3.5.18 \(\int \frac {e^{-\coth ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [418]

Optimal. Leaf size=73 \[ \frac {2 \sqrt {1-\frac {1}{a^2 x^2}} x}{c^2}-\frac {a \sqrt {1-\frac {1}{a^2 x^2}} x}{c^2 \left (a-\frac {1}{x}\right )}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c^2} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {6312, 871, 821, 272, 65, 214} \begin {gather*} \frac {2 x \sqrt {1-\frac {1}{a^2 x^2}}}{c^2}-\frac {a x \sqrt {1-\frac {1}{a^2 x^2}}}{c^2 \left (a-\frac {1}{x}\right )}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g
))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e
^2), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0
] && EqQ[Simplify[m + 2*p + 3], 0]

Rule 871

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

Rule 6312

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[-c^n, Subst[Int[(c + d*x)^(p -
n)*((1 - x^2/a^2)^(n/2)/x^2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] && IntegerQ[(n - 1)
/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1]) && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 69, normalized size = 0.95 \begin {gather*} \frac {-2-a x+a^2 x^2+a \sqrt {1-\frac {1}{a^2 x^2}} x \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a^2 c^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(254\) vs. \(2(67)=134\).
time = 0.15, size = 255, normalized size = 3.49

method	result	size
risch	\(\frac {\left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}{a \,c^{2}}+\frac {\left (\frac {\ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right )}{a^{2} \sqrt {a^{2}}}-\frac {\sqrt {a^{2} \left (x -\frac {1}{a}\right )^{2}+2 a \left (x -\frac {1}{a}\right )}}{a^{4} \left (x -\frac {1}{a}\right )}\right ) a^{2} \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{c^{2} \left (a x -1\right )}\)	\(144\)
default	\(-\frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) \left (-3 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a^{2} x^{2}-2 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{3} x^{2}+\left (\left (a x +1\right ) \left (a x -1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}+6 \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, \sqrt {a^{2}}\, a x +4 \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right ) a^{2} x -3 \sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}-2 a \ln \left (\frac {a^{2} x +\sqrt {a^{2}}\, \sqrt {\left (a x +1\right ) \left (a x -1\right )}}{\sqrt {a^{2}}}\right )\right )}{2 a \sqrt {\left (a x +1\right ) \left (a x -1\right )}\, c^{2} \left (a x -1\right )^{2} \sqrt {a^{2}}}\)	\(255\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)/a*(-3*((a*x+1)*(a*x-1))^(1/2)*(a^2)^(1/2)*a^2*x^2-2*ln((a^2*x+(a^2)^(1/2)
*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^3*x^2+((a*x+1)*(a*x-1))^(3/2)*(a^2)^(1/2)+6*((a*x+1)*(a*x-1))^(1/2)*(
a^2)^(1/2)*a*x+4*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2))*a^2*x-3*(a^2)^(1/2)*((a*x+1)*(a*x
-1))^(1/2)-2*a*ln((a^2*x+(a^2)^(1/2)*((a*x+1)*(a*x-1))^(1/2))/(a^2)^(1/2)))/((a*x+1)*(a*x-1))^(1/2)/c^2/(a*x-1
)^2/(a^2)^(1/2)

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Maxima [A]
time = 0.25, size = 120, normalized size = 1.64 \begin {gather*} -a {\left (\frac {\frac {3 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - a^{2} c^{2} \sqrt {\frac {a x - 1}{a x + 1}}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c^{2}} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*((3*(a*x - 1)/(a*x + 1) - 1)/(a^2*c^2*((a*x - 1)/(a*x + 1))^(3/2) - a^2*c^2*sqrt((a*x - 1)/(a*x + 1))) - lo
g(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c^2) + log(sqrt((a*x - 1)/(a*x + 1)) - 1)/(a^2*c^2))

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Fricas [A]
time = 0.34, size = 97, normalized size = 1.33 \begin {gather*} \frac {{\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} x^{2} - a x - 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{2} x - a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {a^{2} \int \frac {x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{2} x^{2} - 2 a x + 1}\, dx}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

undef

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Mupad [B]
time = 1.23, size = 62, normalized size = 0.85 \begin {gather*} \frac {2\,a\,x+4\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}-4}{2\,a\,c^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x + 4*atanh(((a*x - 1)/(a*x + 1))^(1/2))*((a*x - 1)/(a*x + 1))^(1/2) - 4)/(2*a*c^2*((a*x - 1)/(a*x + 1))^
(1/2))

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