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

Optimal. Leaf size=49 \[ c x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{2 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}-\frac{c \csc ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.145407, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {6177, 1807, 844, 216, 266, 63, 208} \[ c x \sqrt{1-\frac{1}{a^2 x^2}}-\frac{2 c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{a}-\frac{c \csc ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))/E^ArcCoth[a*x],x]

[Out]

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

Rule 6177

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]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 266

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 63

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

Mathematica [A]  time = 0.101654, size = 73, normalized size = 1.49 \[ \frac{c \left (a x \sqrt{1-\frac{1}{a^2 x^2}}-2 \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )-2 \sin ^{-1}\left (\frac{\sqrt{1-\frac{1}{a x}}}{\sqrt{2}}\right )-2 \sin ^{-1}\left (\frac{1}{a x}\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.128, size = 136, normalized size = 2.8 \begin{align*} -{\frac{c \left ( ax+1 \right ) }{a}\sqrt{{\frac{ax-1}{ax+1}}} \left ( \sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}+\arctan \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}-1}}} \right ) \sqrt{{a}^{2}}+2\,a\ln \left ({\frac{{a}^{2}x+\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}{\sqrt{{a}^{2}}}} \right ) -2\,\sqrt{{a}^{2}}\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) } \right ){\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((c-c/a/x)*((a*x-1)/(a*x+1))^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.59328, size = 154, normalized size = 3.14 \begin{align*} -2 \, a{\left (\frac{c \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} - \frac{c \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right )}{a^{2}} + \frac{c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac{c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.00251, size = 223, normalized size = 4.55 \begin{align*} \frac{2 \, c \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) - 2 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) + 2 \, c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) +{\left (a c x + c\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18115, size = 115, normalized size = 2.35 \begin{align*} \frac{2 \, c \arctan \left (-x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1}\right ) \mathrm{sgn}\left (a x + 1\right )}{a} + \frac{2 \, c \log \left ({\left | -x{\left | a \right |} + \sqrt{a^{2} x^{2} - 1} \right |}\right ) \mathrm{sgn}\left (a x + 1\right )}{{\left | a \right |}} + \frac{\sqrt{a^{2} x^{2} - 1} c \mathrm{sgn}\left (a x + 1\right )}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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