\(\int \frac {1}{\sqrt {d+e x^2} (d^2-e^2 x^4)} \, dx\) [197]

Optimal result

Integrand size = 26, antiderivative size = 61 \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}} \]

[Out]

1/4*arctanh(x*2^(1/2)*e^(1/2)/(e*x^2+d)^(1/2))/d^2*2^(1/2)/e^(1/2)+1/2*x/d^2/(e*x^2+d)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1164, 390, 385, 214} \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}}+\frac {x}{2 d^2 \sqrt {d+e x^2}} \]

[In]

Int[1/(Sqrt[d + e*x^2]*(d^2 - e^2*x^4)),x]

[Out]

x/(2*d^2*Sqrt[d + e*x^2]) + ArcTanh[(Sqrt[2]*Sqrt[e]*x)/Sqrt[d + e*x^2]]/(2*Sqrt[2]*d^2*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 390

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a
*d)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && Eq
Q[n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

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 ) \left (d+e x^2\right )^{3/2}} \, dx \\ & = \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\int \frac {1}{\left (d-e x^2\right ) \sqrt {d+e x^2}} \, dx}{2 d} \\ & = \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\text {Subst}\left (\int \frac {1}{d-2 d e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{2 d} \\ & = \frac {x}{2 d^2 \sqrt {d+e x^2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 \sqrt {2} d^2 \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.13 \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\frac {\frac {2 x}{\sqrt {d+e x^2}}+\frac {\sqrt {2} \text {arctanh}\left (\frac {d-e x^2+\sqrt {e} x \sqrt {d+e x^2}}{\sqrt {2} d}\right )}{\sqrt {e}}}{4 d^2} \]

[In]

Integrate[1/(Sqrt[d + e*x^2]*(d^2 - e^2*x^4)),x]

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {e \,x^{2}+d}\, \sqrt {2}}{2 x \sqrt {e}}\right ) \sqrt {e \,x^{2}+d}+2 x \sqrt {e}}{4 \sqrt {e \,x^{2}+d}\, \sqrt {e}\, d^{2}}\)	\(59\)
default	\(\frac {e \sqrt {2}\, \ln \left (\frac {4 d +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x -\frac {\sqrt {e d}}{e}\right )^{2} e +2 \sqrt {e d}\, \left (x -\frac {\sqrt {e d}}{e}\right )+2 d}}{x -\frac {\sqrt {e d}}{e}}\right )}{4 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}\, \sqrt {d}}-\frac {e \sqrt {2}\, \ln \left (\frac {4 d -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 \sqrt {2}\, \sqrt {d}\, \sqrt {\left (x +\frac {\sqrt {e d}}{e}\right )^{2} e -2 \sqrt {e d}\, \left (x +\frac {\sqrt {e d}}{e}\right )+2 d}}{x +\frac {\sqrt {e d}}{e}}\right )}{4 \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \sqrt {e d}\, \sqrt {d}}+\frac {\sqrt {\left (x +\frac {\sqrt {-e d}}{e}\right )^{2} e -2 \sqrt {-e d}\, \left (x +\frac {\sqrt {-e d}}{e}\right )}}{2 d \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \left (x +\frac {\sqrt {-e d}}{e}\right )}+\frac {\sqrt {\left (x -\frac {\sqrt {-e d}}{e}\right )^{2} e +2 \sqrt {-e d}\, \left (x -\frac {\sqrt {-e d}}{e}\right )}}{2 d \left (\sqrt {e d}-\sqrt {-e d}\right ) \left (\sqrt {e d}+\sqrt {-e d}\right ) \left (x -\frac {\sqrt {-e d}}{e}\right )}\)	\(441\)

[In]

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

[Out]

1/4*(2^(1/2)*arctanh(1/2*(e*x^2+d)^(1/2)/x*2^(1/2)/e^(1/2))*(e*x^2+d)^(1/2)+2*x*e^(1/2))/(e*x^2+d)^(1/2)/e^(1/
2)/d^2

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (45) = 90\).

Time = 0.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 3.43 \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\left [\frac {\sqrt {2} {\left (e x^{2} + d\right )} \sqrt {e} \log \left (\frac {17 \, e^{2} x^{4} + 14 \, d e x^{2} + 4 \, \sqrt {2} {\left (3 \, e x^{3} + d x\right )} \sqrt {e x^{2} + d} \sqrt {e} + d^{2}}{e^{2} x^{4} - 2 \, d e x^{2} + d^{2}}\right ) + 8 \, \sqrt {e x^{2} + d} e x}{16 \, {\left (d^{2} e^{2} x^{2} + d^{3} e\right )}}, -\frac {\sqrt {2} {\left (e x^{2} + d\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {2} {\left (3 \, e x^{2} + d\right )} \sqrt {e x^{2} + d} \sqrt {-e}}{4 \, {\left (e^{2} x^{3} + d e x\right )}}\right ) - 4 \, \sqrt {e x^{2} + d} e x}{8 \, {\left (d^{2} e^{2} x^{2} + d^{3} e\right )}}\right ] \]

[In]

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

[Out]

[1/16*(sqrt(2)*(e*x^2 + d)*sqrt(e)*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)*sq
rt(e) + d^2)/(e^2*x^4 - 2*d*e*x^2 + d^2)) + 8*sqrt(e*x^2 + d)*e*x)/(d^2*e^2*x^2 + d^3*e), -1/8*(sqrt(2)*(e*x^2
+ d)*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)) - 4*sqrt(e*x^2 + d
)*e*x)/(d^2*e^2*x^2 + d^3*e)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (45) = 90\).

Time = 0.33 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.66 \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\frac {\sqrt {2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} - 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}{{\left | 2 \, {\left (\sqrt {e} x - \sqrt {e x^{2} + d}\right )}^{2} + 4 \, \sqrt {2} {\left | d \right |} - 6 \, d \right |}}\right )}{8 \, d \sqrt {e} {\left | d \right |}} + \frac {x}{2 \, \sqrt {e x^{2} + d} d^{2}} \]

[In]

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

[Out]

1/8*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))/(d*sqrt(e)*abs(d)) + 1/2*x/(sqrt(e*x^2 + d)*d^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x^2} \left (d^2-e^2 x^4\right )} \, dx=\int \frac {1}{\left (d^2-e^2\,x^4\right )\,\sqrt {e\,x^2+d}} \,d x \]

[In]

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

[Out]

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