Integrand size = 37, antiderivative size = 109 \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=-\frac {B \arctan \left (\frac {\sqrt {e} x \sqrt {d-e x^2}}{\sqrt {-d^2+e^2 x^4}}\right )}{e^{3/2}}+\frac {(B d+A e) \arctan \left (\frac {\sqrt {2} \sqrt {e} x \sqrt {d-e x^2}}{\sqrt {-d^2+e^2 x^4}}\right )}{\sqrt {2} d e^{3/2}} \] Output:
-B*arctan(e^(1/2)*x*(-e*x^2+d)^(1/2)/(e^2*x^4-d^2)^(1/2))/e^(3/2)+1/2*(A*e +B*d)*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^(3/2)
Result contains complex when optimal does not.
Time = 5.74 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.28 \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {\frac {(B d+A e) \sqrt {-2 d^2+2 e^2 x^4} \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{d \sqrt {-d-e x^2} \sqrt {d-e x^2}}-2 i B \log \left (-2 i \sqrt {e} x+\frac {2 \sqrt {-d^2+e^2 x^4}}{\sqrt {d-e x^2}}\right )}{2 e^{3/2}} \] Input:
Integrate[(A + B*x^2)/(Sqrt[d - e*x^2]*Sqrt[-d^2 + e^2*x^4]),x]
Output:
(((B*d + A*e)*Sqrt[-2*d^2 + 2*e^2*x^4]*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[-d - e*x^2]])/(d*Sqrt[-d - e*x^2]*Sqrt[d - e*x^2]) - (2*I)*B*Log[(-2*I)*Sqrt[ e]*x + (2*Sqrt[-d^2 + e^2*x^4])/Sqrt[d - e*x^2]])/(2*e^(3/2))
Time = 0.27 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {1396, 398, 224, 216, 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.
\(\displaystyle \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {e^2 x^4-d^2}} \, dx\) |
\(\Big \downarrow \) 1396 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \int \frac {B x^2+A}{\sqrt {-e x^2-d} \left (d-e x^2\right )}dx}{\sqrt {e^2 x^4-d^2}}\) |
\(\Big \downarrow \) 398 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {-e x^2-d} \left (d-e x^2\right )}dx}{e}-\frac {B \int \frac {1}{\sqrt {-e x^2-d}}dx}{e}\right )}{\sqrt {e^2 x^4-d^2}}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {-e x^2-d} \left (d-e x^2\right )}dx}{e}-\frac {B \int \frac {1}{\frac {e x^2}{-e x^2-d}+1}d\frac {x}{\sqrt {-e x^2-d}}}{e}\right )}{\sqrt {e^2 x^4-d^2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {(A e+B d) \int \frac {1}{\sqrt {-e x^2-d} \left (d-e x^2\right )}dx}{e}-\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{e^{3/2}}\right )}{\sqrt {e^2 x^4-d^2}}\) |
\(\Big \downarrow \) 291 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {(A e+B d) \int \frac {1}{\frac {2 d e x^2}{-e x^2-d}+d}d\frac {x}{\sqrt {-e x^2-d}}}{e}-\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{e^{3/2}}\right )}{\sqrt {e^2 x^4-d^2}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\sqrt {-d-e x^2} \sqrt {d-e x^2} \left (\frac {(A e+B d) \arctan \left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{\sqrt {2} d e^{3/2}}-\frac {B \arctan \left (\frac {\sqrt {e} x}{\sqrt {-d-e x^2}}\right )}{e^{3/2}}\right )}{\sqrt {e^2 x^4-d^2}}\) |
Input:
Int[(A + B*x^2)/(Sqrt[d - e*x^2]*Sqrt[-d^2 + e^2*x^4]),x]
Output:
(Sqrt[-d - e*x^2]*Sqrt[d - e*x^2]*(-((B*ArcTan[(Sqrt[e]*x)/Sqrt[-d - e*x^2 ]])/e^(3/2)) + ((B*d + A*e)*ArcTan[(Sqrt[2]*Sqrt[e]*x)/Sqrt[-d - e*x^2]])/ (Sqrt[2]*d*e^(3/2))))/Sqrt[-d^2 + e^2*x^4]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]) , x_Symbol] :> Simp[f/b Int[1/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/ b Int[1/((a + b*x^2)*Sqrt[c + d*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f} , x]
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. \(479\) vs. \(2(90)=180\).
