3.1.94 \(\int \frac {e^{\text {sech}^{-1}(c x)}}{x (1-c^2 x^2)} \, dx\) [94]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 42, normalized size of antiderivative = 1.00, 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 = {6476, 1972, 97, 331, 212} \begin {gather*} -\frac {\sqrt {1-c x}}{c x \sqrt {\frac {1}{c x+1}}}-\frac {1}{c x}+\tanh ^{-1}(c x) \end {gather*}

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 97

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 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 331

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 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*} \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*}

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Mathematica [A]
time = 0.13, size = 59, normalized size = 1.40 \begin {gather*} -\frac {1}{c x}-\left (1+\frac {1}{c x}\right ) \sqrt {\frac {1-c x}{1+c x}}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x) \end {gather*}

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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.30, size = 65, normalized size = 1.55

method	result	size
default	\(-\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \mathrm {csgn}\left (c \right )^{2}+\frac {\frac {c \ln \left (c x +1\right )}{2}-\frac {1}{x}-\frac {c \ln \left (c x -1\right )}{2}}{c}\)	\(65\)

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,method=_RETURNVERBOSE)

[Out]

-(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*csgn(c)^2+1/c*(1/2*c*ln(c*x+1)-1/x-1/2*c*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))/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 = 0.56, size = 62, normalized size = 1.48 \begin {gather*} -\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 {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))/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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {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))/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 = 0.00, 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))/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)

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Mupad [B]
time = 2.53, size = 37, normalized size = 0.88 \begin {gather*} \mathrm {atanh}\left (c\,x\right )-\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}-\frac {1}{c\,x} \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))/(x*(c^2*x^2 - 1)),x)

[Out]

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

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