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

Optimal. Leaf size=45 \[ -\frac {\sqrt {1-c x}}{c^3 \sqrt {\frac {1}{c x+1}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^3} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6341, 1956, 74, 260} \[ -\frac {\log \left (1-c^2 x^2\right )}{2 c^3}-\frac {\sqrt {1-c x}}{c^3 \sqrt {\frac {1}{c x+1}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 74

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1956

Int[(x_)^(m_.)*((e_.)*((a_) + (b_.)*(x_)^(n_.))^(r_.))^(p_)*((f_.)*((c_) + (d_.)*(x_)^(n_.))^(s_))^(q_), x_Sym
bol] :> Dist[((e*(a + b*x^n)^r)^p*(f*(c + d*x^n)^s)^q)/((a + b*x^n)^(p*r)*(c + d*x^n)^(q*s)), Int[x^m*(a + b*x
^n)^(p*r)*(c + d*x^n)^(q*s), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r, s}, x]

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]

Rubi steps

\begin {align*} \int \frac {e^{\text {sech}^{-1}(c x)} x^2}{1-c^2 x^2} \, dx &=\frac {\int \frac {x \sqrt {\frac {1}{1+c x}}}{\sqrt {1-c x}} \, dx}{c}+\frac {\int \frac {x}{1-c^2 x^2} \, dx}{c}\\ &=-\frac {\log \left (1-c^2 x^2\right )}{2 c^3}+\frac {\left (\sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {x}{\sqrt {1-c x} \sqrt {1+c x}} \, dx}{c}\\ &=-\frac {\sqrt {1-c x}}{c^3 \sqrt {\frac {1}{1+c x}}}-\frac {\log \left (1-c^2 x^2\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 44, normalized size = 0.98 \[ -\frac {\log \left (1-c^2 x^2\right )+2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{2 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 49, normalized size = 1.09 \[ -\frac {2 \, c x \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + \log \left (c^{2} x^{2} - 1\right )}{2 \, c^{3}} \]

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^2/(-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)) + log(c^2*x^2 - 1))/c^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{2} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1}\,{d x} \]

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^2/(-c^2*x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 52, normalized size = 1.16 \[ -\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}}{c^{2}}-\frac {\ln \left (c^{2} x^{2}-1\right )}{2 c^{3}} \]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (c x + 1\right )}{2 \, c^{3}} - \frac {\log \left (c x - 1\right )}{2 \, c^{3}} - \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x}{c^{3} x^{2} - c}\,{d x} \]

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

[Out]

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

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mupad [B]  time = 2.70, size = 44, normalized size = 0.98 \[ -\frac {\ln \left (c^2\,x^2-1\right )}{2\,c^3}-\frac {x\,\sqrt {\frac {1}{c\,x}-1}\,\sqrt {\frac {1}{c\,x}+1}}{c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {x}{c^{2} x^{2} - 1}\, dx + \int \frac {c x^{2} \sqrt {-1 + \frac {1}{c x}} \sqrt {1 + \frac {1}{c x}}}{c^{2} x^{2} - 1}\, dx}{c} \]

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

[Out]

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

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