Time = 0.07 (sec) , antiderivative size = 480, normalized size of antiderivative = 4.40
method | result | size |
default | \(-\frac {\sqrt {e^{2} x^{4}-d^{2}}\, \left (A \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {-d e}\right ) \left (e x +\sqrt {-d e}\right )}{e}}}\right ) \sqrt {2}\, \sqrt {-d}\, e -A \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}-d}}\right ) \sqrt {2}\, \sqrt {-d}\, e +A \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-\sqrt {d e}\, x -d \right )}{e x -\sqrt {d e}}\right ) e^{\frac {3}{2}} d -A \ln \left (\frac {2 e \left (\sqrt {d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-d \right )}{e x +\sqrt {d e}}\right ) e^{\frac {3}{2}} d -B \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {\frac {\left (-e x +\sqrt {-d e}\right ) \left (e x +\sqrt {-d e}\right )}{e}}}\right ) \sqrt {2}\, \sqrt {-d}\, d -B \sqrt {d e}\, \arctan \left (\frac {\sqrt {e}\, x}{\sqrt {-e \,x^{2}-d}}\right ) \sqrt {2}\, \sqrt {-d}\, d +B \ln \left (\frac {2 e \left (\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-\sqrt {d e}\, x -d \right )}{e x -\sqrt {d e}}\right ) \sqrt {e}\, d^{2}-B \ln \left (\frac {2 e \left (\sqrt {d e}\, x +\sqrt {2}\, \sqrt {-d}\, \sqrt {-e \,x^{2}-d}-d \right )}{e x +\sqrt {d e}}\right ) \sqrt {e}\, d^{2}\right ) \sqrt {2}}{2 \sqrt {-e \,x^{2}+d}\, \sqrt {-e \,x^{2}-d}\, \left (\sqrt {-d e}-\sqrt {d e}\right ) \left (\sqrt {-d e}+\sqrt {d e}\right ) \sqrt {e}\, \sqrt {d e}\, \sqrt {-d}}\) | \(480\) |
Input:
int((B*x^2+A)/(-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)*(A*(d*e)^(1/2)*arctan(e^(1/2)*x/(1/e*(-e*x+(-d*e) ^(1/2))*(e*x+(-d*e)^(1/2)))^(1/2))*2^(1/2)*(-d)^(1/2)*e-A*(d*e)^(1/2)*arct an(e^(1/2)*x/(-e*x^2-d)^(1/2))*2^(1/2)*(-d)^(1/2)*e+A*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^(3/2)*d-A*ln (2*e*((d*e)^(1/2)*x+2^(1/2)*(-d)^(1/2)*(-e*x^2-d)^(1/2)-d)/(e*x+(d*e)^(1/2 )))*e^(3/2)*d-B*(d*e)^(1/2)*arctan(e^(1/2)*x/(1/e*(-e*x+(-d*e)^(1/2))*(e*x +(-d*e)^(1/2)))^(1/2))*2^(1/2)*(-d)^(1/2)*d-B*(d*e)^(1/2)*arctan(e^(1/2)*x /(-e*x^2-d)^(1/2))*2^(1/2)*(-d)^(1/2)*d+B*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^2-B*ln(2*e*((d*e )^(1/2)*x+2^(1/2)*(-d)^(1/2)*(-e*x^2-d)^(1/2)-d)/(e*x+(d*e)^(1/2)))*e^(1/2 )*d^2)*2^(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))/e^(1/2)/(d*e)^(1/2)/(-d)^(1/2)
Time = 0.10 (sec) , antiderivative size = 278, normalized size of antiderivative = 2.55 \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\left [-\frac {2 \, B d \sqrt {-e} \log \left (\frac {2 \, e^{2} x^{4} - d e x^{2} - 2 \, \sqrt {e^{2} x^{4} - d^{2}} \sqrt {-e x^{2} + d} \sqrt {-e} x - d^{2}}{e x^{2} - d}\right ) + \sqrt {2} {\left (B d + A e\right )} \sqrt {-e} \log \left (-\frac {3 \, e^{2} x^{4} - 2 \, d e x^{2} + 2 \, \sqrt {2} \sqrt {e^{2} x^{4} - d^{2}} \sqrt {-e x^{2} + d} \sqrt {-e} x - d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right )}{4 \, d e^{2}}, -\frac {2 \, B d \sqrt {e} \arctan \left (\frac {\sqrt {-e x^{2} + d} \sqrt {e} x}{\sqrt {e^{2} x^{4} - d^{2}}}\right ) - \sqrt {2} {\left (B d + A e\right )} \sqrt {e} \arctan \left (\frac {\sqrt {2} \sqrt {-e x^{2} + d} \sqrt {e} x}{\sqrt {e^{2} x^{4} - d^{2}}}\right )}{2 \, d e^{2}}\right ] \] Input:
integrate((B*x^2+A)/(-e*x^2+d)^(1/2)/(e^2*x^4-d^2)^(1/2),x, algorithm="fri cas")
Output:
