqrt {b^2 x^4+1} \left (5-3 b x^2\right )\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2}{15} \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt {b} x\right ),\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}+3 \left (\frac {\left (b x^2+1\right ) \sqrt {\frac {b^2 x^4+1}{\left (b x^2+1\right )^2}} E\left (2 \arctan \left (\sqrt {b} x\right )|\frac {1}{2}\right )}{\sqrt {b} \sqrt {b^2 x^4+1}}-\frac {x \sqrt {b^2 x^4+1}}{b x^2+1}\right )\right )+\frac {1}{15} x \sqrt {b^2 x^4+1} \left (5-3 b x^2\right )\) |
Input:
Int[(1 - b*x^2)*Sqrt[1 + b^2*x^4],x]
Output:
(x*(5 - 3*b*x^2)*Sqrt[1 + b^2*x^4])/15 + (2*(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*Ar cTan[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]*EllipticF[2*ArcTan[Sqrt[b]*x], 1/2])/(Sqrt[b]*S qrt[1 + b^2*x^4])))/15
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*(4*p + 3) + e*(4*p + 1)*x^2)*((a + c*x^4)^p/((4*p + 1)*(4*p + 3))), x] + Simp[2*(p/((4*p + 1)*(4*p + 3))) Int[Simp[2*a*d*(4*p + 3) + (2*a*e*(4*p + 1))*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] && GtQ[p, 0] && FractionQ[p] && IntegerQ[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 = 1.33 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.20
method | result | size |
meijerg | \(x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], -b^{2} x^{4}\right )-\frac {b \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -b^{2} x^{4}\right )}{3}\) | \(38\) |
risch | \(-\frac {x \left (3 b \,x^{2}-5\right ) \sqrt {b^{2} x^{4}+1}}{15}+\frac {2 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{3 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}-\frac {2 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{5 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\) | \(143\) |
elliptic | \(-\frac {b \,x^{3} \sqrt {b^{2} x^{4}+1}}{5}+\frac {x \sqrt {b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{3 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}-\frac {2 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{5 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\) | \(152\) |
default | \(-b \left (\frac {x^{3} \sqrt {b^{2} x^{4}+1}}{5}+\frac {2 i \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i b}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i b}, i\right )\right )}{5 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}\, b}\right )+\frac {x \sqrt {b^{2} x^{4}+1}}{3}+\frac {2 \sqrt {-i b \,x^{2}+1}\, \sqrt {i b \,x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i b}, i\right )}{3 \sqrt {i b}\, \sqrt {b^{2} x^{4}+1}}\) | \(158\) |
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 = 0.07 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.51 \[ \int \left (1-b x^2\right ) \sqrt {1+b^2 x^4} \, dx=-\frac {6 \, b x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 2 \, {\left (5 \, b^{2} + 3 \, b\right )} x \left (-\frac {1}{b^{2}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{b^{2}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (3 \, b^{2} x^{4} - 5 \, b x^{2} + 6\right )} \sqrt {b^{2} x^{4} + 1}}{15 \, b x} \] Input:
integrate((-b*x^2+1)*(b^2*x^4+1)^(1/2),x, algorithm="fricas")
Output:
-1/15*(6*b*x*(-1/b^2)^(3/4)*elliptic_e(arcsin((-1/b^2)^(1/4)/x), -1) - 2*( 5*b^2 + 3*b)*x*(-1/b^2)^(3/4)*elliptic_f(arcsin((-1/b^2)^(1/4)/x), -1) + ( 3*b^2*x^4 - 5*b*x^2 + 6)*sqrt(b^2*x^4 + 1))/(b*x)
Result contains complex when optimal does not.
Time = 0.90 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.38 \[ \int \left (1-b x^2\right ) \sqrt {1+b^2 x^4} \, dx=- \frac {b x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {b^{2} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] 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(I*pi))/( 4*gamma(7/4)) + x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), b**2*x**4*exp_pola r(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. \[ \int \left (1-b x^2\right ) \sqrt {1+b^2 x^4} \, dx=-\int \sqrt {b^2\,x^4+1}\,\left (b\,x^2-1\right ) \,d x \] Input:
int(-(b^2*x^4 + 1)^(1/2)*(b*x^2 - 1),x)
Output:
-int((b^2*x^4 + 1)^(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(sqrt(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