\(\int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 (1-c^2 x^2)} \, dx\) [96]

Optimal result

Integrand size = 22, antiderivative size = 85 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=-\frac {1}{3 c x^3}-\frac {c}{x}-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{1+c x}}}-\frac {2 c \sqrt {1-c x}}{3 x \sqrt {\frac {1}{1+c x}}}+c^2 \text {arctanh}(c x) \]

[Out]

-1/3/c/x^3-c/x+c^2*arctanh(c*x)-1/3*(-c*x+1)^(1/2)/c/x^3/(1/(c*x+1))^(1/2)-2/3*c*(-c*x+1)^(1/2)/x/(1/(c*x+1))^
(1/2)

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6476, 1972, 105, 12, 97, 331, 212} \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=c^2 \text {arctanh}(c x)-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{c x+1}}}-\frac {1}{3 c x^3}-\frac {2 c \sqrt {1-c x}}{3 x \sqrt {\frac {1}{c x+1}}}-\frac {c}{x} \]

[In]

Int[E^ArcSech[c*x]/(x^3*(1 - c^2*x^2)),x]

[Out]

-1/3*1/(c*x^3) - c/x - Sqrt[1 - c*x]/(3*c*x^3*Sqrt[(1 + c*x)^(-1)]) - (2*c*Sqrt[1 - c*x])/(3*x*Sqrt[(1 + c*x)^
(-1)]) + c^2*ArcTanh[c*x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 105

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] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 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 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*} \text {integral}& = \frac {\int \frac {\sqrt {\frac {1}{1+c x}}}{x^4 \sqrt {1-c x}} \, dx}{c}+\frac {\int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx}{c} \\ & = -\frac {1}{3 c x^3}+c \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x^4 \sqrt {1-c x} \sqrt {1+c x}} \, dx}{c} \\ & = -\frac {1}{3 c x^3}-\frac {c}{x}-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{1+c x}}}+c^3 \int \frac {1}{1-c^2 x^2} \, dx-\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {2 c^2}{x^2 \sqrt {1-c x} \sqrt {1+c x}} \, dx}{3 c} \\ & = -\frac {1}{3 c x^3}-\frac {c}{x}-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{1+c x}}}+c^2 \text {arctanh}(c x)+\frac {1}{3} \left (2 c \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 \\ & = -\frac {1}{3 c x^3}-\frac {c}{x}-\frac {\sqrt {1-c x}}{3 c x^3 \sqrt {\frac {1}{1+c x}}}-\frac {2 c \sqrt {1-c x}}{3 x \sqrt {\frac {1}{1+c x}}}+c^2 \text {arctanh}(c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=-\frac {2+6 c^2 x^2+2 \sqrt {\frac {1-c x}{1+c x}} \left (1+c x+2 c^2 x^2+2 c^3 x^3\right )+3 c^3 x^3 \log (1-c x)-3 c^3 x^3 \log (1+c x)}{6 c x^3} \]

[In]

Integrate[E^ArcSech[c*x]/(x^3*(1 - c^2*x^2)),x]

[Out]

-1/6*(2 + 6*c^2*x^2 + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2*c^3*x^3) + 3*c^3*x^3*Log[1 - c*x] -
3*c^3*x^3*Log[1 + c*x])/(c*x^3)

Maple [C] (verified)

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

Time = 0.73 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06

method	result	size
default	\(-\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \operatorname {csgn}\left (c \right )^{2} \left (2 c^{2} x^{2}+1\right )}{3 x^{2}}+\frac {\frac {c^{3} \ln \left (c x +1\right )}{2}-\frac {1}{3 x^{3}}-\frac {c^{2}}{x}-\frac {c^{3} \ln \left (c x -1\right )}{2}}{c}\)	\(90\)

[In]

int((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))/x^3/(-c^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

-1/3*(-(c*x-1)/c/x)^(1/2)/x^2*((c*x+1)/c/x)^(1/2)*csgn(c)^2*(2*c^2*x^2+1)+1/c*(1/2*c^3*ln(c*x+1)-1/3/x^3-c^2/x
-1/2*c^3*ln(c*x-1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\frac {3 \, c^{3} x^{3} \log \left (c x + 1\right ) - 3 \, c^{3} x^{3} \log \left (c x - 1\right ) - 6 \, c^{2} x^{2} - 2 \, {\left (2 \, c^{3} x^{3} + c x\right )} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} - 2}{6 \, c x^{3}} \]

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))/x^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

1/6*(3*c^3*x^3*log(c*x + 1) - 3*c^3*x^3*log(c*x - 1) - 6*c^2*x^2 - 2*(2*c^3*x^3 + c*x)*sqrt((c*x + 1)/(c*x))*s
qrt(-(c*x - 1)/(c*x)) - 2)/(c*x^3)

Sympy [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=- \frac {\int \frac {c x \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{6} - x^{4}}\, dx + \int \frac {1}{c^{2} x^{6} - x^{4}}\, dx}{c} \]

[In]

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

[Out]

-(Integral(c*x*sqrt(-1 + 1/(c*x))*sqrt(1 + 1/(c*x))/(c**2*x**6 - x**4), x) + Integral(1/(c**2*x**6 - x**4), x)
)/c

Maxima [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{{\left (c^{2} x^{2} - 1\right )} x^{3}} \,d x } \]

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))/x^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*c^2*log(c*x + 1) - 1/2*c^2*log(c*x - 1) + c*integrate(x^(-2), x) + integrate(x^(-4), x)/c - integrate(sqrt
(c*x + 1)*sqrt(-c*x + 1)/(c^3*x^6 - c*x^4), x)

Giac [F]

\[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=\int { -\frac {\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}}{{\left (c^{2} x^{2} - 1\right )} x^{3}} \,d x } \]

[In]

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

Mupad [B] (verification not implemented)

Time = 5.48 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\text {sech}^{-1}(c x)}}{x^3 \left (1-c^2 x^2\right )} \, dx=c^2\,\mathrm {atanh}\left (c\,x\right )-\frac {\left (\frac {\sqrt {\frac {1}{c\,x}+1}}{3}+\frac {2\,c^2\,x^2\,\sqrt {\frac {1}{c\,x}+1}}{3}\right )\,\sqrt {\frac {1}{c\,x}-1}}{x^2}-\frac {c^2\,x^2+\frac {1}{3}}{c\,x^3} \]

[In]

int(-((1/(c*x) - 1)^(1/2)*(1/(c*x) + 1)^(1/2) + 1/(c*x))/(x^3*(c^2*x^2 - 1)),x)

[Out]

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