3.1.27 \(\int e^{4 \coth ^{-1}(a x)} x \, dx\) [27]

Optimal. Leaf size=39 \[ \frac {4 x}{a}+\frac {x^2}{2}+\frac {4}{a^2 (1-a x)}+\frac {8 \log (1-a x)}{a^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6302, 6261, 78} \begin {gather*} \frac {4}{a^2 (1-a x)}+\frac {8 \log (1-a x)}{a^2}+\frac {4 x}{a}+\frac {x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6261

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[x^m*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rule 6302

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

________________________________________________________________________________________

Mathematica [A]
time = 0.02, size = 39, normalized size = 1.00 \begin {gather*} \frac {4 x}{a}+\frac {x^2}{2}+\frac {4}{a^2 (1-a x)}+\frac {8 \log (1-a x)}{a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 39, normalized size = 1.00

method	result	size
risch	\(\frac {x^{2}}{2}+\frac {4 x}{a}-\frac {4}{a^{2} \left (a x -1\right )}+\frac {8 \ln \left (a x -1\right )}{a^{2}}\)	\(36\)
default	\(\frac {\frac {1}{2} a \,x^{2}+4 x}{a}-\frac {4}{a^{2} \left (a x -1\right )}+\frac {8 \ln \left (a x -1\right )}{a^{2}}\)	\(39\)
norman	\(\frac {\frac {7 x^{2}}{2}+\frac {a \,x^{3}}{2}-\frac {8 x}{a}}{a x -1}+\frac {8 \ln \left (a x -1\right )}{a^{2}}\)	\(39\)
meijerg	\(\frac {\frac {a x \left (-2 a^{2} x^{2}-6 a x +12\right )}{-4 a x +4}+3 \ln \left (-a x +1\right )}{a^{2}}-\frac {2 \left (-\frac {a x \left (-3 a x +6\right )}{3 \left (-a x +1\right )}-2 \ln \left (-a x +1\right )\right )}{a^{2}}+\frac {\frac {a x}{-a x +1}+\ln \left (-a x +1\right )}{a^{2}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x-1)^2*(a*x+1)^2*x,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.26, size = 41, normalized size = 1.05 \begin {gather*} \frac {a x^{2} + 8 \, x}{2 \, a} - \frac {4}{a^{3} x - a^{2}} + \frac {8 \, \log \left (a x - 1\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a*x^2 + 8*x)/a - 4/(a^3*x - a^2) + 8*log(a*x - 1)/a^2

________________________________________________________________________________________

Fricas [A]
time = 0.35, size = 49, normalized size = 1.26 \begin {gather*} \frac {a^{3} x^{3} + 7 \, a^{2} x^{2} - 8 \, a x + 16 \, {\left (a x - 1\right )} \log \left (a x - 1\right ) - 8}{2 \, {\left (a^{3} x - a^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(a^3*x^3 + 7*a^2*x^2 - 8*a*x + 16*(a*x - 1)*log(a*x - 1) - 8)/(a^3*x - a^2)

________________________________________________________________________________________

Sympy [A]
time = 0.06, size = 31, normalized size = 0.79 \begin {gather*} \frac {x^{2}}{2} - \frac {4}{a^{3} x - a^{2}} + \frac {4 x}{a} + \frac {8 \log {\left (a x - 1 \right )}}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2/2 - 4/(a**3*x - a**2) + 4*x/a + 8*log(a*x - 1)/a**2

________________________________________________________________________________________

Giac [A]
time = 0.40, size = 64, normalized size = 1.64 \begin {gather*} \frac {\frac {{\left (a x - 1\right )}^{2} {\left (\frac {10}{a x - 1} + 1\right )}}{a} - \frac {16 \, \log \left (\frac {{\left | a x - 1 \right |}}{{\left (a x - 1\right )}^{2} {\left | a \right |}}\right )}{a} - \frac {8}{{\left (a x - 1\right )} a}}{2 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x - 1)^2*(10/(a*x - 1) + 1)/a - 16*log(abs(a*x - 1)/((a*x - 1)^2*abs(a)))/a - 8/((a*x - 1)*a))/a

________________________________________________________________________________________

Mupad [B]
time = 0.04, size = 38, normalized size = 0.97 \begin {gather*} \frac {8\,\ln \left (a\,x-1\right )}{a^2}+\frac {4\,x}{a}+\frac {x^2}{2}+\frac {4}{a\,\left (a-a^2\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a*x + 1)^2)/(a*x - 1)^2,x)

[Out]

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

________________________________________________________________________________________