3.12 \(\int \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=40 \[ x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{\sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {6217, 261} \[ x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{\sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 6217

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]], x_Symbol] :> Simp[x*ArcTanh[(c*x)/Sqrt[a + b*x^2]], x] -
Dist[c, Int[x/Sqrt[a + b*x^2], x], x] /; FreeQ[{a, b, c}, x] && EqQ[b, c^2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \[ x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{\sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.77, size = 51, normalized size = 1.28 \[ \frac {e x \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, \sqrt {e x^{2} + d} \sqrt {e}}{2 \, e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(x*ln((1+(sqrt(exp(1)*x^2+d))^-1*exp
(1/2)*x)/(1-(sqrt(exp(1)*x^2+d))^-1*exp(1/2)*x))-2*d*exp(1/2)/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2)*atan((sqrt
(d+x^2*exp(1))*exp(1)-sqrt(d+x^2*exp(1))*exp(1/2)^2)/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2)))

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maple [B]  time = 0.03, size = 76, normalized size = 1.90 \[ x \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )+\frac {e^{\frac {3}{2}} \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{d}-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{3 \sqrt {e}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+e^(3/2)/d*(1/3*x^2/e*(e*x^2+d)^(1/2)-2/3*d/e^2*(e*x^2+d)^(1/2))-1/3/e^(1/
2)/d*(e*x^2+d)^(3/2)

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maxima [B]  time = 0.34, size = 65, normalized size = 1.62 \[ x \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}}}{3 \, d \sqrt {e}} + \frac {{\left (e x^{2} + d\right )}^{\frac {3}{2}} - 3 \, \sqrt {e x^{2} + d} d}{3 \, d \sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)) - 1/3*(e*x^2 + d)^(3/2)/(d*sqrt(e)) + 1/3*((e*x^2 + d)^(3/2) - 3*sqrt(e*x
^2 + d)*d)/(d*sqrt(e))

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mupad [B]  time = 1.06, size = 32, normalized size = 0.80 \[ x\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )-\frac {\sqrt {e\,x^2+d}}{\sqrt {e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.63, size = 36, normalized size = 0.90 \[ \begin {cases} x \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )} - \frac {\sqrt {d + e x^{2}}}{\sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*atanh(sqrt(e)*x/sqrt(d + e*x**2)) - sqrt(d + e*x**2)/sqrt(e), Ne(e, 0)), (0, True))

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