3.1027 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^4} \, dx\)

Optimal. Leaf size=15 \[ -\frac{c (a x+1)^3}{3 x^3} \]

[Out]

-(c*(1 + a*x)^3)/(3*x^3)

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Rubi [A]  time = 0.0448639, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {6150, 37} \[ -\frac{c (a x+1)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2))/x^4,x]

[Out]

-(c*(1 + a*x)^3)/(3*x^3)

Rule 6150

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

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0087305, size = 15, normalized size = 1. \[ -\frac{c (a x+1)^3}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2))/x^4,x]

[Out]

-(c*(1 + a*x)^3)/(3*x^3)

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Maple [A]  time = 0.03, size = 23, normalized size = 1.5 \begin{align*} c \left ( -{\frac{{a}^{2}}{x}}-{\frac{a}{{x}^{2}}}-{\frac{1}{3\,{x}^{3}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)/x^4,x)

[Out]

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

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Maxima [A]  time = 0.940828, size = 28, normalized size = 1.87 \begin{align*} -\frac{3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)/x^4,x, algorithm="maxima")

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3

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Fricas [A]  time = 1.66385, size = 51, normalized size = 3.4 \begin{align*} -\frac{3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3

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Sympy [A]  time = 0.330752, size = 24, normalized size = 1.6 \begin{align*} - \frac{3 a^{2} c x^{2} + 3 a c x + c}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*(-a**2*c*x**2+c)/x**4,x)

[Out]

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

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Giac [A]  time = 1.13783, size = 28, normalized size = 1.87 \begin{align*} -\frac{3 \, a^{2} c x^{2} + 3 \, a c x + c}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*(-a^2*c*x^2+c)/x^4,x, algorithm="giac")

[Out]

-1/3*(3*a^2*c*x^2 + 3*a*c*x + c)/x^3