3.1036 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x^4} \, dx\)

Optimal. Leaf size=39 \[ a^4 \left (-c^2\right ) x-2 a^3 c^2 \log (x)-\frac{a c^2}{x^2}-\frac{c^2}{3 x^3} \]

[Out]

-c^2/(3*x^3) - (a*c^2)/x^2 - a^4*c^2*x - 2*a^3*c^2*Log[x]

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Rubi [A]  time = 0.0802476, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {6150, 75} \[ a^4 \left (-c^2\right ) x-2 a^3 c^2 \log (x)-\frac{a c^2}{x^2}-\frac{c^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c^2/(3*x^3) - (a*c^2)/x^2 - a^4*c^2*x - 2*a^3*c^2*Log[x]

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 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
 + p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

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

Mathematica [A]  time = 0.0150206, size = 39, normalized size = 1. \[ a^4 \left (-c^2\right ) x-2 a^3 c^2 \log (x)-\frac{a c^2}{x^2}-\frac{c^2}{3 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c^2/(3*x^3) - (a*c^2)/x^2 - a^4*c^2*x - 2*a^3*c^2*Log[x]

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Maple [A]  time = 0.031, size = 38, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{3\,{x}^{3}}}-{\frac{a{c}^{2}}{{x}^{2}}}-{a}^{4}{c}^{2}x-2\,{a}^{3}{c}^{2}\ln \left ( x \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)^2/x^4,x)

[Out]

-1/3*c^2/x^3-a*c^2/x^2-a^4*c^2*x-2*a^3*c^2*ln(x)

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Maxima [A]  time = 0.948565, size = 49, normalized size = 1.26 \begin{align*} -a^{4} c^{2} x - 2 \, a^{3} c^{2} \log \left (x\right ) - \frac{3 \, a c^{2} x + c^{2}}{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)^2/x^4,x, algorithm="maxima")

[Out]

-a^4*c^2*x - 2*a^3*c^2*log(x) - 1/3*(3*a*c^2*x + c^2)/x^3

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Fricas [A]  time = 1.67061, size = 90, normalized size = 2.31 \begin{align*} -\frac{3 \, a^{4} c^{2} x^{4} + 6 \, a^{3} c^{2} x^{3} \log \left (x\right ) + 3 \, a c^{2} x + c^{2}}{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)^2/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*a^4*c^2*x^4 + 6*a^3*c^2*x^3*log(x) + 3*a*c^2*x + c^2)/x^3

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Sympy [A]  time = 0.35641, size = 37, normalized size = 0.95 \begin{align*} - a^{4} c^{2} x - 2 a^{3} c^{2} \log{\left (x \right )} - \frac{3 a c^{2} x + c^{2}}{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)**2/x**4,x)

[Out]

-a**4*c**2*x - 2*a**3*c**2*log(x) - (3*a*c**2*x + c**2)/(3*x**3)

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Giac [A]  time = 1.13319, size = 50, normalized size = 1.28 \begin{align*} -a^{4} c^{2} x - 2 \, a^{3} c^{2} \log \left ({\left | x \right |}\right ) - \frac{3 \, a c^{2} x + c^{2}}{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)^2/x^4,x, algorithm="giac")

[Out]

-a^4*c^2*x - 2*a^3*c^2*log(abs(x)) - 1/3*(3*a*c^2*x + c^2)/x^3