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

Optimal. Leaf size=25 \[ -\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a}+c x \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6167, 6131, 6129, 72} \[ -\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a}+c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6129

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

Rule 6131

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

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 25, normalized size = 1.00 \[ -\frac {c \log (x)}{a}+\frac {4 c \log (1-a x)}{a}+c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.66, size = 23, normalized size = 0.92 \[ \frac {a c x + 4 \, c \log \left (a x - 1\right ) - c \log \relax (x)}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.12, size = 55, normalized size = 2.20 \[ \frac {{\left (a x - 1\right )} c}{a} - \frac {3 \, c \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {c \log \left ({\left | -\frac {1}{a x - 1} - 1 \right |}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 25, normalized size = 1.00 \[ c x +\frac {4 c \ln \left (a x -1\right )}{a}-\frac {c \ln \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 24, normalized size = 0.96 \[ c x + \frac {4 \, c \log \left (a x - 1\right )}{a} - \frac {c \log \relax (x)}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x + 4*c*log(a*x - 1)/a - c*log(x)/a

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 24, normalized size = 0.96 \[ c\,x-\frac {c\,\ln \relax (x)}{a}+\frac {4\,c\,\ln \left (a\,x-1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.21, size = 17, normalized size = 0.68 \[ c x + \frac {c \left (- \log {\relax (x )} + 4 \log {\left (x - \frac {1}{a} \right )}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x + c*(-log(x) + 4*log(x - 1/a))/a

________________________________________________________________________________________