3.1038 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x^6} \, dx\)

Optimal. Leaf size=42 \[ \frac{a^3 c^2}{x^2}+\frac{a^4 c^2}{x}-\frac{a c^2}{2 x^4}-\frac{c^2}{5 x^5} \]

[Out]

-c^2/(5*x^5) - (a*c^2)/(2*x^4) + (a^3*c^2)/x^2 + (a^4*c^2)/x

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Rubi [A]  time = 0.080575, antiderivative size = 42, 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} \[ \frac{a^3 c^2}{x^2}+\frac{a^4 c^2}{x}-\frac{a c^2}{2 x^4}-\frac{c^2}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^2)/x^6,x]

[Out]

-c^2/(5*x^5) - (a*c^2)/(2*x^4) + (a^3*c^2)/x^2 + (a^4*c^2)/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^6} \, dx &=c^2 \int \frac{(1-a x) (1+a x)^3}{x^6} \, dx\\ &=c^2 \int \left (\frac{1}{x^6}+\frac{2 a}{x^5}-\frac{2 a^3}{x^3}-\frac{a^4}{x^2}\right ) \, dx\\ &=-\frac{c^2}{5 x^5}-\frac{a c^2}{2 x^4}+\frac{a^3 c^2}{x^2}+\frac{a^4 c^2}{x}\\ \end{align*}

Mathematica [A]  time = 0.010401, size = 23, normalized size = 0.55 \[ \frac{c^2 (a x+1)^4 (3 a x-2)}{10 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*(c - a^2*c*x^2)^2)/x^6,x]

[Out]

(c^2*(1 + a*x)^4*(-2 + 3*a*x))/(10*x^5)

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Maple [A]  time = 0.032, size = 31, normalized size = 0.7 \begin{align*}{c}^{2} \left ( -{\frac{a}{2\,{x}^{4}}}+{\frac{{a}^{4}}{x}}-{\frac{1}{5\,{x}^{5}}}+{\frac{{a}^{3}}{{x}^{2}}} \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^6,x)

[Out]

c^2*(-1/2*a/x^4+a^4/x-1/5/x^5+a^3/x^2)

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Maxima [A]  time = 0.95437, size = 54, normalized size = 1.29 \begin{align*} \frac{10 \, a^{4} c^{2} x^{4} + 10 \, a^{3} c^{2} x^{3} - 5 \, a c^{2} x - 2 \, c^{2}}{10 \, x^{5}} \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^6,x, algorithm="maxima")

[Out]

1/10*(10*a^4*c^2*x^4 + 10*a^3*c^2*x^3 - 5*a*c^2*x - 2*c^2)/x^5

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Fricas [A]  time = 1.6516, size = 86, normalized size = 2.05 \begin{align*} \frac{10 \, a^{4} c^{2} x^{4} + 10 \, a^{3} c^{2} x^{3} - 5 \, a c^{2} x - 2 \, c^{2}}{10 \, x^{5}} \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^6,x, algorithm="fricas")

[Out]

1/10*(10*a^4*c^2*x^4 + 10*a^3*c^2*x^3 - 5*a*c^2*x - 2*c^2)/x^5

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Sympy [A]  time = 0.412481, size = 41, normalized size = 0.98 \begin{align*} \frac{10 a^{4} c^{2} x^{4} + 10 a^{3} c^{2} x^{3} - 5 a c^{2} x - 2 c^{2}}{10 x^{5}} \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**6,x)

[Out]

(10*a**4*c**2*x**4 + 10*a**3*c**2*x**3 - 5*a*c**2*x - 2*c**2)/(10*x**5)

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Giac [A]  time = 1.11447, size = 54, normalized size = 1.29 \begin{align*} \frac{10 \, a^{4} c^{2} x^{4} + 10 \, a^{3} c^{2} x^{3} - 5 \, a c^{2} x - 2 \, c^{2}}{10 \, x^{5}} \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^6,x, algorithm="giac")

[Out]

1/10*(10*a^4*c^2*x^4 + 10*a^3*c^2*x^3 - 5*a*c^2*x - 2*c^2)/x^5