3.94 \(\int \frac{e^{\text{sech}^{-1}(c x)}}{x (1-c^2 x^2)} \, dx\)

Optimal. Leaf size=42 \[ -\frac{\sqrt{1-c x}}{c x \sqrt{\frac{1}{c x+1}}}-\frac{1}{c x}+\tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.142613, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {6341, 1956, 95, 325, 206} \[ -\frac{\sqrt{1-c x}}{c x \sqrt{\frac{1}{c x+1}}}-\frac{1}{c x}+\tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[c*x]/(x*(1 - c^2*x^2)),x]

[Out]

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

Rule 6341

Int[(E^ArcSech[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/(a*c), Int[((d*x)^(m
- 1)*Sqrt[1/(1 + c*x)])/Sqrt[1 - c*x], 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 1956

Int[(x_)^(m_.)*((e_.)*((a_) + (b_.)*(x_)^(n_.))^(r_.))^(p_)*((f_.)*((c_) + (d_.)*(x_)^(n_.))^(s_))^(q_), x_Sym
bol] :> Dist[((e*(a + b*x^n)^r)^p*(f*(c + d*x^n)^s)^q)/((a + b*x^n)^(p*r)*(c + d*x^n)^(q*s)), Int[x^m*(a + b*x
^n)^(p*r)*(c + d*x^n)^(q*s), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r, s}, x]

Rule 95

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -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 206

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

Rubi steps

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

Mathematica [A]  time = 0.163771, size = 59, normalized size = 1.4 \[ -\sqrt{\frac{1-c x}{c x+1}} \left (\frac{1}{c x}+1\right )-\frac{1}{c x}-\frac{1}{2} \log (1-c x)+\frac{1}{2} \log (c x+1) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSech[c*x]/(x*(1 - c^2*x^2)),x]

[Out]

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

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Maple [C]  time = 0.21, size = 61, normalized size = 1.5 \begin{align*} -\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ({\it csgn} \left ( c \right ) \right ) ^{2}-{\frac{\ln \left ( cx-1 \right ) }{2}}-{\frac{1}{cx}}+{\frac{\ln \left ( cx+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{-\frac{1}{x}}{c} - \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1}}{c^{3} x^{4} - c x^{2}}\,{d x} + \frac{1}{2} \, \log \left (c x + 1\right ) - \frac{1}{2} \, \log \left (c x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(-2), x)/c - integrate(sqrt(c*x + 1)*sqrt(-c*x + 1)/(c^3*x^4 - c*x^2), x) + 1/2*log(c*x + 1) - 1/2
*log(c*x - 1)

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Fricas [A]  time = 2.16174, size = 144, normalized size = 3.43 \begin{align*} -\frac{2 \, c x \sqrt{\frac{c x + 1}{c x}} \sqrt{-\frac{c x - 1}{c x}} - c x \log \left (c x + 1\right ) + c x \log \left (c x - 1\right ) + 2}{2 \, c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 x}} \sqrt{1 + \frac{1}{c x}}}{c^{2} x^{4} - x^{2}}\, dx + \int \frac{1}{c^{2} x^{4} - x^{2}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(Integral(c*x*sqrt(-1 + 1/(c*x))*sqrt(1 + 1/(c*x))/(c**2*x**4 - x**2), x) + Integral(1/(c**2*x**4 - x**2), x)
)/c

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}}{{\left (c^{2} x^{2} - 1\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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