\(\int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx\) [159]

   Optimal result
   Rubi [A] (verified)
   Mathematica [C] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 20, antiderivative size = 124 \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\frac {a^{3/4} e \sqrt {1-\frac {c x^4}{a}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt {a-c x^4}}+\frac {a^{3/4} \left (\frac {\sqrt {c} d}{\sqrt {a}}-e\right ) \sqrt {1-\frac {c x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{c^{3/4} \sqrt {a-c x^4}} \]

[Out]

a^(3/4)*e*EllipticE(c^(1/4)*x/a^(1/4),I)*(1-c*x^4/a)^(1/2)/c^(3/4)/(-c*x^4+a)^(1/2)+a^(3/4)*EllipticF(c^(1/4)*
x/a^(1/4),I)*(-e+d*c^(1/2)/a^(1/2))*(1-c*x^4/a)^(1/2)/c^(3/4)/(-c*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1215, 230, 227, 1214, 1213, 435} \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\frac {a^{3/4} \sqrt {1-\frac {c x^4}{a}} \left (\frac {\sqrt {c} d}{\sqrt {a}}-e\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),-1\right )}{c^{3/4} \sqrt {a-c x^4}}+\frac {a^{3/4} e \sqrt {1-\frac {c x^4}{a}} E\left (\left .\arcsin \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt {a-c x^4}} \]

[In]

Int[(d + e*x^2)/Sqrt[a - c*x^4],x]

[Out]

(a^(3/4)*e*Sqrt[1 - (c*x^4)/a]*EllipticE[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(c^(3/4)*Sqrt[a - c*x^4]) + (a^(3/4
)*((Sqrt[c]*d)/Sqrt[a] - e)*Sqrt[1 - (c*x^4)/a]*EllipticF[ArcSin[(c^(1/4)*x)/a^(1/4)], -1])/(c^(3/4)*Sqrt[a -
c*x^4])

Rule 227

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[EllipticF[ArcSin[Rt[-b, 4]*(x/Rt[a, 4])], -1]/(Rt[a, 4]*Rt[
-b, 4]), x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

Rule 435

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*Ell
ipticE[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
]

Rule 1213

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + e*(x^2/d)]/Sqrt
[1 - e*(x^2/d)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1214

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + c*(x^4/a)]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + c*(x^4/a)], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rule 1215

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-c/a, 2]}, Dist[(d*q - e)/q, In
t[1/Sqrt[a + c*x^4], x], x] + Dist[e/q, Int[(1 + q*x^2)/Sqrt[a + c*x^4], x], x]] /; FreeQ[{a, c, d, e}, x] &&
NegQ[c/a] && NeQ[c*d^2 + a*e^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {a} e\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a-c x^4}} \, dx}{\sqrt {c}}+\left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {a-c x^4}} \, dx \\ & = \frac {\left (\sqrt {a} e \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {1-\frac {c x^4}{a}}} \, dx}{\sqrt {c} \sqrt {a-c x^4}}+\frac {\left (\left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {c x^4}{a}}} \, dx}{\sqrt {a-c x^4}} \\ & = \frac {\sqrt [4]{a} \left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} \sqrt {a-c x^4}}+\frac {\left (\sqrt {a} e \sqrt {1-\frac {c x^4}{a}}\right ) \int \frac {\sqrt {1+\frac {\sqrt {c} x^2}{\sqrt {a}}}}{\sqrt {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}} \, dx}{\sqrt {c} \sqrt {a-c x^4}} \\ & = \frac {a^{3/4} e \sqrt {1-\frac {c x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{c^{3/4} \sqrt {a-c x^4}}+\frac {\sqrt [4]{a} \left (d-\frac {\sqrt {a} e}{\sqrt {c}}\right ) \sqrt {1-\frac {c x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )\right |-1\right )}{\sqrt [4]{c} \sqrt {a-c x^4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.62 \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\frac {\sqrt {1-\frac {c x^4}{a}} \left (3 d x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {c x^4}{a}\right )+e x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},\frac {c x^4}{a}\right )\right )}{3 \sqrt {a-c x^4}} \]

[In]

Integrate[(d + e*x^2)/Sqrt[a - c*x^4],x]

[Out]

(Sqrt[1 - (c*x^4)/a]*(3*d*x*Hypergeometric2F1[1/4, 1/2, 5/4, (c*x^4)/a] + e*x^3*Hypergeometric2F1[1/2, 3/4, 7/
4, (c*x^4)/a]))/(3*Sqrt[a - c*x^4])

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.24

method result size
default \(\frac {d \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {e \sqrt {a}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, \sqrt {c}}\) \(154\)
elliptic \(\frac {d \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}}-\frac {e \sqrt {a}\, \sqrt {1-\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {\sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {\sqrt {c}}{\sqrt {a}}}, i\right )\right )}{\sqrt {\frac {\sqrt {c}}{\sqrt {a}}}\, \sqrt {-c \,x^{4}+a}\, \sqrt {c}}\) \(154\)

[In]

int((e*x^2+d)/(-c*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

d/(1/a^(1/2)*c^(1/2))^(1/2)*(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(-c*x^4+a)^(1/2)*E
llipticF(x*(1/a^(1/2)*c^(1/2))^(1/2),I)-e*a^(1/2)/(1/a^(1/2)*c^(1/2))^(1/2)*(1-1/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1
+1/a^(1/2)*c^(1/2)*x^2)^(1/2)/(-c*x^4+a)^(1/2)/c^(1/2)*(EllipticF(x*(1/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(
1/a^(1/2)*c^(1/2))^(1/2),I))

Fricas [A] (verification not implemented)

none

Time = 0.10 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=-\frac {a \sqrt {-c} e x \left (\frac {a}{c}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - {\left (c d + a e\right )} \sqrt {-c} x \left (\frac {a}{c}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{c}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {-c x^{4} + a} a e}{a c x} \]

[In]

integrate((e*x^2+d)/(-c*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

-(a*sqrt(-c)*e*x*(a/c)^(3/4)*elliptic_e(arcsin((a/c)^(1/4)/x), -1) - (c*d + a*e)*sqrt(-c)*x*(a/c)^(3/4)*ellipt
ic_f(arcsin((a/c)^(1/4)/x), -1) + sqrt(-c*x^4 + a)*a*e)/(a*c*x)

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\frac {d x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {5}{4}\right )} + \frac {e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {c x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {7}{4}\right )} \]

[In]

integrate((e*x**2+d)/(-c*x**4+a)**(1/2),x)

[Out]

d*x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(5/4)) + e*x**3*gamma(3/4
)*hyper((1/2, 3/4), (7/4,), c*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(7/4))

Maxima [F]

\[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {-c x^{4} + a}} \,d x } \]

[In]

integrate((e*x^2+d)/(-c*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)/sqrt(-c*x^4 + a), x)

Giac [F]

\[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\int { \frac {e x^{2} + d}{\sqrt {-c x^{4} + a}} \,d x } \]

[In]

integrate((e*x^2+d)/(-c*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)/sqrt(-c*x^4 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x^2}{\sqrt {a-c x^4}} \, dx=\int \frac {e\,x^2+d}{\sqrt {a-c\,x^4}} \,d x \]

[In]

int((d + e*x^2)/(a - c*x^4)^(1/2),x)

[Out]

int((d + e*x^2)/(a - c*x^4)^(1/2), x)