\(\int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx\) [17]

Optimal result
Mathematica [A] (verified)
Rubi [C] (warning: unable to verify)
Maple [F]
Fricas [F]
Sympy [C] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 118 \[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\frac {d x \sqrt {1-\frac {e^2 x^8}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},\frac {e^2 x^8}{d^2}\right )}{\sqrt {d^2-e^2 x^8}}+\frac {e x^5 \sqrt {1-\frac {e^2 x^8}{d^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},\frac {e^2 x^8}{d^2}\right )}{5 \sqrt {d^2-e^2 x^8}} \] Output:

d*x*(1-e^2*x^8/d^2)^(1/2)*hypergeom([1/8, 1/2],[9/8],e^2*x^8/d^2)/(-e^2*x^ 
8+d^2)^(1/2)+1/5*e*x^5*(1-e^2*x^8/d^2)^(1/2)*hypergeom([1/2, 5/8],[13/8],e 
^2*x^8/d^2)/(-e^2*x^8+d^2)^(1/2)
 

Mathematica [A] (verified)

Time = 10.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74 \[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\frac {\sqrt {1-\frac {e^2 x^8}{d^2}} \left (5 d x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},\frac {e^2 x^8}{d^2}\right )+e x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},\frac {e^2 x^8}{d^2}\right )\right )}{5 \sqrt {d^2-e^2 x^8}} \] Input:

Integrate[(d + e*x^4)/Sqrt[d^2 - e^2*x^8],x]
 

Output:

(Sqrt[1 - (e^2*x^8)/d^2]*(5*d*x*Hypergeometric2F1[1/8, 1/2, 9/8, (e^2*x^8) 
/d^2] + e*x^5*Hypergeometric2F1[1/2, 5/8, 13/8, (e^2*x^8)/d^2]))/(5*Sqrt[d 
^2 - e^2*x^8])
 

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.71, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1396, 937, 937, 936}

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.

\(\displaystyle \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx\)

\(\Big \downarrow \) 1396

\(\displaystyle \frac {\sqrt {d-e x^4} \sqrt {d+e x^4} \int \frac {\sqrt {e x^4+d}}{\sqrt {d-e x^4}}dx}{\sqrt {d^2-e^2 x^8}}\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {\sqrt {d+e x^4} \sqrt {1-\frac {e x^4}{d}} \int \frac {\sqrt {e x^4+d}}{\sqrt {1-\frac {e x^4}{d}}}dx}{\sqrt {d^2-e^2 x^8}}\)

\(\Big \downarrow \) 937

\(\displaystyle \frac {\left (d+e x^4\right ) \sqrt {1-\frac {e x^4}{d}} \int \frac {\sqrt {\frac {e x^4}{d}+1}}{\sqrt {1-\frac {e x^4}{d}}}dx}{\sqrt {\frac {e x^4}{d}+1} \sqrt {d^2-e^2 x^8}}\)

\(\Big \downarrow \) 936

\(\displaystyle \frac {x \left (d+e x^4\right ) \sqrt {1-\frac {e x^4}{d}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},-\frac {1}{2},\frac {5}{4},\frac {e x^4}{d},-\frac {e x^4}{d}\right )}{\sqrt {\frac {e x^4}{d}+1} \sqrt {d^2-e^2 x^8}}\)

Input:

Int[(d + e*x^4)/Sqrt[d^2 - e^2*x^8],x]
 

Output:

(x*(d + e*x^4)*Sqrt[1 - (e*x^4)/d]*AppellF1[1/4, 1/2, -1/2, 5/4, (e*x^4)/d 
, -((e*x^4)/d)])/(Sqrt[1 + (e*x^4)/d]*Sqrt[d^2 - e^2*x^8])
 

Defintions of rubi rules used

rule 936
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 937
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) 
  Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, p, q 
}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1396
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x 
_Symbol] :> Simp[(a + c*x^(2*n))^FracPart[p]/((d + e*x^n)^FracPart[p]*(a/d 
+ c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, 
x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a* 
e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 1] && EqQ[n, 2])
 
Maple [F]

\[\int \frac {x^{4} e +d}{\sqrt {-e^{2} x^{8}+d^{2}}}d x\]

Input:

int((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x)
 

Output:

int((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x)
 

Fricas [F]

\[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-e^{2} x^{8} + d^{2}}} \,d x } \] Input:

integrate((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(-e^2*x^8 + d^2)/(e*x^4 - d), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.08 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\frac {x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, \frac {1}{2} \\ \frac {9}{8} \end {matrix}\middle | {\frac {e^{2} x^{8} e^{2 i \pi }}{d^{2}}} \right )}}{8 \Gamma \left (\frac {9}{8}\right )} + \frac {e x^{5} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{8} \\ \frac {13}{8} \end {matrix}\middle | {\frac {e^{2} x^{8} e^{2 i \pi }}{d^{2}}} \right )}}{8 d \Gamma \left (\frac {13}{8}\right )} \] Input:

integrate((e*x**4+d)/(-e**2*x**8+d**2)**(1/2),x)
 

Output:

x*gamma(1/8)*hyper((1/8, 1/2), (9/8,), e**2*x**8*exp_polar(2*I*pi)/d**2)/( 
8*gamma(9/8)) + e*x**5*gamma(5/8)*hyper((1/2, 5/8), (13/8,), e**2*x**8*exp 
_polar(2*I*pi)/d**2)/(8*d*gamma(13/8))
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-e^{2} x^{8} + d^{2}}} \,d x } \] Input:

integrate((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x, algorithm="maxima")
 

Output:

integrate((e*x^4 + d)/sqrt(-e^2*x^8 + d^2), x)
 

Giac [F]

\[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-e^{2} x^{8} + d^{2}}} \,d x } \] Input:

integrate((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x, algorithm="giac")
 

Output:

integrate((e*x^4 + d)/sqrt(-e^2*x^8 + d^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\int \frac {e\,x^4+d}{\sqrt {d^2-e^2\,x^8}} \,d x \] Input:

int((d + e*x^4)/(d^2 - e^2*x^8)^(1/2),x)
 

Output:

int((d + e*x^4)/(d^2 - e^2*x^8)^(1/2), x)
 

Reduce [F]

\[ \int \frac {d+e x^4}{\sqrt {d^2-e^2 x^8}} \, dx=\int \frac {\sqrt {-e^{2} x^{8}+d^{2}}}{-e \,x^{4}+d}d x \] Input:

int((e*x^4+d)/(-e^2*x^8+d^2)^(1/2),x)
 

Output:

int(sqrt(d**2 - e**2*x**8)/(d - e*x**4),x)