### 3.65 $$\int \frac{e^{\text{csch}^{-1}(c x)}}{1+c^2 x^2} \, dx$$

Optimal. Leaf size=33 $-\frac{\log \left (c^2 x^2+1\right )}{2 c}+\frac{\log (x)}{c}-\frac{\text{csch}^{-1}(c x)}{c}$

[Out]

-(ArcCsch[c*x]/c) + Log[x]/c - Log[1 + c^2*x^2]/(2*c)

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Rubi [A]  time = 0.0428856, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.389, Rules used = {6340, 335, 215, 266, 36, 29, 31} $-\frac{\log \left (c^2 x^2+1\right )}{2 c}+\frac{\log (x)}{c}-\frac{\text{csch}^{-1}(c x)}{c}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCsch[c*x]/(1 + c^2*x^2),x]

[Out]

-(ArcCsch[c*x]/c) + Log[x]/c - Log[1 + c^2*x^2]/(2*c)

Rule 6340

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

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 215

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

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0331613, size = 37, normalized size = 1.12 $-\frac{\log \left (c^2 x^2+1\right )}{2 c}+\frac{\log (x)}{c}-\frac{\sinh ^{-1}\left (\frac{1}{c x}\right )}{c}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(ArcSinh[1/(c*x)]/c) + Log[x]/c - Log[1 + c^2*x^2]/(2*c)

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Maple [B]  time = 0.15, size = 172, normalized size = 5.2 \begin{align*}{\frac{x}{{c}^{2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}} \left ( \sqrt{{c}^{-2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{c}^{2}-\sqrt{-{\frac{1}{{c}^{4}} \left ( x{c}^{2}+\sqrt{-{c}^{2}} \right ) \left ( -x{c}^{2}+\sqrt{-{c}^{2}} \right ) }}{c}^{2}\sqrt{{c}^{-2}}-\ln \left ( 2\,{\frac{1}{x{c}^{2}} \left ( \sqrt{{c}^{-2}}\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{c}^{2}+1 \right ) } \right ) \right ){\frac{1}{\sqrt{{c}^{-2}}}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}+{\frac{\ln \left ( x \right ) }{c}}-{\frac{\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,c}} \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))/(c^2*x^2+1),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\log \left (c^{2} x^{2} + 1\right )}{2 \, c} + \frac{\log \left (x\right )}{c} + \int \frac{\sqrt{c^{2} x^{2} + 1}}{c^{3} x^{3} + c x}\,{d x} \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))/(c^2*x^2+1),x, algorithm="maxima")

[Out]

-1/2*log(c^2*x^2 + 1)/c + log(x)/c + integrate(sqrt(c^2*x^2 + 1)/(c^3*x^3 + c*x), x)

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Fricas [B]  time = 2.67739, size = 194, normalized size = 5.88 \begin{align*} -\frac{\log \left (c^{2} x^{2} + 1\right ) + 2 \, \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - 2 \, \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) - 2 \, \log \left (x\right )}{2 \, 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))/(c^2*x^2+1),x, algorithm="fricas")

[Out]

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

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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^{3} + x}\, dx + \int \frac{1}{c^{2} x^{3} + x}\, 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))/(c**2*x**2+1),x)

[Out]

(Integral(c*x*sqrt(1 + 1/(c**2*x**2))/(c**2*x**3 + x), x) + Integral(1/(c**2*x**3 + x), x))/c

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Giac [B]  time = 1.15889, size = 95, normalized size = 2.88 \begin{align*} -\frac{\log \left (c^{2} x^{2} + 1\right )}{2 \, c} - \frac{{\left ({\left | c \right |} \mathrm{sgn}\left (x\right ) - c\right )} \log \left (\sqrt{c^{2} x^{2} + 1} + 1\right )}{2 \, c^{2}} + \frac{{\left ({\left | c \right |} \mathrm{sgn}\left (x\right ) + c\right )} \log \left (\sqrt{c^{2} x^{2} + 1} - 1\right )}{2 \, 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))/(c^2*x^2+1),x, algorithm="giac")

[Out]

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