3.801 \(\int e^{4 \coth ^{-1}(a x)} (c-\frac{c}{a^2 x^2}) \, dx\)

Optimal. Leaf size=33 \[ \frac{c}{a^2 x}-\frac{4 c \log (x)}{a}+\frac{8 c \log (1-a x)}{a}+c x \]

[Out]

c/(a^2*x) + c*x - (4*c*Log[x])/a + (8*c*Log[1 - a*x])/a

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Rubi [A]  time = 0.0933374, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6167, 6157, 6150, 88} \[ \frac{c}{a^2 x}-\frac{4 c \log (x)}{a}+\frac{8 c \log (1-a x)}{a}+c x \]

Antiderivative was successfully verified.

[In]

Int[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2)),x]

[Out]

c/(a^2*x) + c*x - (4*c*Log[x])/a + (8*c*Log[1 - a*x])/a

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rule 6157

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[d^p, Int[(u*(1 - a^2*x^
2)^p*E^(n*ArcTanh[a*x]))/x^(2*p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c + a^2*d, 0] && IntegerQ[p]

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int e^{4 \coth ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right ) \, dx &=\int e^{4 \tanh ^{-1}(a x)} \left (c-\frac{c}{a^2 x^2}\right ) \, dx\\ &=-\frac{c \int \frac{e^{4 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )}{x^2} \, dx}{a^2}\\ &=-\frac{c \int \frac{(1+a x)^3}{x^2 (1-a x)} \, dx}{a^2}\\ &=-\frac{c \int \left (-a^2+\frac{1}{x^2}+\frac{4 a}{x}-\frac{8 a^2}{-1+a x}\right ) \, dx}{a^2}\\ &=\frac{c}{a^2 x}+c x-\frac{4 c \log (x)}{a}+\frac{8 c \log (1-a x)}{a}\\ \end{align*}

Mathematica [A]  time = 0.0199034, size = 33, normalized size = 1. \[ \frac{c}{a^2 x}-\frac{4 c \log (x)}{a}+\frac{8 c \log (1-a x)}{a}+c x \]

Antiderivative was successfully verified.

[In]

Integrate[E^(4*ArcCoth[a*x])*(c - c/(a^2*x^2)),x]

[Out]

c/(a^2*x) + c*x - (4*c*Log[x])/a + (8*c*Log[1 - a*x])/a

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Maple [A]  time = 0.055, size = 33, normalized size = 1. \begin{align*} cx+{\frac{c}{{a}^{2}x}}-4\,{\frac{c\ln \left ( x \right ) }{a}}+8\,{\frac{c\ln \left ( ax-1 \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2),x)

[Out]

c*x+c/a^2/x-4*c*ln(x)/a+8*c/a*ln(a*x-1)

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Maxima [A]  time = 1.02735, size = 43, normalized size = 1.3 \begin{align*} c x + \frac{8 \, c \log \left (a x - 1\right )}{a} - \frac{4 \, c \log \left (x\right )}{a} + \frac{c}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2),x, algorithm="maxima")

[Out]

c*x + 8*c*log(a*x - 1)/a - 4*c*log(x)/a + c/(a^2*x)

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Fricas [A]  time = 1.57583, size = 88, normalized size = 2.67 \begin{align*} \frac{a^{2} c x^{2} + 8 \, a c x \log \left (a x - 1\right ) - 4 \, a c x \log \left (x\right ) + c}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2),x, algorithm="fricas")

[Out]

(a^2*c*x^2 + 8*a*c*x*log(a*x - 1) - 4*a*c*x*log(x) + c)/(a^2*x)

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Sympy [A]  time = 0.457601, size = 26, normalized size = 0.79 \begin{align*} c x + \frac{4 c \left (- \log{\left (x \right )} + 2 \log{\left (x - \frac{1}{a} \right )}\right )}{a} + \frac{c}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)**2*(a*x+1)**2*(c-c/a**2/x**2),x)

[Out]

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

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Giac [A]  time = 1.11255, size = 89, normalized size = 2.7 \begin{align*} -\frac{4 \, c \log \left (\frac{{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2}{\left | a \right |}}\right )}{a} - \frac{4 \, c \log \left ({\left | -\frac{1}{a x - 1} - 1 \right |}\right )}{a} + \frac{{\left (a x - 1\right )} c}{a{\left (\frac{1}{a x - 1} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x-1)^2*(a*x+1)^2*(c-c/a^2/x^2),x, algorithm="giac")

[Out]

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