\(\int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx\) [196]

Optimal result

Integrand size = 26, antiderivative size = 38 \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {2} d \sqrt {e}} \]

[Out]

1/2*arctanh(x*2^(1/2)*e^(1/2)/(e*x^2+d)^(1/2))/d*2^(1/2)/e^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1164, 385, 214} \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {2} d \sqrt {e}} \]

[In]

Int[Sqrt[d + e*x^2]/(d^2 - e^2*x^4),x]

[Out]

ArcTanh[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d + e*x^2]]/(Sqrt[2]*d*Sqrt[e])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 1164

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[(d + e*x^2)^(p + q)*(a/d + (c/e)
*x^2)^p, x] /; FreeQ[{a, c, d, e, q}, x] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{\left (d-e x^2\right ) \sqrt {d+e x^2}} \, dx \\ & = \text {Subst}\left (\int \frac {1}{d-2 d e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right ) \\ & = \frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {2} d \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {d-e x^2+\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {2} d}\right )}{\sqrt {2} d \sqrt {e}} \]

[In]

Integrate[Sqrt[d + e*x^2]/(d^2 - e^2*x^4),x]

[Out]

ArcTanh[(d - e*x^2 + Sqrt[e]*x*Sqrt[d + e*x^2])/(Sqrt[2]*d)]/(Sqrt[2]*d*Sqrt[e])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, \sqrt {2}}{2 x \sqrt {e}}\right )}{2 d \sqrt {e}}\)	\(33\)
default	\(-\frac {e \left (\sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}+\frac {\sqrt {e d}\, \ln \left (\frac {\sqrt {e d}+e \left (x -\frac {\sqrt {e d}}{e}\right )}{\sqrt {e}}+\sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}\right )}{\sqrt {e}}-\sqrt {d}\, \sqrt {2}\, \ln \left (\frac {4 d +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}}{x -\frac {\sqrt {e d}}{e}}\right )\right )}{2 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}}+\frac {e \left (\sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}-\frac {\sqrt {e d}\, \ln \left (\frac {-\sqrt {e d}+e \left (x +\frac {\sqrt {e d}}{e}\right )}{\sqrt {e}}+\sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}\right )}{\sqrt {e}}-\sqrt {d}\, \sqrt {2}\, \ln \left (\frac {4 d -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}}{x +\frac {\sqrt {e d}}{e}}\right )\right )}{2 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}}-\frac {e \left (\sqrt {\left (x +\frac {\sqrt {-e d}}{e}\right )^{2} e -2 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right )}-\frac {\sqrt {-e d}\, \ln \left (\frac {-\sqrt {-e d}+e \left (x +\frac {\sqrt {-e d}}{e}\right )}{\sqrt {e}}+\sqrt {\left (x +\frac {\sqrt {-e d}}{e}\right )^{2} e -2 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right )}\right )}{\sqrt {e}}\right )}{2 \sqrt {-e d}\, \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right )}+\frac {e \left (\sqrt {\left (x -\frac {\sqrt {-e d}}{e}\right )^{2} e +2 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right )}+\frac {\sqrt {-e d}\, \ln \left (\frac {\sqrt {-e d}+e \left (x -\frac {\sqrt {-e d}}{e}\right )}{\sqrt {e}}+\sqrt {\left (x -\frac {\sqrt {-e d}}{e}\right )^{2} e +2 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right )}\right )}{\sqrt {e}}\right )}{2 \sqrt {-e d}\, \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right )}\)	\(818\)

[In]

int((e*x^2+d)^(1/2)/(-e^2*x^4+d^2),x,method=_RETURNVERBOSE)

[Out]

1/2/d*2^(1/2)/e^(1/2)*arctanh(1/2*(e*x^2+d)^(1/2)/x*2^(1/2)/e^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 3.63 \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\left [\frac {\sqrt {2} \log \left (\frac {17 \, e^{2} x^{4} + 14 \, d e x^{2} + 4 \, \sqrt {2} {\left (3 \, e x^{3} + d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right )}{8 \, d \sqrt {e}}, -\frac {\sqrt {2} \sqrt {-e} \arctan \left (\frac {\sqrt {2} {\left (3 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{4 \, {\left (e^{2} x^{3} + d e x\right )}}\right )}{4 \, d e}\right ] \]

[In]

integrate((e*x^2+d)^(1/2)/(-e^2*x^4+d^2),x, algorithm="fricas")

[Out]

[1/8*sqrt(2)*log((17*e^2*x^4 + 14*d*e*x^2 + 4*sqrt(2)*(3*e*x^3 + d*x)*sqrt(e*x^2 + d)*sqrt(e) + d^2)/(e^2*x^4
- 2*d*e*x^2 + d^2))/(d*sqrt(e)), -1/4*sqrt(2)*sqrt(-e)*arctan(1/4*sqrt(2)*(3*e*x^2 + d)*sqrt(e*x^2 + d)*sqrt(-
e)/(e^2*x^3 + d*e*x))/(d*e)]

Sympy [F]

\[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=- \int \frac {1}{- d \sqrt {d + e x^{2}} + e x^{2} \sqrt {d + e x^{2}}}\, dx \]

[In]

integrate((e*x**2+d)**(1/2)/(-e**2*x**4+d**2),x)

[Out]

-Integral(1/(-d*sqrt(d + e*x**2) + e*x**2*sqrt(d + e*x**2)), x)

Maxima [F]

\[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\int { -\frac {\sqrt {e x^{2} + d}}{e^{2} x^{4} - d^{2}} \,d x } \]

[In]

integrate((e*x^2+d)^(1/2)/(-e^2*x^4+d^2),x, algorithm="maxima")

[Out]

-integrate(sqrt(e*x^2 + d)/(e^2*x^4 - d^2), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (29) = 58\).

Time = 0.32 (sec) , antiderivative size = 82, normalized size of antiderivative = 2.16 \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\frac {\sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{4 \, \sqrt {e} {\left | d \right |}} \]

[In]

integrate((e*x^2+d)^(1/2)/(-e^2*x^4+d^2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*log(abs(2*(sqrt(e)*x - sqrt(e*x^2 + d))^2 - 4*sqrt(2)*abs(d) - 6*d)/abs(2*(sqrt(e)*x - sqrt(e*x^2
+ d))^2 + 4*sqrt(2)*abs(d) - 6*d))/(sqrt(e)*abs(d))

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+e x^2}}{d^2-e^2 x^4} \, dx=\int \frac {\sqrt {e\,x^2+d}}{d^2-e^2\,x^4} \,d x \]

[In]

int((d + e*x^2)^(1/2)/(d^2 - e^2*x^4),x)

[Out]

int((d + e*x^2)^(1/2)/(d^2 - e^2*x^4), x)