3.810 \(\int \frac{e^{-\coth ^{-1}(a x)}}{c-\frac{c}{a^2 x^2}} \, dx\)

Optimal. Leaf size=105 \[ \frac{x \sqrt{1-\frac{1}{a x}}}{c \sqrt{\frac{1}{a x}+1}}+\frac{2 \sqrt{1-\frac{1}{a x}}}{a c \sqrt{\frac{1}{a x}+1}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.0773693, antiderivative size = 105, normalized size of antiderivative = 1., 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 = {6194, 103, 21, 94, 92, 208} \[ \frac{x \sqrt{1-\frac{1}{a x}}}{c \sqrt{\frac{1}{a x}+1}}+\frac{2 \sqrt{1-\frac{1}{a x}}}{a c \sqrt{\frac{1}{a x}+1}}-\frac{\tanh ^{-1}\left (\sqrt{1-\frac{1}{a x}} \sqrt{\frac{1}{a x}+1}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6194

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

Rule 103

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] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 94

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

Rule 92

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.113981, size = 57, normalized size = 0.54 \[ \frac{\frac{x \sqrt{1-\frac{1}{a^2 x^2}} (a x+2)}{a x+1}-\frac{\log \left (x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )}{a}}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.157, size = 250, normalized size = 2.4 \begin{align*} -{\frac{1}{2\,a \left ( ax+1 \right ) c} \left ( 2\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ){x}^{2}{a}^{3}-3\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }{x}^{2}{a}^{2}+4\,\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) x{a}^{2}+ \left ( \left ( ax-1 \right ) \left ( ax+1 \right ) \right ) ^{{\frac{3}{2}}}\sqrt{{a}^{2}}-6\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }xa+2\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) -3\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ) \sqrt{{\frac{ax-1}{ax+1}}}{\frac{1}{\sqrt{{a}^{2}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03992, size = 163, normalized size = 1.55 \begin{align*} -a{\left (\frac{2 \, \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{{\left (a x - 1\right )} a^{2} c}{a x + 1} - a^{2} c} + \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac{\log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2} c} - \frac{\sqrt{\frac{a x - 1}{a x + 1}}}{a^{2} c}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.46311, size = 161, normalized size = 1.53 \begin{align*} \frac{{\left (a x + 2\right )} \sqrt{\frac{a x - 1}{a x + 1}} - \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) + \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef