\(\int \frac {d+e x^2}{d^2-e^2 x^4} \, dx\) [192]

Optimal result

Integrand size = 22, antiderivative size = 24 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1164, 214} \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}} \]

[In]

Int[(d + e*x^2)/(d^2 - e^2*x^4),x]

[Out]

ArcTanh[(Sqrt[e]*x)/Sqrt[d]]/(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 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}{d-e x^2} \, dx \\ & = \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {e}} \]

[In]

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

[Out]

ArcTanh[(Sqrt[e]*x)/Sqrt[d]]/(Sqrt[d]*Sqrt[e])

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
default	\(\frac {\operatorname {arctanh}\left (\frac {e x}{\sqrt {e d}}\right )}{\sqrt {e d}}\)	\(16\)
risch	\(\frac {\ln \left (e x +\sqrt {e d}\right )}{2 \sqrt {e d}}-\frac {\ln \left (-e x +\sqrt {e d}\right )}{2 \sqrt {e d}}\)	\(37\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.83 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=\left [\frac {\sqrt {d e} \log \left (\frac {e x^{2} + 2 \, \sqrt {d e} x + d}{e x^{2} - d}\right )}{2 \, d e}, -\frac {\sqrt {-d e} \arctan \left (\frac {\sqrt {-d e} x}{d}\right )}{d e}\right ] \]

[In]

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

[Out]

[1/2*sqrt(d*e)*log((e*x^2 + 2*sqrt(d*e)*x + d)/(e*x^2 - d))/(d*e), -sqrt(-d*e)*arctan(sqrt(-d*e)*x/d)/(d*e)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=- \frac {\sqrt {\frac {1}{d e}} \log {\left (- d \sqrt {\frac {1}{d e}} + x \right )}}{2} + \frac {\sqrt {\frac {1}{d e}} \log {\left (d \sqrt {\frac {1}{d e}} + x \right )}}{2} \]

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate((e*x^2+d)/(-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 = 18, normalized size of antiderivative = 0.75 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=-\frac {\arctan \left (\frac {e x}{\sqrt {-d e}}\right )}{\sqrt {-d e}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {d+e x^2}{d^2-e^2 x^4} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {d}}\right )}{\sqrt {d}\,\sqrt {e}} \]

[In]

int((d + e*x^2)/(d^2 - e^2*x^4),x)

[Out]

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