3.1024 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x} \, dx\)

Optimal. Leaf size=21 \[ \frac{1}{2} a^2 c x^2+2 a c x+c \log (x) \]

[Out]

2*a*c*x + (a^2*c*x^2)/2 + c*Log[x]

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Rubi [A]  time = 0.0493203, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {6150, 43} \[ \frac{1}{2} a^2 c x^2+2 a c x+c \log (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*c*x + (a^2*c*x^2)/2 + c*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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0147296, size = 19, normalized size = 0.9 \[ c \left (\frac{a^2 x^2}{2}+2 a x+\log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c*(2*a*x + (a^2*x^2)/2 + Log[x])

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Maple [A]  time = 0.026, size = 20, normalized size = 1. \begin{align*} 2\,acx+{\frac{{a}^{2}c{x}^{2}}{2}}+c\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)/x,x)

[Out]

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

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Maxima [A]  time = 0.944807, size = 26, normalized size = 1.24 \begin{align*} \frac{1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \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)/x,x, algorithm="maxima")

[Out]

1/2*a^2*c*x^2 + 2*a*c*x + c*log(x)

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Fricas [A]  time = 2.01512, size = 49, normalized size = 2.33 \begin{align*} \frac{1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \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)/x,x, algorithm="fricas")

[Out]

1/2*a^2*c*x^2 + 2*a*c*x + c*log(x)

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Sympy [A]  time = 0.10553, size = 20, normalized size = 0.95 \begin{align*} \frac{a^{2} c x^{2}}{2} + 2 a c x + c \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)/x,x)

[Out]

a**2*c*x**2/2 + 2*a*c*x + c*log(x)

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Giac [A]  time = 1.16146, size = 27, normalized size = 1.29 \begin{align*} \frac{1}{2} \, a^{2} c x^{2} + 2 \, a c x + c \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)/x,x, algorithm="giac")

[Out]

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