3.1033 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)^2}{x} \, dx\)

Optimal. Leaf size=40 \[ -\frac{1}{4} a^4 c^2 x^4-\frac{2}{3} a^3 c^2 x^3+2 a c^2 x+c^2 \log (x) \]

[Out]

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

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Rubi [A]  time = 0.0746111, antiderivative size = 40, 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} \[ -\frac{1}{4} a^4 c^2 x^4-\frac{2}{3} a^3 c^2 x^3+2 a c^2 x+c^2 \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0150586, size = 33, normalized size = 0.82 \[ -\frac{1}{12} c^2 \left (3 a^4 x^4+8 a^3 x^3-24 a x-12 \log (x)+3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(3 - 24*a*x + 8*a^3*x^3 + 3*a^4*x^4 - 12*Log[x]))/12

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Maple [A]  time = 0.026, size = 37, normalized size = 0.9 \begin{align*} 2\,a{c}^{2}x-{\frac{2\,{a}^{3}{c}^{2}{x}^{3}}{3}}-{\frac{{a}^{4}{c}^{2}{x}^{4}}{4}}+{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,x)

[Out]

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

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Maxima [A]  time = 0.944819, size = 49, normalized size = 1.22 \begin{align*} -\frac{1}{4} \, a^{4} c^{2} x^{4} - \frac{2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left (x\right ) \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,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77666, size = 82, normalized size = 2.05 \begin{align*} -\frac{1}{4} \, a^{4} c^{2} x^{4} - \frac{2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left (x\right ) \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,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.291865, size = 39, normalized size = 0.98 \begin{align*} - \frac{a^{4} c^{2} x^{4}}{4} - \frac{2 a^{3} c^{2} x^{3}}{3} + 2 a c^{2} x + c^{2} \log{\left (x \right )} \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,x)

[Out]

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

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Giac [A]  time = 1.12561, size = 50, normalized size = 1.25 \begin{align*} -\frac{1}{4} \, a^{4} c^{2} x^{4} - \frac{2}{3} \, a^{3} c^{2} x^{3} + 2 \, a c^{2} x + c^{2} \log \left ({\left | x \right |}\right ) \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,x, algorithm="giac")

[Out]

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