\(\int \frac {1}{d^2-e^2 x^2} \, dx\) [554]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right )}{d e} \]

[Out]

arctanh(e*x/d)/d/e

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {214} \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right )}{d e} \]

[In]

Int[(d^2 - e^2*x^2)^(-1),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \frac {\tanh ^{-1}\left (\frac {e x}{d}\right )}{d e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\text {arctanh}\left (\frac {e x}{d}\right )}{d e} \]

[In]

Integrate[(d^2 - e^2*x^2)^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.86

method	result	size
parallelrisch	\(-\frac {\ln \left (e x -d \right )-\ln \left (e x +d \right )}{2 d e}\)	\(26\)
default	\(\frac {\ln \left (e x +d \right )}{2 d e}-\frac {\ln \left (-e x +d \right )}{2 d e}\)	\(31\)
norman	\(\frac {\ln \left (e x +d \right )}{2 d e}-\frac {\ln \left (-e x +d \right )}{2 d e}\)	\(31\)
risch	\(\frac {\ln \left (e x +d \right )}{2 d e}-\frac {\ln \left (-e x +d \right )}{2 d e}\)	\(31\)

[In]

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

[Out]

-1/2*(ln(e*x-d)-ln(e*x+d))/d/e

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.79 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\log \left (e x + d\right ) - \log \left (e x - d\right )}{2 \, d e} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (8) = 16\).

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=- \frac {\frac {\log {\left (- \frac {d}{e} + x \right )}}{2} - \frac {\log {\left (\frac {d}{e} + x \right )}}{2}}{d e} \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 31 vs. \(2 (14) = 28\).

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 2.21 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\log \left (e x + d\right )}{2 \, d e} - \frac {\log \left (e x - d\right )}{2 \, d e} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (14) = 28\).

Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.36 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\log \left ({\left | e x + d \right |}\right )}{2 \, d e} - \frac {\log \left ({\left | e x - d \right |}\right )}{2 \, d e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1}{d^2-e^2 x^2} \, dx=\frac {\mathrm {atanh}\left (\frac {e\,x}{d}\right )}{d\,e} \]

[In]

int(1/(d^2 - e^2*x^2),x)

[Out]

atanh((e*x)/d)/(d*e)