Integrand size = 25, antiderivative size = 52 \[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=d x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},\frac {e^2 x^8}{d^2}\right )+\frac {1}{5} e x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},\frac {e^2 x^8}{d^2}\right ) \] Output:
d*x*hypergeom([1/8, 1/2],[9/8],e^2*x^8/d^2)+1/5*e*x^5*hypergeom([1/2, 5/8] ,[13/8],e^2*x^8/d^2)
Time = 10.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=d x \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},\frac {e^2 x^8}{d^2}\right )+\frac {1}{5} e x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},\frac {e^2 x^8}{d^2}\right ) \] Input:
Integrate[(d + e*x^4)/Sqrt[1 - (e^2*x^8)/d^2],x]
Output:
d*x*Hypergeometric2F1[1/8, 1/2, 9/8, (e^2*x^8)/d^2] + (e*x^5*Hypergeometri c2F1[1/2, 5/8, 13/8, (e^2*x^8)/d^2])/5
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.26 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.63, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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.
\(\displaystyle \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {d+e x^4} \sqrt {\frac {1}{d}-\frac {e x^4}{d^2}} \int \frac {\sqrt {e x^4+d}}{\sqrt {\frac {1}{d}-\frac {e x^4}{d^2}}}dx}{\sqrt {1-\frac {e^2 x^8}{d^2}}}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\left (d+e x^4\right ) \sqrt {\frac {1}{d}-\frac {e x^4}{d^2}} \int \frac {\sqrt {\frac {e x^4}{d}+1}}{\sqrt {\frac {1}{d}-\frac {e x^4}{d^2}}}dx}{\sqrt {\frac {e x^4}{d}+1} \sqrt {1-\frac {e^2 x^8}{d^2}}}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\left (d+e x^4\right ) \sqrt {1-\frac {e x^4}{d}} \int \frac {\sqrt {\frac {e x^4}{d}+1}}{\sqrt {1-\frac {e x^4}{d}}}dx}{\sqrt {\frac {e x^4}{d}+1} \sqrt {1-\frac {e^2 x^8}{d^2}}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \left (d+e x^4\right ) \sqrt {1-\frac {e x^4}{d}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},-\frac {1}{2},\frac {5}{4},\frac {e x^4}{d},-\frac {e x^4}{d}\right )}{\sqrt {\frac {e x^4}{d}+1} \sqrt {1-\frac {e^2 x^8}{d^2}}}\) |
Input:
Int[(d + e*x^4)/Sqrt[1 - (e^2*x^8)/d^2],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[1 - (e^2*x^8)/d^2])
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])
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])
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])
Time = 3.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.83
method | result | size |
meijerg | \(d x \operatorname {hypergeom}\left (\left [\frac {1}{8}, \frac {1}{2}\right ], \left [\frac {9}{8}\right ], \frac {e^{2} x^{8}}{d^{2}}\right )+\frac {e \,x^{5} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {5}{8}\right ], \left [\frac {13}{8}\right ], \frac {e^{2} x^{8}}{d^{2}}\right )}{5}\) | \(43\) |
Input:
int((e*x^4+d)/(1-e^2*x^8/d^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
d*x*hypergeom([1/8,1/2],[9/8],e^2*x^8/d^2)+1/5*e*x^5*hypergeom([1/2,5/8],[ 13/8],e^2*x^8/d^2)
\[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-\frac {e^{2} x^{8}}{d^{2}} + 1}} \,d x } \] Input:
integrate((e*x^4+d)/(1-e^2*x^8/d^2)^(1/2),x, algorithm="fricas")
Output:
integral(-d^2*sqrt(-(e^2*x^8 - d^2)/d^2)/(e*x^4 - d), x)
Result contains complex when optimal does not.
Time = 1.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.50 \[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\frac {d x \Gamma \left (\frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{8}, \frac {1}{2} \\ \frac {9}{8} \end {matrix}\middle | {\frac {e^{2} x^{8} e^{2 i \pi }}{d^{2}}} \right )}}{8 \Gamma \left (\frac {9}{8}\right )} + \frac {e x^{5} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{8} \\ \frac {13}{8} \end {matrix}\middle | {\frac {e^{2} x^{8} e^{2 i \pi }}{d^{2}}} \right )}}{8 \Gamma \left (\frac {13}{8}\right )} \] Input:
integrate((e*x**4+d)/(1-e**2*x**8/d**2)**(1/2),x)
Output:
d*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*e xp_polar(2*I*pi)/d**2)/(8*gamma(13/8))
\[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-\frac {e^{2} x^{8}}{d^{2}} + 1}} \,d x } \] Input:
integrate((e*x^4+d)/(1-e^2*x^8/d^2)^(1/2),x, algorithm="maxima")
Output:
integrate((e*x^4 + d)/sqrt(-e^2*x^8/d^2 + 1), x)
\[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\int { \frac {e x^{4} + d}{\sqrt {-\frac {e^{2} x^{8}}{d^{2}} + 1}} \,d x } \] Input:
integrate((e*x^4+d)/(1-e^2*x^8/d^2)^(1/2),x, algorithm="giac")
Output:
integrate((e*x^4 + d)/sqrt(-e^2*x^8/d^2 + 1), x)
Timed out. \[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\int \frac {e\,x^4+d}{\sqrt {1-\frac {e^2\,x^8}{d^2}}} \,d x \] Input:
int((d + e*x^4)/(1 - (e^2*x^8)/d^2)^(1/2),x)
Output:
int((d + e*x^4)/(1 - (e^2*x^8)/d^2)^(1/2), x)
\[ \int \frac {d+e x^4}{\sqrt {1-\frac {e^2 x^8}{d^2}}} \, dx=\left (\int \frac {\sqrt {-e^{2} x^{8}+d^{2}}}{-e \,x^{4}+d}d x \right ) d \] Input:
int((e*x^4+d)/(1-e^2*x^8/d^2)^(1/2),x)
Output:
int(sqrt(d**2 - e**2*x**8)/(d - e*x**4),x)*d