### 3.63 $$\int e^{\frac{1}{2} \coth ^{-1}(a x)} \, dx$$

Optimal. Leaf size=96 $x \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{\frac{1}{a x}+1}+\frac{\tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}$

[Out]

(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x + ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/a + ArcTanh[(1 + 1
/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/a

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Rubi [A]  time = 0.0368659, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {6170, 94, 93, 212, 206, 203} $x \left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{\frac{1}{a x}+1}+\frac{\tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x + ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/a + ArcTanh[(1 + 1
/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)]/a

Rule 6170

Int[E^(ArcCoth[(a_.)*(x_)]*(n_)), x_Symbol] :> -Subst[Int[(1 + x/a)^(n/2)/(x^2*(1 - x/a)^(n/2)), x], x, 1/x] /
; FreeQ[{a, n}, x] &&  !IntegerQ[n]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int e^{\frac{1}{2} \coth ^{-1}(a x)} \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt [4]{1+\frac{x}{a}}}{x^2 \sqrt [4]{1-\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )\\ &=\left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}} x-\frac{\operatorname{Subst}\left (\int \frac{1}{x \sqrt [4]{1-\frac{x}{a}} \left (1+\frac{x}{a}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )}{2 a}\\ &=\left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}} x-\frac{2 \operatorname{Subst}\left (\int \frac{1}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}\\ &=\left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}} x+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}+\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}\\ &=\left (1-\frac{1}{a x}\right )^{3/4} \sqrt [4]{1+\frac{1}{a x}} x+\frac{\tan ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}+\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{a}\\ \end{align*}

Mathematica [A]  time = 0.0882639, size = 51, normalized size = 0.53 $\frac{\frac{2 e^{\frac{1}{2} \coth ^{-1}(a x)}}{e^{2 \coth ^{-1}(a x)}-1}+\tan ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )+\tanh ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )}{a}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*E^(ArcCoth[a*x]/2))/(-1 + E^(2*ArcCoth[a*x])) + ArcTan[E^(ArcCoth[a*x]/2)] + ArcTanh[E^(ArcCoth[a*x]/2)])/
a

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Maple [F]  time = 0.138, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{\sqrt [4]{{\frac{ax-1}{ax+1}}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/4),x)

[Out]

int(1/((a*x-1)/(a*x+1))^(1/4),x)

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Maxima [A]  time = 1.47326, size = 150, normalized size = 1.56 \begin{align*} -\frac{1}{2} \, a{\left (\frac{4 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{\frac{{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac{2 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{2}} - \frac{\log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{2}} + \frac{\log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{a^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.60895, size = 225, normalized size = 2.34 \begin{align*} \frac{2 \,{\left (a x + 1\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}} - 2 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) + \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right ) - \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{2 \, a} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt [4]{\frac{a x - 1}{a x + 1}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/4),x)

[Out]

Integral(((a*x - 1)/(a*x + 1))**(-1/4), x)

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Giac [A]  time = 1.24035, size = 146, normalized size = 1.52 \begin{align*} -\frac{1}{2} \, a{\left (\frac{2 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{2}} - \frac{\log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{2}} + \frac{\log \left ({\left | \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1 \right |}\right )}{a^{2}} + \frac{4 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}{a^{2}{\left (\frac{a x - 1}{a x + 1} - 1\right )}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/4),x, algorithm="giac")

[Out]

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