3.1.4 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x} \, dx\) [4]

Optimal. Leaf size=238 \[ -\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d+e x^2}} \]

[Out]

arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*ln(x)-1/2*arcsinh(x*e^(1/2)/d^(1/2))^2*d^(1/2)*(1+e*x^2/d)^(1/2)/(e*x^2+d)^
(1/2)+arcsinh(x*e^(1/2)/d^(1/2))*ln(1-(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*d^(1/2)*(1+e*x^2/d)^(1/2)/(e*x^
2+d)^(1/2)-arcsinh(x*e^(1/2)/d^(1/2))*ln(x)*d^(1/2)*(1+e*x^2/d)^(1/2)/(e*x^2+d)^(1/2)+1/2*polylog(2,(x*e^(1/2)
/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*d^(1/2)*(1+e*x^2/d)^(1/2)/(e*x^2+d)^(1/2)

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Rubi [A]
time = 0.10, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {6354, 2364, 2362, 5775, 3797, 2221, 2317, 2438} \begin {gather*} \frac {\sqrt {d} \sqrt {\frac {e x^2}{d}+1} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d+e x^2}}-\frac {\sqrt {d} \log (x) \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d+e x^2}}+\log (x) \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Sqrt[d]*Sqrt[1 + (e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/Sqrt[d + e*x^2] + (Sqrt[d]*Sqrt[1 + (e*x^2)/
d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/Sqrt[d + e*x^2] - (Sqrt[d]*Sqrt[1
+ (e*x^2)/d]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[x])/Sqrt[d + e*x^2] + ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]*Log[
x] + (Sqrt[d]*Sqrt[1 + (e*x^2)/d]*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(2*Sqrt[d + e*x^2])

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2362

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[e, 2]*(x/Sqr
t[d])]*((a + b*Log[c*x^n])/Rt[e, 2]), x] - Dist[b*(n/Rt[e, 2]), Int[ArcSinh[Rt[e, 2]*(x/Sqrt[d])]/x, x], x] /;
FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && PosQ[e]

Rule 2364

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (e/d)*x^2]/Sqr
t[d + e*x^2], Int[(a + b*Log[c*x^n])/Sqrt[1 + (e/d)*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] &&  !GtQ[d, 0
]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 6354

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

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x} \, dx &=\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)-\sqrt {e} \int \frac {\log (x)}{\sqrt {d+e x^2}} \, dx\\ &=\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)-\frac {\left (\sqrt {e} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {\log (x)}{\sqrt {1+\frac {e x^2}{d}}} \, dx}{\sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)+\frac {\left (\sqrt {d} \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{\sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)+\frac {\left (\sqrt {d} \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{\sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)-\frac {\left (2 \sqrt {d} \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{\sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)-\frac {\left (\sqrt {d} \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{\sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)-\frac {\left (\sqrt {d} \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d+e x^2}}\\ &=-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{2 \sqrt {d+e x^2}}+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{\sqrt {d+e x^2}}-\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)}{\sqrt {d+e x^2}}+\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \log (x)+\frac {\sqrt {d} \sqrt {1+\frac {e x^2}{d}} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [F]
time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 209, normalized size = 0.88

method	result	size
default	\(-\frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )^{2}}{2}+\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right ) \ln \left (1+\frac {\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}+1}{\sqrt {-\frac {x^{2} e}{e \,x^{2}+d}+1}}\right )+\polylog \left (2, -\frac {\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}+1}{\sqrt {-\frac {x^{2} e}{e \,x^{2}+d}+1}}\right )+\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right ) \ln \left (1-\frac {\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}+1}{\sqrt {-\frac {x^{2} e}{e \,x^{2}+d}+1}}\right )+\polylog \left (2, \frac {\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}+1}{\sqrt {-\frac {x^{2} e}{e \,x^{2}+d}+1}}\right )\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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