3.5 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^3} \, dx\)

Optimal. Leaf size=53 \[ -\frac {\sqrt {e} \sqrt {d+e x^2}}{2 d x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6221, 264} \[ -\frac {\sqrt {e} \sqrt {d+e x^2}}{2 d x}-\frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^3,x]

[Out]

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

Rule 264

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

Rule 6221

Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*ArcT
anh[(c*x)/Sqrt[a + b*x^2]])/(d*(m + 1)), x] - Dist[c/(d*(m + 1)), Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /;
FreeQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 50, normalized size = 0.94 \[ -\frac {\sqrt {e} x \sqrt {d+e x^2}+d \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 d x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^3,x]

[Out]

-1/2*(Sqrt[e]*x*Sqrt[d + e*x^2] + d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(d*x^2)

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fricas [A]  time = 0.60, size = 54, normalized size = 1.02 \[ -\frac {2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{4 \, d x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 71, normalized size = 1.34 \[ \frac {e}{{\left (x e^{\frac {1}{2}} - \sqrt {x^{2} e + d}\right )}^{2} - d} - \frac {\log \left (-\frac {\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} + 1}{\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}} - 1}\right )}{4 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 60, normalized size = 1.13 \[ -\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{2 x^{2}}-\frac {\sqrt {e}\, \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{2 d^{2} x}+\frac {e^{\frac {3}{2}} x \sqrt {e \,x^{2}+d}}{2 d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 51, normalized size = 0.96 \[ -\frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{2 \, x^{2}} - \frac {e^{\frac {3}{2}} x^{2} + d \sqrt {e}}{2 \, \sqrt {e x^{2} + d} d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^3,x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**3,x)

[Out]

Integral(atanh(sqrt(e)*x/sqrt(d + e*x**2))/x**3, x)

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