Integrand size = 19, antiderivative size = 73 \[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\frac {1}{15} x \left (5 d+3 e x^2\right ) \sqrt {9-x^4}+\frac {18}{5} \sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )+\frac {2}{5} \sqrt {3} (5 d-9 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right ) \] Output:
1/15*x*(3*e*x^2+5*d)*(-x^4+9)^(1/2)+18/5*3^(1/2)*e*EllipticE(1/3*x*3^(1/2) ,I)+2/5*3^(1/2)*(5*d-9*e)*EllipticF(1/3*x*3^(1/2),I)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.61 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.60 \[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=3 d x \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{4},\frac {5}{4},\frac {x^4}{9}\right )+e x^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {x^4}{9}\right ) \] Input:
Integrate[(d + e*x^2)*Sqrt[9 - x^4],x]
Output:
3*d*x*Hypergeometric2F1[-1/2, 1/4, 5/4, x^4/9] + e*x^3*Hypergeometric2F1[- 1/2, 3/4, 7/4, x^4/9]
Time = 0.39 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {1491, 27, 1495, 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.
\(\displaystyle \int \sqrt {9-x^4} \left (d+e x^2\right ) \, dx\) |
\(\Big \downarrow \) 1491 |
\(\displaystyle \frac {1}{15} \int \frac {18 \left (3 e x^2+5 d\right )}{\sqrt {9-x^4}}dx+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {6}{5} \int \frac {3 e x^2+5 d}{\sqrt {9-x^4}}dx+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 1495 |
\(\displaystyle \frac {6}{5} \int \frac {3 e x^2+5 d}{\sqrt {3-x^2} \sqrt {x^2+3}}dx+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {6}{5} \left ((5 d-9 e) \int \frac {1}{\sqrt {3-x^2} \sqrt {x^2+3}}dx+3 e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\right )+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 284 |
\(\displaystyle \frac {6}{5} \left ((5 d-9 e) \int \frac {1}{\sqrt {9-x^4}}dx+3 e \int \frac {\sqrt {x^2+3}}{\sqrt {3-x^2}}dx\right )+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {6}{5} \left ((5 d-9 e) \int \frac {1}{\sqrt {9-x^4}}dx+3 \sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\right )+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {6}{5} \left (\frac {(5 d-9 e) \operatorname {EllipticF}\left (\arcsin \left (\frac {x}{\sqrt {3}}\right ),-1\right )}{\sqrt {3}}+3 \sqrt {3} e E\left (\left .\arcsin \left (\frac {x}{\sqrt {3}}\right )\right |-1\right )\right )+\frac {1}{15} x \sqrt {9-x^4} \left (5 d+3 e x^2\right )\) |
Input:
Int[(d + e*x^2)*Sqrt[9 - x^4],x]
Output:
(x*(5*d + 3*e*x^2)*Sqrt[9 - x^4])/15 + (6*(3*Sqrt[3]*e*EllipticE[ArcSin[x/ Sqrt[3]], -1] + ((5*d - 9*e)*EllipticF[ArcSin[x/Sqrt[3]], -1])/Sqrt[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[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[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[((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[(-a)*c, 2]}, Simp[Sqrt[-c] Int[(d + e*x^2)/(Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c, 0]
Result contains higher order function than in optimal. Order 5 vs. order 4.
