### 3.64 $$\int \frac{e^{\text{csch}^{-1}(c x)} x}{1+c^2 x^2} \, dx$$

Optimal. Leaf size=27 $\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{c^2}+\frac{\tan ^{-1}(c x)}{c^2}$

[Out]

ArcTan[c*x]/c^2 + ArcTanh[Sqrt[1 + 1/(c^2*x^2)]]/c^2

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Rubi [A]  time = 0.0592035, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.263, Rules used = {6342, 266, 63, 208, 203} $\frac{\tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{c^2}+\frac{\tan ^{-1}(c x)}{c^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[(E^ArcCsch[c*x]*x)/(1 + c^2*x^2),x]

[Out]

ArcTan[c*x]/c^2 + ArcTanh[Sqrt[1 + 1/(c^2*x^2)]]/c^2

Rule 6342

Int[(E^ArcCsch[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d^2/(a*c^2), Int[(d*x)
^(m - 2)/Sqrt[1 + 1/(c^2*x^2)], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d,
m}, x] && EqQ[b - a*c^2, 0]

Rule 266

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{\text{csch}^{-1}(c x)} x}{1+c^2 x^2} \, dx &=\frac{\int \frac{1}{\sqrt{1+\frac{1}{c^2 x^2}} x} \, dx}{c^2}+\frac{\int \frac{1}{1+c^2 x^2} \, dx}{c}\\ &=\frac{\tan ^{-1}(c x)}{c^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{2 c^2}\\ &=\frac{\tan ^{-1}(c x)}{c^2}-\operatorname{Subst}\left (\int \frac{1}{-c^2+c^2 x^2} \, dx,x,\sqrt{1+\frac{1}{c^2 x^2}}\right )\\ &=\frac{\tan ^{-1}(c x)}{c^2}+\frac{\tanh ^{-1}\left (\sqrt{1+\frac{1}{c^2 x^2}}\right )}{c^2}\\ \end{align*}

Mathematica [A]  time = 0.0486192, size = 38, normalized size = 1.41 $\frac{\log \left (x \left (\sqrt{\frac{c^2 x^2+1}{c^2 x^2}}+1\right )\right )}{c^2}+\frac{\tan ^{-1}(c x)}{c^2}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcCsch[c*x]*x)/(1 + c^2*x^2),x]

[Out]

ArcTan[c*x]/c^2 + Log[x*(1 + Sqrt[(1 + c^2*x^2)/(c^2*x^2)])]/c^2

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Maple [B]  time = 0.14, size = 85, normalized size = 3.2 \begin{align*}{\frac{x}{{c}^{2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}\ln \left ( x+\sqrt{-{\frac{1}{{c}^{4}} \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}+{\frac{\arctan \left ( cx \right ) }{{c}^{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(1+1/c^2/x^2)^(1/2))*x/(c^2*x^2+1),x)

[Out]

((c^2*x^2+1)/c^2/x^2)^(1/2)*x*ln(x+(-(x*c^2+(-c^2)^(1/2))*(-x*c^2+(-c^2)^(1/2))/c^4)^(1/2))/((c^2*x^2+1)/c^2)^
(1/2)/c^2+arctan(c*x)/c^2

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Maxima [B]  time = 1.55592, size = 80, normalized size = 2.96 \begin{align*} \frac{\log \left (\sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1\right ) - \log \left (\sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - 1\right )}{2 \, c^{2}} + \frac{\arctan \left (c x\right )}{c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x/(c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*(log(sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1) - log(sqrt((c^2*x^2 + 1)/(c^2*x^2)) - 1))/c^2 + arctan(c*x)/c^2

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Fricas [A]  time = 2.55209, size = 89, normalized size = 3.3 \begin{align*} \frac{\arctan \left (c x\right ) - \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right )}{c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x/(c^2*x^2+1),x, algorithm="fricas")

[Out]

(arctan(c*x) - log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x))/c^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{c x \sqrt{1 + \frac{1}{c^{2} x^{2}}}}{c^{2} x^{2} + 1}\, dx + \int \frac{1}{c^{2} x^{2} + 1}\, dx}{c} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c**2/x**2)**(1/2))*x/(c**2*x**2+1),x)

[Out]

(Integral(c*x*sqrt(1 + 1/(c**2*x**2))/(c**2*x**2 + 1), x) + Integral(1/(c**2*x**2 + 1), x))/c

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Giac [A]  time = 1.15771, size = 46, normalized size = 1.7 \begin{align*} -\frac{\log \left (-x{\left | c \right |} + \sqrt{c^{2} x^{2} + 1}\right ) \mathrm{sgn}\left (x\right )}{c^{2}} + \frac{\arctan \left (c x\right )}{c^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(1+1/c^2/x^2)^(1/2))*x/(c^2*x^2+1),x, algorithm="giac")

[Out]

-log(-x*abs(c) + sqrt(c^2*x^2 + 1))*sgn(x)/c^2 + arctan(c*x)/c^2