Integrand size = 23, antiderivative size = 150 \[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\frac {x \sqrt {1+e^4 x^4}}{1+e^2 x^2}-\frac {\left (1+e^2 x^2\right ) \sqrt {\frac {1+e^4 x^4}{\left (1+e^2 x^2\right )^2}} E\left (2 \arctan (e x)\left |\frac {1}{2}\right .\right )}{e \sqrt {1+e^4 x^4}}+\frac {\left (1+e^2 x^2\right ) \sqrt {\frac {1+e^4 x^4}{\left (1+e^2 x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (e x),\frac {1}{2}\right )}{e \sqrt {1+e^4 x^4}} \] Output:
x*(e^4*x^4+1)^(1/2)/(e^2*x^2+1)-(e^2*x^2+1)*((e^4*x^4+1)/(e^2*x^2+1)^2)^(1 /2)*EllipticE(sin(2*arctan(e*x)),1/2*2^(1/2))/e/(e^4*x^4+1)^(1/2)+(e^2*x^2 +1)*((e^4*x^4+1)/(e^2*x^2+1)^2)^(1/2)*InverseJacobiAM(2*arctan(e*x),1/2*2^ (1/2))/e/(e^4*x^4+1)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.33 \[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^4 x^4\right )+\frac {1}{3} e^2 x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^4 x^4\right ) \] Input:
Integrate[(1 + e^2*x^2)/Sqrt[1 + e^4*x^4],x]
Output:
x*Hypergeometric2F1[1/4, 1/2, 5/4, -(e^4*x^4)] + (e^2*x^3*Hypergeometric2F 1[1/2, 3/4, 7/4, -(e^4*x^4)])/3
Time = 0.41 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1512, 761, 1510}
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.
\(\displaystyle \int \frac {e^2 x^2+1}{\sqrt {e^4 x^4+1}} \, dx\) |
\(\Big \downarrow \) 1512 |
\(\displaystyle 2 \int \frac {1}{\sqrt {e^4 x^4+1}}dx-\int \frac {1-e^2 x^2}{\sqrt {e^4 x^4+1}}dx\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (e^2 x^2+1\right ) \sqrt {\frac {e^4 x^4+1}{\left (e^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (e x),\frac {1}{2}\right )}{e \sqrt {e^4 x^4+1}}-\int \frac {1-e^2 x^2}{\sqrt {e^4 x^4+1}}dx\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {\left (e^2 x^2+1\right ) \sqrt {\frac {e^4 x^4+1}{\left (e^2 x^2+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan (e x),\frac {1}{2}\right )}{e \sqrt {e^4 x^4+1}}-\frac {\left (e^2 x^2+1\right ) \sqrt {\frac {e^4 x^4+1}{\left (e^2 x^2+1\right )^2}} E\left (2 \arctan (e x)\left |\frac {1}{2}\right .\right )}{e \sqrt {e^4 x^4+1}}+\frac {x \sqrt {e^4 x^4+1}}{e^2 x^2+1}\) |
Input:
Int[(1 + e^2*x^2)/Sqrt[1 + e^4*x^4],x]
Output:
(x*Sqrt[1 + e^4*x^4])/(1 + e^2*x^2) - ((1 + e^2*x^2)*Sqrt[(1 + e^4*x^4)/(1 + e^2*x^2)^2]*EllipticE[2*ArcTan[e*x], 1/2])/(e*Sqrt[1 + e^4*x^4]) + ((1 + e^2*x^2)*Sqrt[(1 + e^4*x^4)/(1 + e^2*x^2)^2]*EllipticF[2*ArcTan[e*x], 1/ 2])/(e*Sqrt[1 + e^4*x^4])
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Time = 0.69 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.27
method | result | size |
meijerg | \(\frac {e^{2} x^{3} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -e^{4} x^{4}\right )}{3}+x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {5}{4}\right ], -e^{4} x^{4}\right )\) | \(40\) |
default | \(\frac {\sqrt {-i e^{2} x^{2}+1}\, \sqrt {i e^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i e^{2}}, i\right )}{\sqrt {i e^{2}}\, \sqrt {e^{4} x^{4}+1}}+\frac {i \sqrt {-i e^{2} x^{2}+1}\, \sqrt {i e^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i e^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i e^{2}}, i\right )\right )}{\sqrt {i e^{2}}\, \sqrt {e^{4} x^{4}+1}}\) | \(138\) |
