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

Optimal. Leaf size=30 \[ -\sqrt{\frac{1}{c^2 x^2}+1}-\frac{1}{c x}-\tan ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.071148, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {6342, 261, 325, 203} \[ -\sqrt{\frac{1}{c^2 x^2}+1}-\frac{1}{c x}-\tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 325

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] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 0.101257, size = 30, normalized size = 1. \[ -\sqrt{\frac{1}{c^2 x^2}+1}-\frac{1}{c x}-\tan ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.24, size = 154, normalized size = 5.1 \begin{align*} -{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}} \left ( \left ({\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}} \right ) ^{{\frac{3}{2}}}{c}^{2}-\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}{x}^{2}{c}^{2}-\ln \left ( x+\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}} \right ) x+\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 ) x \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}}}}}}}-{\frac{1}{cx}}-\arctan \left ( cx \right ) \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)*(((c^2*x^2+1)/c^2)^(3/2)*c^2-((c^2*x^2+1)/c^2)^(1/2)*x^2*c^2-ln(x+((c^2*x^2+1)/c^
2)^(1/2))*x+ln(x+(-(x*c^2+(-c^2)^(1/2))*(-x*c^2+(-c^2)^(1/2))/c^4)^(1/2))*x)/((c^2*x^2+1)/c^2)^(1/2)-1/c/x-arc
tan(c*x)

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Maxima [A]  time = 1.52835, size = 46, normalized size = 1.53 \begin{align*} -\frac{\sqrt{c^{2} x^{2} + 1}}{c x} - \frac{1}{c x} - \arctan \left (c x\right ) \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]

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

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Fricas [A]  time = 2.58744, size = 97, normalized size = 3.23 \begin{align*} -\frac{c x \arctan \left (c x\right ) + c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + c x + 1}{c 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))/x/(c^2*x^2+1),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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]

Exception raised: TypeError

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Giac [A]  time = 1.1543, size = 58, normalized size = 1.93 \begin{align*} \frac{2 \, \mathrm{sgn}\left (x\right )}{{\left (x{\left | c \right |} - \sqrt{c^{2} x^{2} + 1}\right )}^{2} - 1} - \frac{1}{c x} - \arctan \left (c x\right ) \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]

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