\(\int \frac {(d+e x^2)^2}{d^2-e^2 x^4} \, dx\) [191]

Optimal result

Integrand size = 24, antiderivative size = 29 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=-x+\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1164, 396, 214} \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}}-x \]

[In]

Int[(d + e*x^2)^2/(d^2 - e^2*x^4),x]

[Out]

-x + (2*Sqrt[d]*ArcTanh[(Sqrt[e]*x)/Sqrt[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 396

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=-x+\frac {2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {e}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76

method	result	size
default	\(-x +\frac {2 d \,\operatorname {arctanh}\left (\frac {e x}{\sqrt {e d}}\right )}{\sqrt {e d}}\)	\(22\)
risch	\(-x -\frac {\sqrt {e d}\, \ln \left (\sqrt {e d}\, x -d \right )}{e}+\frac {\sqrt {e d}\, \ln \left (-\sqrt {e d}\, x -d \right )}{e}\)	\(49\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.52 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=\left [\sqrt {\frac {d}{e}} \log \left (\frac {e x^{2} + 2 \, e x \sqrt {\frac {d}{e}} + d}{e x^{2} - d}\right ) - x, -2 \, \sqrt {-\frac {d}{e}} \arctan \left (\frac {e x \sqrt {-\frac {d}{e}}}{d}\right ) - x\right ] \]

[In]

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

[Out]

[sqrt(d/e)*log((e*x^2 + 2*e*x*sqrt(d/e) + d)/(e*x^2 - d)) - x, -2*sqrt(-d/e)*arctan(e*x*sqrt(-d/e)/d) - x]

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=- x - \sqrt {\frac {d}{e}} \log {\left (x - \sqrt {\frac {d}{e}} \right )} + \sqrt {\frac {d}{e}} \log {\left (x + \sqrt {\frac {d}{e}} \right )} \]

[In]

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

[Out]

-x - sqrt(d/e)*log(x - sqrt(d/e)) + sqrt(d/e)*log(x + sqrt(d/e))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=-\frac {2 \, d \arctan \left (\frac {e x}{\sqrt {-d e}}\right )}{\sqrt {-d e}} - x \]

[In]

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

[Out]

-2*d*arctan(e*x/sqrt(-d*e))/sqrt(-d*e) - x

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {\left (d+e x^2\right )^2}{d^2-e^2 x^4} \, dx=\frac {2\,\sqrt {d}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {e}}-x \]

[In]

int((d + e*x^2)^2/(d^2 - e^2*x^4),x)

[Out]

(2*d^(1/2)*atanh((e^(1/2)*x)/d^(1/2)))/e^(1/2) - x