Time = 1.22 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.45
method | result | size |
meijerg | \(e \,x^{3} \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], \frac {x^{4}}{9}\right )+3 d x \operatorname {hypergeom}\left (\left [-\frac {1}{2}, \frac {1}{4}\right ], \left [\frac {5}{4}\right ], \frac {x^{4}}{9}\right )\) | \(33\) |
risch | \(-\frac {x \left (3 e \,x^{2}+5 d \right ) \left (x^{4}-9\right )}{15 \sqrt {-x^{4}+9}}+\frac {2 d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{3 \sqrt {-x^{4}+9}}-\frac {6 e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{5 \sqrt {-x^{4}+9}}\) | \(125\) |
elliptic | \(\frac {e \,x^{3} \sqrt {-x^{4}+9}}{5}+\frac {d x \sqrt {-x^{4}+9}}{3}+\frac {2 d \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{3 \sqrt {-x^{4}+9}}-\frac {6 e \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{5 \sqrt {-x^{4}+9}}\) | \(126\) |
default | \(d \left (\frac {x \sqrt {-x^{4}+9}}{3}+\frac {2 \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )}{3 \sqrt {-x^{4}+9}}\right )+e \left (\frac {x^{3} \sqrt {-x^{4}+9}}{5}-\frac {6 \sqrt {3}\, \sqrt {-3 x^{2}+9}\, \sqrt {3 x^{2}+9}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3}\, x}{3}, i\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3}\, x}{3}, i\right )\right )}{5 \sqrt {-x^{4}+9}}\right )\) | \(128\) |
Input:
int((e*x^2+d)*(-x^4+9)^(1/2),x,method=_RETURNVERBOSE)
Output:
e*x^3*hypergeom([-1/2,3/4],[7/4],1/9*x^4)+3*d*x*hypergeom([-1/2,1/4],[5/4] ,1/9*x^4)
Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\frac {-162 i \, \sqrt {3} e x E(\arcsin \left (\frac {\sqrt {3}}{x}\right )\,|\,-1) + 6 i \, \sqrt {3} {\left (5 \, d + 27 \, e\right )} x F(\arcsin \left (\frac {\sqrt {3}}{x}\right )\,|\,-1) + {\left (3 \, e x^{4} + 5 \, d x^{2} - 54 \, e\right )} \sqrt {-x^{4} + 9}}{15 \, x} \] Input:
integrate((e*x^2+d)*(-x^4+9)^(1/2),x, algorithm="fricas")
Output:
1/15*(-162*I*sqrt(3)*e*x*elliptic_e(arcsin(sqrt(3)/x), -1) + 6*I*sqrt(3)*( 5*d + 27*e)*x*elliptic_f(arcsin(sqrt(3)/x), -1) + (3*e*x^4 + 5*d*x^2 - 54* e)*sqrt(-x^4 + 9))/x
Time = 0.85 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\frac {3 d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {3 e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {x^{4} e^{2 i \pi }}{9}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \] Input:
integrate((e*x**2+d)*(-x**4+9)**(1/2),x)
Output:
3*d*x*gamma(1/4)*hyper((-1/2, 1/4), (5/4,), x**4*exp_polar(2*I*pi)/9)/(4*g amma(5/4)) + 3*e*x**3*gamma(3/4)*hyper((-1/2, 3/4), (7/4,), x**4*exp_polar (2*I*pi)/9)/(4*gamma(7/4))
\[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\int { \sqrt {-x^{4} + 9} {\left (e x^{2} + d\right )} \,d x } \] Input:
integrate((e*x^2+d)*(-x^4+9)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(-x^4 + 9)*(e*x^2 + d), x)
\[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\int { \sqrt {-x^{4} + 9} {\left (e x^{2} + d\right )} \,d x } \] Input:
integrate((e*x^2+d)*(-x^4+9)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(-x^4 + 9)*(e*x^2 + d), x)
Timed out. \[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\int \sqrt {9-x^4}\,\left (e\,x^2+d\right ) \,d x \] Input:
int((9 - x^4)^(1/2)*(d + e*x^2),x)
Output:
int((9 - x^4)^(1/2)*(d + e*x^2), x)
\[ \int \left (d+e x^2\right ) \sqrt {9-x^4} \, dx=\frac {\sqrt {-x^{4}+9}\, d x}{3}+\frac {\sqrt {-x^{4}+9}\, e \,x^{3}}{5}-6 \left (\int \frac {\sqrt {-x^{4}+9}}{x^{4}-9}d x \right ) d -\frac {18 \left (\int \frac {\sqrt {-x^{4}+9}\, x^{2}}{x^{4}-9}d x \right ) e}{5} \] Input:
int((e*x^2+d)*(-x^4+9)^(1/2),x)
Output:
(5*sqrt( - x**4 + 9)*d*x + 3*sqrt( - x**4 + 9)*e*x**3 - 90*int(sqrt( - x** 4 + 9)/(x**4 - 9),x)*d - 54*int((sqrt( - x**4 + 9)*x**2)/(x**4 - 9),x)*e)/ 15