[-1/4*(2*B*d*sqrt(-e)*log((2*e^2*x^4 - d*e*x^2 - 2*sqrt(e^2*x^4 - d^2)*sqr t(-e*x^2 + d)*sqrt(-e)*x - d^2)/(e*x^2 - d)) + sqrt(2)*(B*d + A*e)*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^2), -1/2*(2*B*d* sqrt(e)*arctan(sqrt(-e*x^2 + d)*sqrt(e)*x/sqrt(e^2*x^4 - d^2)) - sqrt(2)*( B*d + A*e)*sqrt(e)*arctan(sqrt(2)*sqrt(-e*x^2 + d)*sqrt(e)*x/sqrt(e^2*x^4 - d^2)))/(d*e^2)]
\[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\int \frac {A + B x^{2}}{\sqrt {\left (- d + e x^{2}\right ) \left (d + e x^{2}\right )} \sqrt {d - e x^{2}}}\, dx \] Input:
integrate((B*x**2+A)/(-e*x**2+d)**(1/2)/(e**2*x**4-d**2)**(1/2),x)
Output:
Integral((A + B*x**2)/(sqrt((-d + e*x**2)*(d + e*x**2))*sqrt(d - e*x**2)), x)
\[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\int { \frac {B x^{2} + A}{\sqrt {e^{2} x^{4} - d^{2}} \sqrt {-e x^{2} + d}} \,d x } \] Input:
integrate((B*x^2+A)/(-e*x^2+d)^(1/2)/(e^2*x^4-d^2)^(1/2),x, algorithm="max ima")
Output:
integrate((B*x^2 + A)/(sqrt(e^2*x^4 - d^2)*sqrt(-e*x^2 + d)), x)
Time = 0.21 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.27 \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {B \log \left ({\left | -\sqrt {-e} x + \sqrt {-e x^{2} - d} \right |}\right )}{\sqrt {-e} e} + \frac {\sqrt {2} {\left (B d + A e\right )} \sqrt {-e} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} - d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {-e} x - \sqrt {-e x^{2} - d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} + 6 \, d \right |}}\right )}{4 \, e^{2} {\left | d \right |}} \] Input:
integrate((B*x^2+A)/(-e*x^2+d)^(1/2)/(e^2*x^4-d^2)^(1/2),x, algorithm="gia c")
Output:
B*log(abs(-sqrt(-e)*x + sqrt(-e*x^2 - d)))/(sqrt(-e)*e) + 1/4*sqrt(2)*(B*d + A*e)*sqrt(-e)*log(abs(2*(sqrt(-e)*x - sqrt(-e*x^2 - d))^2 - 4*sqrt(2)*a bs(d) + 6*d)/abs(2*(sqrt(-e)*x - sqrt(-e*x^2 - d))^2 + 4*sqrt(2)*abs(d) + 6*d))/(e^2*abs(d))
Timed out. \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\int \frac {B\,x^2+A}{\sqrt {e^2\,x^4-d^2}\,\sqrt {d-e\,x^2}} \,d x \] Input:
int((A + B*x^2)/((e^2*x^4 - d^2)^(1/2)*(d - e*x^2)^(1/2)),x)
Output:
int((A + B*x^2)/((e^2*x^4 - d^2)^(1/2)*(d - e*x^2)^(1/2)), x)
Time = 0.15 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.65 \[ \int \frac {A+B x^2}{\sqrt {d-e x^2} \sqrt {-d^2+e^2 x^4}} \, dx=\frac {\sqrt {e}\, i \left (4 \mathit {asinh} \left (\frac {\sqrt {e}\, x}{\sqrt {d}}\right ) b d +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}-\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e -\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}-\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) a e +\sqrt {2}\, \mathrm {log}\left (\frac {\sqrt {e \,x^{2}+d}+\sqrt {d}\, \sqrt {2}+\sqrt {d}+\sqrt {e}\, x}{\sqrt {d}}\right ) b d \right )}{4 d \,e^{2}} \] Input:
int((B*x^2+A)/(-e*x^2+d)^(1/2)/(e^2*x^4-d^2)^(1/2),x)
Output:
(sqrt(e)*i*(4*asinh((sqrt(e)*x)/sqrt(d))*b*d + sqrt(2)*log((sqrt(d + e*x** 2) - sqrt(d)*sqrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e + sqrt(2)*log((sq rt(d + e*x**2) - sqrt(d)*sqrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d - sqr t(2)*log((sqrt(d + e*x**2) - sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqrt(d ))*a*e - sqrt(2)*log((sqrt(d + e*x**2) - sqrt(d)*sqrt(2) + sqrt(d) + sqrt( e)*x)/sqrt(d))*b*d - sqrt(2)*log((sqrt(d + e*x**2) + sqrt(d)*sqrt(2) - sqr t(d) + sqrt(e)*x)/sqrt(d))*a*e - sqrt(2)*log((sqrt(d + e*x**2) + sqrt(d)*s qrt(2) - sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d + sqrt(2)*log((sqrt(d + e*x**2) + sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*a*e + sqrt(2)*log((sqrt (d + e*x**2) + sqrt(d)*sqrt(2) + sqrt(d) + sqrt(e)*x)/sqrt(d))*b*d))/(4*d* e**2)