elliptic | \(\frac {\sqrt {-i e^{2} x^{2}+1}\, \sqrt {i e^{2} x^{2}+1}\, \operatorname {EllipticF}\left (x \sqrt {i e^{2}}, i\right )}{\sqrt {i e^{2}}\, \sqrt {e^{4} x^{4}+1}}+\frac {i \sqrt {-i e^{2} x^{2}+1}\, \sqrt {i e^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {i e^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {i e^{2}}, i\right )\right )}{\sqrt {i e^{2}}\, \sqrt {e^{4} x^{4}+1}}\) | \(138\) |
Input:
int((e^2*x^2+1)/(e^4*x^4+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/3*e^2*x^3*hypergeom([1/2,3/4],[7/4],-e^4*x^4)+x*hypergeom([1/4,1/2],[5/4 ],-e^4*x^4)
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.51 \[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\frac {e^{2} x \left (-\frac {1}{e^{4}}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {1}{e^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (e^{4} - e^{2}\right )} x \left (-\frac {1}{e^{4}}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {1}{e^{4}}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + \sqrt {e^{4} x^{4} + 1}}{e^{2} x} \] Input:
integrate((e^2*x^2+1)/(e^4*x^4+1)^(1/2),x, algorithm="fricas")
Output:
(e^2*x*(-1/e^4)^(3/4)*elliptic_e(arcsin((-1/e^4)^(1/4)/x), -1) + (e^4 - e^ 2)*x*(-1/e^4)^(3/4)*elliptic_f(arcsin((-1/e^4)^(1/4)/x), -1) + sqrt(e^4*x^ 4 + 1))/(e^2*x)
Result contains complex when optimal does not.
Time = 0.80 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.45 \[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\frac {e^{2} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {e^{4} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {5}{4} \end {matrix}\middle | {e^{4} x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate((e**2*x**2+1)/(e**4*x**4+1)**(1/2),x)
Output:
e**2*x**3*gamma(3/4)*hyper((1/2, 3/4), (7/4,), e**4*x**4*exp_polar(I*pi))/ (4*gamma(7/4)) + x*gamma(1/4)*hyper((1/4, 1/2), (5/4,), e**4*x**4*exp_pola r(I*pi))/(4*gamma(5/4))
\[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\int { \frac {e^{2} x^{2} + 1}{\sqrt {e^{4} x^{4} + 1}} \,d x } \] Input:
integrate((e^2*x^2+1)/(e^4*x^4+1)^(1/2),x, algorithm="maxima")
Output:
integrate((e^2*x^2 + 1)/sqrt(e^4*x^4 + 1), x)
\[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\int { \frac {e^{2} x^{2} + 1}{\sqrt {e^{4} x^{4} + 1}} \,d x } \] Input:
integrate((e^2*x^2+1)/(e^4*x^4+1)^(1/2),x, algorithm="giac")
Output:
integrate((e^2*x^2 + 1)/sqrt(e^4*x^4 + 1), x)
Timed out. \[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\int \frac {e^2\,x^2+1}{\sqrt {e^4\,x^4+1}} \,d x \] Input:
int((e^2*x^2 + 1)/(e^4*x^4 + 1)^(1/2),x)
Output:
int((e^2*x^2 + 1)/(e^4*x^4 + 1)^(1/2), x)
\[ \int \frac {1+e^2 x^2}{\sqrt {1+e^4 x^4}} \, dx=\int \frac {\sqrt {e^{4} x^{4}+1}}{e^{4} x^{4}+1}d x +\left (\int \frac {\sqrt {e^{4} x^{4}+1}\, x^{2}}{e^{4} x^{4}+1}d x \right ) e^{2} \] Input:
int((e^2*x^2+1)/(e^4*x^4+1)^(1/2),x)
Output:
int(sqrt(e**4*x**4 + 1)/(e**4*x**4 + 1),x) + int((sqrt(e**4*x**4 + 1)*x**2 )/(e**4*x**4 + 1),x)*e**2