Integrand size = 24, antiderivative size = 132 \[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\frac {c x \sqrt {1-\frac {d^2 x^8}{c^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{2},\frac {9}{8},\frac {d^2 x^8}{c^2}\right )}{\sqrt {c-d x^4} \sqrt {c+d x^4}}+\frac {d x^5 \sqrt {1-\frac {d^2 x^8}{c^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},\frac {d^2 x^8}{c^2}\right )}{5 \sqrt {c-d x^4} \sqrt {c+d x^4}} \] Output:
c*x*(1-d^2*x^8/c^2)^(1/2)*hypergeom([1/8, 1/2],[9/8],d^2*x^8/c^2)/(-d*x^4+ c)^(1/2)/(d*x^4+c)^(1/2)+1/5*d*x^5*(1-d^2*x^8/c^2)^(1/2)*hypergeom([1/2, 5 /8],[13/8],d^2*x^8/c^2)/(-d*x^4+c)^(1/2)/(d*x^4+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.75 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.52 \[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\frac {5 c^2 x \sqrt {c+d x^4} \sqrt {1-\frac {d x^4}{c}} \sqrt {1-\frac {d^2 x^8}{c^2}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},-\frac {1}{2},\frac {5}{4},\frac {d x^4}{c},-\frac {d x^4}{c}\right )}{\left (c-d x^4\right )^{3/2} \sqrt {1+\frac {d x^4}{c}} \left (5 c \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},-\frac {1}{2},\frac {5}{4},\frac {d x^4}{c},-\frac {d x^4}{c}\right )+2 d x^4 \left (\operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},-\frac {1}{2},\frac {9}{4},\frac {d x^4}{c},-\frac {d x^4}{c}\right )+\, _2F_1\left (\frac {1}{2},\frac {5}{8};\frac {13}{8};\frac {d^2 x^8}{c^2}\right )\right )\right )} \] Input:
Integrate[Sqrt[c + d*x^4]/Sqrt[c - d*x^4],x]
Output:
(5*c^2*x*Sqrt[c + d*x^4]*Sqrt[1 - (d*x^4)/c]*Sqrt[1 - (d^2*x^8)/c^2]*Appel lF1[1/4, 1/2, -1/2, 5/4, (d*x^4)/c, -((d*x^4)/c)])/((c - d*x^4)^(3/2)*Sqrt [1 + (d*x^4)/c]*(5*c*AppellF1[1/4, 1/2, -1/2, 5/4, (d*x^4)/c, -((d*x^4)/c) ] + 2*d*x^4*(AppellF1[5/4, 3/2, -1/2, 9/4, (d*x^4)/c, -((d*x^4)/c)] + Hype rgeometricPFQ[{1/2, 5/8}, {13/8}, (d^2*x^8)/c^2])))
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.37 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {1-\frac {d x^4}{c}} \int \frac {\sqrt {d x^4+c}}{\sqrt {1-\frac {d x^4}{c}}}dx}{\sqrt {c-d x^4}}\) |
\(\Big \downarrow \) 937 |
\(\displaystyle \frac {\sqrt {c+d x^4} \sqrt {1-\frac {d x^4}{c}} \int \frac {\sqrt {\frac {d x^4}{c}+1}}{\sqrt {1-\frac {d x^4}{c}}}dx}{\sqrt {c-d x^4} \sqrt {\frac {d x^4}{c}+1}}\) |
\(\Big \downarrow \) 936 |
\(\displaystyle \frac {x \sqrt {c+d x^4} \sqrt {1-\frac {d x^4}{c}} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},-\frac {1}{2},\frac {5}{4},\frac {d x^4}{c},-\frac {d x^4}{c}\right )}{\sqrt {c-d x^4} \sqrt {\frac {d x^4}{c}+1}}\) |
Input:
Int[Sqrt[c + d*x^4]/Sqrt[c - d*x^4],x]
Output:
(x*Sqrt[c + d*x^4]*Sqrt[1 - (d*x^4)/c]*AppellF1[1/4, 1/2, -1/2, 5/4, (d*x^ 4)/c, -((d*x^4)/c)])/(Sqrt[c - d*x^4]*Sqrt[1 + (d*x^4)/c])
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])
\[\int \frac {\sqrt {d \,x^{4}+c}}{\sqrt {-d \,x^{4}+c}}d x\]
Input:
int((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x)
Output:
int((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x)
\[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int { \frac {\sqrt {d x^{4} + c}}{\sqrt {-d x^{4} + c}} \,d x } \] Input:
integrate((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x, algorithm="fricas")
Output:
integral(-sqrt(d*x^4 + c)*sqrt(-d*x^4 + c)/(d*x^4 - c), x)
\[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int \frac {\sqrt {c + d x^{4}}}{\sqrt {c - d x^{4}}}\, dx \] Input:
integrate((d*x**4+c)**(1/2)/(-d*x**4+c)**(1/2),x)
Output:
Integral(sqrt(c + d*x**4)/sqrt(c - d*x**4), x)
\[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int { \frac {\sqrt {d x^{4} + c}}{\sqrt {-d x^{4} + c}} \,d x } \] Input:
integrate((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^4 + c)/sqrt(-d*x^4 + c), x)
\[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int { \frac {\sqrt {d x^{4} + c}}{\sqrt {-d x^{4} + c}} \,d x } \] Input:
integrate((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(d*x^4 + c)/sqrt(-d*x^4 + c), x)
Timed out. \[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int \frac {\sqrt {d\,x^4+c}}{\sqrt {c-d\,x^4}} \,d x \] Input:
int((c + d*x^4)^(1/2)/(c - d*x^4)^(1/2),x)
Output:
int((c + d*x^4)^(1/2)/(c - d*x^4)^(1/2), x)
\[ \int \frac {\sqrt {c+d x^4}}{\sqrt {c-d x^4}} \, dx=\int \frac {\sqrt {d \,x^{4}+c}\, \sqrt {-d \,x^{4}+c}}{-d \,x^{4}+c}d x \] Input:
int((d*x^4+c)^(1/2)/(-d*x^4+c)^(1/2),x)
Output:
int((sqrt(c + d*x**4)*sqrt(c - d*x**4))/(c - d*x**4),x)