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

Optimal. Leaf size=47 \[ \frac{c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0706478, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6175, 6178, 266, 47, 63, 208} \[ \frac{c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}-\frac{1}{2} a c x^2 \sqrt{1-\frac{1}{a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*(c - a*c*x),x]

[Out]

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

Rule 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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)} (c-a c x) \, dx &=-\left ((a c) \int e^{\coth ^{-1}(a x)} \left (1-\frac{1}{a x}\right ) x \, dx\right )\\ &=(a c) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} (a c) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x}{a^2}}}{x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2-\frac{c \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-\frac{x}{a^2}}} \, dx,x,\frac{1}{x^2}\right )}{4 a}\\ &=-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{1}{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 )\\ &=-\frac{1}{2} a c \sqrt{1-\frac{1}{a^2 x^2}} x^2+\frac{c \tanh ^{-1}\left (\sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.0502585, size = 51, normalized size = 1.09 \[ \frac{c \left (\log \left (a x \left (\sqrt{1-\frac{1}{a^2 x^2}}+1\right )\right )-a^2 x^2 \sqrt{1-\frac{1}{a^2 x^2}}\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.135, size = 93, normalized size = 2. \begin{align*} -{\frac{c \left ( ax-1 \right ) }{2} \left ( x\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}}-\ln \left ({ \left ({a}^{2}x+\sqrt{{a}^{2}{x}^{2}-1}\sqrt{{a}^{2}} \right ){\frac{1}{\sqrt{{a}^{2}}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{\frac{1}{\sqrt{ \left ( ax-1 \right ) \left ( ax+1 \right ) }}}{\frac{1}{\sqrt{{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.0445, size = 178, normalized size = 3.79 \begin{align*} \frac{1}{2} \, a{\left (\frac{2 \,{\left (c \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + c \sqrt{\frac{a x - 1}{a x + 1}}\right )}}{\frac{2 \,{\left (a x - 1\right )} a^{2}}{a x + 1} - \frac{{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - 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(1/((a*x-1)/(a*x+1))^(1/2)*(-a*c*x+c),x, algorithm="maxima")

[Out]

1/2*a*(2*(c*((a*x - 1)/(a*x + 1))^(3/2) + c*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - (a*x - 1)^
2*a^2/(a*x + 1)^2 - 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 = 1.57671, size = 180, normalized size = 3.83 \begin{align*} \frac{c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + 1\right ) - c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} - 1\right ) -{\left (a^{2} c x^{2} + a c x\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.26605, size = 211, normalized size = 4.49 \begin{align*} \frac{1}{4} \, a{\left (\frac{c \log \left (\sqrt{\frac{a x - 1}{a x + 1}} + \frac{1}{\sqrt{\frac{a x - 1}{a x + 1}}} + 2\right )}{a^{2}} - \frac{c \log \left ({\left | \sqrt{\frac{a x - 1}{a x + 1}} + \frac{1}{\sqrt{\frac{a x - 1}{a x + 1}}} - 2 \right |}\right )}{a^{2}} - \frac{4 \, c{\left (\sqrt{\frac{a x - 1}{a x + 1}} + \frac{1}{\sqrt{\frac{a x - 1}{a x + 1}}}\right )}}{{\left ({\left (\sqrt{\frac{a x - 1}{a x + 1}} + \frac{1}{\sqrt{\frac{a x - 1}{a x + 1}}}\right )}^{2} - 4\right )} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*(c*log(sqrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)) + 2)/a^2 - c*log(abs(sqrt((a*x - 1)/(a*x
+ 1)) + 1/sqrt((a*x - 1)/(a*x + 1)) - 2))/a^2 - 4*c*(sqrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)))/
(((sqrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)))^2 - 4)*a^2))