3.1025 \(\int \frac{e^{2 \tanh ^{-1}(a x)} (c-a^2 c x^2)}{x^2} \, dx\)

Optimal. Leaf size=19 \[ a^2 c x+2 a c \log (x)-\frac{c}{x} \]

[Out]

-(c/x) + a^2*c*x + 2*a*c*Log[x]

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Rubi [A]  time = 0.0494398, antiderivative size = 19, 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} \[ a^2 c x+2 a c \log (x)-\frac{c}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0125687, size = 18, normalized size = 0.95 \[ c \left (a^2 x+2 a \log (x)-\frac{1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c*(-x^(-1) + a^2*x + 2*a*Log[x])

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Maple [A]  time = 0.032, size = 20, normalized size = 1.1 \begin{align*} -{\frac{c}{x}}+cx{a}^{2}+2\,ac\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^2,x)

[Out]

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

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Maxima [A]  time = 0.94102, size = 26, normalized size = 1.37 \begin{align*} a^{2} c x + 2 \, a c \log \left (x\right ) - \frac{c}{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)/x^2,x, algorithm="maxima")

[Out]

a^2*c*x + 2*a*c*log(x) - c/x

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Fricas [A]  time = 1.71466, size = 49, normalized size = 2.58 \begin{align*} \frac{a^{2} c x^{2} + 2 \, a c x \log \left (x\right ) - c}{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)/x^2,x, algorithm="fricas")

[Out]

(a^2*c*x^2 + 2*a*c*x*log(x) - c)/x

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Sympy [A]  time = 0.277468, size = 17, normalized size = 0.89 \begin{align*} a^{2} c x + 2 a c \log{\left (x \right )} - \frac{c}{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)/x**2,x)

[Out]

a**2*c*x + 2*a*c*log(x) - c/x

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Giac [A]  time = 1.15575, size = 27, normalized size = 1.42 \begin{align*} a^{2} c x + 2 \, a c \log \left ({\left | x \right |}\right ) - \frac{c}{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)/x^2,x, algorithm="giac")

[Out]

a^2*c*x + 2*a*c*log(abs(x)) - c/x