\(\int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx\) [92]

Optimal result

Integrand size = 20, antiderivative size = 37 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=\frac {\sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \arcsin (c x)}{c^2}+\frac {\text {arctanh}(c x)}{c^2} \]

[Out]

arctanh(c*x)/c^2+arcsin(c*x)*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)/c^2

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6476, 1972, 41, 222, 212} \[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=\frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \arcsin (c x)}{c^2}+\frac {\text {arctanh}(c x)}{c^2} \]

[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 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 212

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

Rule 222

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 1972

Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(a + b*x^n)
^(p*q)], Int[u*(a + b*x^n)^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]

Rule 6476

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]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {arctanh}(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 {\text {arctanh}(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} \arcsin (c x)}{c^2}+\frac {\text {arctanh}(c x)}{c^2} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.84 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=-\frac {\log (1-c x)}{2 c^2}+\frac {\log (1+c x)}{2 c^2}+\frac {i \log \left (-2 i c x+2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)\right )}{c^2} \]

[In]

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

[Out]

-1/2*Log[1 - c*x]/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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.69 (sec) , antiderivative size = 97, normalized size of antiderivative = 2.62

method	result	size
default	\(\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \arctan \left (\frac {\operatorname {csgn}\left (c \right ) c x}{\sqrt {-\left (c x -1\right ) \left (c x +1\right )}}\right ) \operatorname {csgn}\left (c \right )}{\sqrt {-c^{2} x^{2}+1}\, c}+\frac {\frac {\ln \left (c x +1\right )}{2 c}-\frac {\ln \left (c x -1\right )}{2 c}}{c}\)	\(97\)

[In]

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

[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/c*(1/2/c*ln(c*x+1)-1/2/c*ln(c*x-1))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=-\frac {2 \, \arctan \left (\sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}}\right ) - \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \, c^{2}} \]

[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

Sympy [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=- \frac {\int \frac {c x \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx + \int \frac {1}{c^{2} x^{2} - 1}\, dx}{c} \]

[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

Maxima [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=\int { -\frac {x {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=\int { -\frac {x {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 6.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.27 \[ \int \frac {e^{\text {sech}^{-1}(c x)} x}{1-c^2 x^2} \, dx=\frac {\mathrm {atanh}\left (c\,x\right )}{c^2}+\frac {\left (\ln \left (\frac {{\left (\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}\right )}^2}{{\left (\sqrt {\frac {1}{c\,x}+1}-1\right )}^2}+1\right )-\ln \left (\frac {\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{c\,x}+1}-1}\right )\right )\,1{}\mathrm {i}}{c^2} \]

[In]

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

[Out]

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