3.1.93 \(\int \frac {e^{\text {sech}^{-1}(c x)}}{1-c^2 x^2} \, dx\) [93]

Optimal. Leaf size=71 \[ -\frac {\sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {1+c x}\right )}{c}+\frac {\log (x)}{c}-\frac {\log \left (1-c^2 x^2\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {6474, 1972, 94, 214, 272, 36, 29, 31} \begin {gather*} -\frac {\log \left (1-c^2 x^2\right )}{2 c}+\frac {\log (x)}{c}-\frac {\sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {c x+1}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^ArcSech[c*x]/(1 - c^2*x^2),x]

[Out]

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

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]

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 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 214

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

Rule 272

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 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 6474

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 73, normalized size = 1.03 \begin {gather*} \frac {2 \log (x)}{c}-\frac {\log \left (1-c^2 x^2\right )}{2 c}-\frac {\log \left (1+\sqrt {\frac {1-c x}{1+c x}}+c x \sqrt {\frac {1-c x}{1+c x}}\right )}{c} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]
time = 0.30, size = 82, normalized size = 1.15

method	result	size
default	\(-\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right )}{\sqrt {-c^{2} x^{2}+1}}+\frac {-\frac {\ln \left (c x +1\right )}{2}+\ln \left (x \right )-\frac {\ln \left (c x -1\right )}{2}}{c}\)	\(82\)

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (45) = 90\).
time = 0.37, size = 92, normalized size = 1.30 \begin {gather*} -\frac {\log \left (c^{2} x^{2} - 1\right ) + \log \left (c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 1\right ) - \log \left (c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} - 1\right ) - 2 \, \log \left (x\right )}{2 \, c} \end {gather*}

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

[Out]

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

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Sympy [A]
time = 9.44, size = 48, normalized size = 0.68 \begin {gather*} - \frac {\log {\left (-1 + \frac {1}{c x} \right )}}{2 c} - \frac {\log {\left (\sqrt {1 + \frac {1}{c x}} \right )}}{c} - \frac {2 \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {1 + \frac {1}{c x}}}{2} \right )}}{c} \end {gather*}

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))/(-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)

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Mupad [B]
time = 2.92, size = 59, normalized size = 0.83 \begin {gather*} \frac {\ln \left (x\right )}{c}-\frac {4\,\mathrm {atanh}\left (\frac {\sqrt {\frac {1}{c\,x}-1}-\mathrm {i}}{\sqrt {\frac {1}{c\,x}+1}-1}\right )}{c}-\frac {\ln \left (3\,c^2\,x^2-3\right )}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/c - (4*atanh(((1/(c*x) - 1)^(1/2) - 1i)/((1/(c*x) + 1)^(1/2) - 1)))/c - log(3*c^2*x^2 - 3)/(2*c)

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