Integrand size = 28, antiderivative size = 54 \[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \sqrt {e} x \sqrt {d+e x^2}}{\sqrt {d^2-e^2 x^4}}\right )}{\sqrt {2} d \sqrt {e}} \] Output:
1/2*arctan(2^(1/2)*e^(1/2)*x*(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2))*2^(1/2) /d/e^(1/2)
Time = 1.63 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\sqrt {d^2-e^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d-e x^2}}\right )}{\sqrt {2} d \sqrt {e} \sqrt {d-e x^2} \sqrt {d+e x^2}} \] Input:
Integrate[1/(Sqrt[d + e*x^2]*Sqrt[d^2 - e^2*x^4]),x]
Output:
(Sqrt[d^2 - e^2*x^4]*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d - e*x^2]])/(Sqrt[2] *d*Sqrt[e]*Sqrt[d - e*x^2]*Sqrt[d + e*x^2])
Time = 0.32 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {1396, 291, 218}
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.
\(\displaystyle \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \int \frac {1}{\sqrt {d-e x^2} \left (e x^2+d\right )}dx}{\sqrt {d^2-e^2 x^4}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \int \frac {1}{\frac {2 d e x^2}{d-e x^2}+d}d\frac {x}{\sqrt {d-e x^2}}}{\sqrt {d^2-e^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {d-e x^2} \sqrt {d+e x^2} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d-e x^2}}\right )}{\sqrt {2} d \sqrt {e} \sqrt {d^2-e^2 x^4}}\) |
Input:
Int[1/(Sqrt[d + e*x^2]*Sqrt[d^2 - e^2*x^4]),x]
Output:
(Sqrt[d - e*x^2]*Sqrt[d + e*x^2]*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d - e*x^2 ]])/(Sqrt[2]*d*Sqrt[e]*Sqrt[d^2 - e^2*x^4])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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])
Leaf count of result is larger than twice the leaf count of optimal. \(248\) vs. \(2(43)=86\).
Time = 0.81 (sec) , antiderivative size = 249, normalized size of antiderivative = 4.61
method | result | size |
default | \(\frac {\sqrt {-e^{2} x^{4}+d^{2}}\, \sqrt {e}\, \left (\sqrt {d}\, \sqrt {2}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}-\sqrt {-d e}\, x +d \right )}{e x -\sqrt {-d e}}\right ) \sqrt {e}-\sqrt {d}\, \sqrt {2}\, \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {d}\, \sqrt {-e \,x^{2}+d}+\sqrt {-d e}\, x +d \right )}{e x +\sqrt {-d e}}\right ) \sqrt {e}-2 \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (e x +\sqrt {d e}\right ) \left (-e x +\sqrt {d e}\right )}{e}}}\right )+2 \sqrt {-d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}+d}}\right )\right )}{2 \sqrt {e \,x^{2}+d}\, \sqrt {-e \,x^{2}+d}\, \left (-\sqrt {d e}+\sqrt {-d e}\right ) \left (\sqrt {d e}+\sqrt {-d e}\right ) \sqrt {-d e}}\) | \(249\) |
Input:
int(1/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/2*(-e^2*x^4+d^2)^(1/2)*e^(1/2)*(d^(1/2)*2^(1/2)*ln(2*e*(2^(1/2)*d^(1/2)* (-e*x^2+d)^(1/2)-(-d*e)^(1/2)*x+d)/(e*x-(-d*e)^(1/2)))*e^(1/2)-d^(1/2)*2^( 1/2)*ln(2*e*(2^(1/2)*d^(1/2)*(-e*x^2+d)^(1/2)+(-d*e)^(1/2)*x+d)/(e*x+(-d*e )^(1/2)))*e^(1/2)-2*(-d*e)^(1/2)*arctan(e^(1/2)*x/((e*x+(d*e)^(1/2))/e*(-e *x+(d*e)^(1/2)))^(1/2))+2*(-d*e)^(1/2)*arctan(e^(1/2)*x/(-e*x^2+d)^(1/2))) /(e*x^2+d)^(1/2)/(-e*x^2+d)^(1/2)/(-(d*e)^(1/2)+(-d*e)^(1/2))/((d*e)^(1/2) +(-d*e)^(1/2))/(-d*e)^(1/2)
Time = 0.08 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.83 \[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\left [-\frac {\sqrt {2} \sqrt {-e} \log \left (-\frac {3 \, e^{2} x^{4} + 2 \, d e x^{2} - 2 \, \sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {-e} x - d^{2}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e}, -\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d} \sqrt {e} x}{e^{2} x^{4} - d^{2}}\right )}{2 \, d \sqrt {e}}\right ] \] Input:
integrate(1/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x, algorithm="fricas")
Output:
[-1/4*sqrt(2)*sqrt(-e)*log(-(3*e^2*x^4 + 2*d*e*x^2 - 2*sqrt(2)*sqrt(-e^2*x ^4 + d^2)*sqrt(e*x^2 + d)*sqrt(-e)*x - d^2)/(e^2*x^4 + 2*d*e*x^2 + d^2))/( d*e), -1/2*sqrt(2)*arctan(sqrt(2)*sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d)*sqr t(e)*x/(e^2*x^4 - d^2))/(d*sqrt(e))]
\[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int \frac {1}{\sqrt {- \left (- d + e x^{2}\right ) \left (d + e x^{2}\right )} \sqrt {d + e x^{2}}}\, dx \] Input:
integrate(1/(e*x**2+d)**(1/2)/(-e**2*x**4+d**2)**(1/2),x)
Output:
Integral(1/(sqrt(-(-d + e*x**2)*(d + e*x**2))*sqrt(d + e*x**2)), x)
\[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int { \frac {1}{\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d}} \,d x } \] Input:
integrate(1/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d)), x)
\[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int { \frac {1}{\sqrt {-e^{2} x^{4} + d^{2}} \sqrt {e x^{2} + d}} \,d x } \] Input:
integrate(1/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x, algorithm="giac")
Output:
integrate(1/(sqrt(-e^2*x^4 + d^2)*sqrt(e*x^2 + d)), x)
Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\int \frac {1}{\sqrt {d^2-e^2\,x^4}\,\sqrt {e\,x^2+d}} \,d x \] Input:
int(1/((d^2 - e^2*x^4)^(1/2)*(d + e*x^2)^(1/2)),x)
Output:
int(1/((d^2 - e^2*x^4)^(1/2)*(d + e*x^2)^(1/2)), x)
Time = 0.24 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {d+e x^2} \sqrt {d^2-e^2 x^4}} \, dx=\frac {\sqrt {e}\, \sqrt {2}\, \left (2 \mathit {atan} \left (\frac {\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )}{\sqrt {2}+1}\right )-\mathrm {log}\left (-\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )+i \right ) i +\mathrm {log}\left (\sqrt {2}\, i +\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right )}{2}\right )-i \right ) i \right )}{4 d e} \] Input:
int(1/(e*x^2+d)^(1/2)/(-e^2*x^4+d^2)^(1/2),x)
Output:
(sqrt(e)*sqrt(2)*(2*atan(tan(asin((sqrt(e)*x)/sqrt(d))/2)/(sqrt(2) + 1)) - log( - sqrt(2)*i + tan(asin((sqrt(e)*x)/sqrt(d))/2) + i)*i + log(sqrt(2)* i + tan(asin((sqrt(e)*x)/sqrt(d))/2) - i)*i))/(4*d*e)