3.293 \(\int \frac{e^{\tanh ^{-1}(a x)} (c-a c x)}{x^3} \, dx\)

Optimal. Leaf size=46 \[ \frac{1}{2} a^2 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{c \sqrt{1-a^2 x^2}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0578259, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {6128, 266, 47, 63, 208} \[ \frac{1}{2} a^2 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-\frac{c \sqrt{1-a^2 x^2}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]

[Out]

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

Rule 6128

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[c^n,
Int[(e + f*x)^m*(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, e, f, m, p}, x] && EqQ[a*c +
d, 0] && IntegerQ[(n - 1)/2] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p - n/2 - 1, 0]) && IntegerQ[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 \frac{e^{\tanh ^{-1}(a x)} (c-a c x)}{x^3} \, dx &=c \int \frac{\sqrt{1-a^2 x^2}}{x^3} \, dx\\ &=\frac{1}{2} c \operatorname{Subst}\left (\int \frac{\sqrt{1-a^2 x}}{x^2} \, dx,x,x^2\right )\\ &=-\frac{c \sqrt{1-a^2 x^2}}{2 x^2}-\frac{1}{4} \left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac{c \sqrt{1-a^2 x^2}}{2 x^2}+\frac{1}{2} c \operatorname{Subst}\left (\int \frac{1}{\frac{1}{a^2}-\frac{x^2}{a^2}} \, dx,x,\sqrt{1-a^2 x^2}\right )\\ &=-\frac{c \sqrt{1-a^2 x^2}}{2 x^2}+\frac{1}{2} a^2 c \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0228295, size = 67, normalized size = 1.46 \[ \frac{c \left (a^2 x^2+a^2 x^2 \sqrt{1-a^2 x^2} \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )-1\right )}{2 x^2 \sqrt{1-a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x))/x^3,x]

[Out]

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

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Maple [A]  time = 0.039, size = 40, normalized size = 0.9 \begin{align*} -c \left ( -{\frac{{a}^{2}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{1}{2\,{x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x)

[Out]

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

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Maxima [A]  time = 1.42605, size = 69, normalized size = 1.5 \begin{align*} \frac{1}{2} \, a^{2} c \log \left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) - \frac{\sqrt{-a^{2} x^{2} + 1} c}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.68195, size = 104, normalized size = 2.26 \begin{align*} -\frac{a^{2} c x^{2} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt{-a^{2} x^{2} + 1} c}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 21.6313, size = 78, normalized size = 1.7 \begin{align*} \frac{a^{2} \left (- \frac{c \log{\left (-1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}} \right )}}{2} + \frac{c \log{\left (1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}} \right )}}{2} - \frac{c}{2 \left (1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}}\right )} - \frac{c}{2 \left (-1 + \frac{1}{\sqrt{- a^{2} x^{2} + 1}}\right )}\right )}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a**2*x**2+1)**(1/2)*(-a*c*x+c)/x**3,x)

[Out]

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

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Giac [A]  time = 1.28067, size = 82, normalized size = 1.78 \begin{align*} -\frac{1}{4} \, a^{2} c{\left (\frac{2 \, \sqrt{-a^{2} x^{2} + 1}}{a^{2} x^{2}} - \log \left (\sqrt{-a^{2} x^{2} + 1} + 1\right ) + \log \left (-\sqrt{-a^{2} x^{2} + 1} + 1\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)/(-a^2*x^2+1)^(1/2)*(-a*c*x+c)/x^3,x, algorithm="giac")

[Out]

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