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

Optimal. Leaf size=37 \[ \frac{\sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c^2}+\frac{\tanh ^{-1}(c x)}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.113641, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6341, 6677, 41, 216, 206} \[ \frac{\sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c^2}+\frac{\tanh ^{-1}(c x)}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 6677

Int[(u_)*((c_.)*((a_.) + (b_.)*(x_))^(n_))^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c*(a + b*x)^n)^FracPart[p])/
(a + b*x)^(n*FracPart[p]), Int[u*(a + b*x)^(n*p), x], x] /; FreeQ[{a, b, c, n, p}, x] &&  !IntegerQ[p] &&  !Ma
tchQ[u, x^(n1_.)*(v_.) /; EqQ[n, n1 + 1]]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

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

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

Mathematica [C]  time = 0.044183, size = 68, normalized size = 1.84 \[ -\frac{\log (1-c x)}{2 c^2}+\frac{\log (c x+1)}{2 c^2}+\frac{i \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{c^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [C]  time = 0.227, size = 92, normalized size = 2.5 \begin{align*}{\frac{x{\it csgn} \left ( c \right ) }{c}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}\arctan \left ({{\it csgn} \left ( c \right ) cx{\frac{1}{\sqrt{- \left ( cx+1 \right ) \left ( cx-1 \right ) }}}} \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}-{\frac{\ln \left ( cx-1 \right ) }{2\,{c}^{2}}}+{\frac{\ln \left ( cx+1 \right ) }{2\,{c}^{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)*x*((c*x+1)/c/x)^(1/2)*arctan(csgn(c)*c*x/(-(c*x+1)*(c*x-1))^(1/2))/(-c^2*x^2+1)^(1/2)*csg
n(c)/c-1/2/c^2*ln(c*x-1)+1/2/c^2*ln(c*x+1)

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

[Out]

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

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Fricas [B]  time = 2.1015, size = 131, normalized size = 3.54 \begin{align*} -\frac{2 \, \arctan \left (\sqrt{\frac{c x + 1}{c x}} \sqrt{-\frac{c x - 1}{c x}}\right ) - \log \left (c x + 1\right ) + \log \left (c x - 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/x)^(1/2)*(1+1/c/x)^(1/2))*x/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

-1/2*(2*arctan(sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x))) - log(c*x + 1) + log(c*x - 1))/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 x}} \sqrt{1 + \frac{1}{c x}}}{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/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**2 - 1), x) + Integral(1/(c**2*x**2 - 1), x))/c

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