3.1034 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x^2} \, dx\)

Optimal. Leaf size=41 \[ -\frac{1}{3} a^4 c^2 x^3-a^3 c^2 x^2+2 a c^2 \log (x)-\frac{c^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.0818491, antiderivative size = 41, 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} \[ -\frac{1}{3} a^4 c^2 x^3-a^3 c^2 x^2+2 a c^2 \log (x)-\frac{c^2}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0132127, size = 41, normalized size = 1. \[ -\frac{1}{3} a^4 c^2 x^3-a^3 c^2 x^2+2 a c^2 \log (x)-\frac{c^2}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 40, normalized size = 1. \begin{align*} -{\frac{{c}^{2}}{x}}-{a}^{3}{c}^{2}{x}^{2}-{\frac{{a}^{4}{c}^{2}{x}^{3}}{3}}+2\,a{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^2,x)

[Out]

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

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Maxima [A]  time = 0.949476, size = 53, normalized size = 1.29 \begin{align*} -\frac{1}{3} \, a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} + 2 \, a c^{2} \log \left (x\right ) - \frac{c^{2}}{x} \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^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67054, size = 88, normalized size = 2.15 \begin{align*} -\frac{a^{4} c^{2} x^{4} + 3 \, a^{3} c^{2} x^{3} - 6 \, a c^{2} x \log \left (x\right ) + 3 \, c^{2}}{3 \, x} \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^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.306966, size = 36, normalized size = 0.88 \begin{align*} - \frac{a^{4} c^{2} x^{3}}{3} - a^{3} c^{2} x^{2} + 2 a c^{2} \log{\left (x \right )} - \frac{c^{2}}{x} \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**2,x)

[Out]

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

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Giac [A]  time = 1.13346, size = 54, normalized size = 1.32 \begin{align*} -\frac{1}{3} \, a^{4} c^{2} x^{3} - a^{3} c^{2} x^{2} + 2 \, a c^{2} \log \left ({\left | x \right |}\right ) - \frac{c^{2}}{x} \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^2,x, algorithm="giac")

[Out]

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