3.294 \(\int \frac{e^{\tanh ^{-1}(a x)} (c-a c x)}{x^4} \, dx\)

Optimal. Leaf size=22 \[ -\frac{c \left (1-a^2 x^2\right )^{3/2}}{3 x^3} \]

[Out]

-(c*(1 - a^2*x^2)^(3/2))/(3*x^3)

________________________________________________________________________________________

Rubi [A]  time = 0.0392372, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {6128, 264} \[ -\frac{c \left (1-a^2 x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcTanh[a*x]*(c - a*c*x))/x^4,x]

[Out]

-(c*(1 - a^2*x^2)^(3/2))/(3*x^3)

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0141034, size = 22, normalized size = 1. \[ -\frac{c \left (1-a^2 x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^ArcTanh[a*x]*(c - a*c*x))/x^4,x]

[Out]

-(c*(1 - a^2*x^2)^(3/2))/(3*x^3)

________________________________________________________________________________________

Maple [A]  time = 0.03, size = 33, normalized size = 1.5 \begin{align*} -{\frac{ \left ( ax+1 \right ) ^{2} \left ( ax-1 \right ) ^{2}c}{3\,{x}^{3}}{\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}}} \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^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.43693, size = 54, normalized size = 2.45 \begin{align*} \frac{\sqrt{-a^{2} x^{2} + 1} a^{2} c}{3 \, x} - \frac{\sqrt{-a^{2} x^{2} + 1} c}{3 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.6253, size = 59, normalized size = 2.68 \begin{align*} \frac{{\left (a^{2} c x^{2} - c\right )} \sqrt{-a^{2} x^{2} + 1}}{3 \, x^{3}} \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^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 3.35489, size = 133, normalized size = 6.05 \begin{align*} - a^{2} c \left (\begin{cases} - \frac{i \sqrt{a^{2} x^{2} - 1}}{x} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{\sqrt{- a^{2} x^{2} + 1}}{x} & \text{otherwise} \end{cases}\right ) + c \left (\begin{cases} - \frac{2 i a^{2} \sqrt{a^{2} x^{2} - 1}}{3 x} - \frac{i \sqrt{a^{2} x^{2} - 1}}{3 x^{3}} & \text{for}\: \left |{a^{2} x^{2}}\right | > 1 \\- \frac{2 a^{2} \sqrt{- a^{2} x^{2} + 1}}{3 x} - \frac{\sqrt{- a^{2} x^{2} + 1}}{3 x^{3}} & \text{otherwise} \end{cases}\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**4,x)

[Out]

-a**2*c*Piecewise((-I*sqrt(a**2*x**2 - 1)/x, Abs(a**2*x**2) > 1), (-sqrt(-a**2*x**2 + 1)/x, True)) + c*Piecewi
se((-2*I*a**2*sqrt(a**2*x**2 - 1)/(3*x) - I*sqrt(a**2*x**2 - 1)/(3*x**3), Abs(a**2*x**2) > 1), (-2*a**2*sqrt(-
a**2*x**2 + 1)/(3*x) - sqrt(-a**2*x**2 + 1)/(3*x**3), True))

________________________________________________________________________________________

Giac [B]  time = 1.29068, size = 167, normalized size = 7.59 \begin{align*} \frac{{\left (a^{4} c - \frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} c}{x^{2}}\right )} a^{6} x^{3}}{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left | a \right |}} + \frac{\frac{3 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4} c}{x} - \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3} c}{x^{3}}}{24 \, a^{2}{\left | a \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^4,x, algorithm="giac")

[Out]

1/24*(a^4*c - 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*c/x^2)*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*abs(a)) +
1/24*(3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4*c/x - (sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*c/x^3)/(a^2*abs(a))