### 3.71 $$\int e^{\frac{3}{2} \coth ^{-1}(a x)} x \, dx$$

Optimal. Leaf size=142 $-\frac{9 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{9 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{1}{2} x^2 \sqrt [4]{1-\frac{1}{a x}} \left (\frac{1}{a x}+1\right )^{7/4}+\frac{3 x \sqrt [4]{1-\frac{1}{a x}} \left (\frac{1}{a x}+1\right )^{3/4}}{4 a}$

[Out]

(3*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x)/(4*a) + ((1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(7/4)*x^2)/2 - (9*Arc
Tan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(4*a^2) + (9*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(
4*a^2)

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Rubi [A]  time = 0.0590618, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.583, Rules used = {6171, 96, 94, 93, 298, 203, 206} $-\frac{9 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{9 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{1}{2} x^2 \sqrt [4]{1-\frac{1}{a x}} \left (\frac{1}{a x}+1\right )^{7/4}+\frac{3 x \sqrt [4]{1-\frac{1}{a x}} \left (\frac{1}{a x}+1\right )^{3/4}}{4 a}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^((3*ArcCoth[a*x])/2)*x,x]

[Out]

(3*(1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(3/4)*x)/(4*a) + ((1 - 1/(a*x))^(1/4)*(1 + 1/(a*x))^(7/4)*x^2)/2 - (9*Arc
Tan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(4*a^2) + (9*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(
4*a^2)

Rule 6171

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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 298

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

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])

Rubi steps

\begin{align*} \int e^{\frac{3}{2} \coth ^{-1}(a x)} x \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^{3/4}}{x^3 \left (1-\frac{x}{a}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{7/4} x^2-\frac{3 \operatorname{Subst}\left (\int \frac{\left (1+\frac{x}{a}\right )^{3/4}}{x^2 \left (1-\frac{x}{a}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )}{4 a}\\ &=\frac{3 \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{3/4} x}{4 a}+\frac{1}{2} \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{7/4} x^2-\frac{9 \operatorname{Subst}\left (\int \frac{1}{x \left (1-\frac{x}{a}\right )^{3/4} \sqrt [4]{1+\frac{x}{a}}} \, dx,x,\frac{1}{x}\right )}{8 a^2}\\ &=\frac{3 \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{3/4} x}{4 a}+\frac{1}{2} \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{7/4} x^2-\frac{9 \operatorname{Subst}\left (\int \frac{x^2}{-1+x^4} \, dx,x,\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{2 a^2}\\ &=\frac{3 \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{3/4} x}{4 a}+\frac{1}{2} \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{7/4} x^2+\frac{9 \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 )}{4 a^2}-\frac{9 \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 )}{4 a^2}\\ &=\frac{3 \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{3/4} x}{4 a}+\frac{1}{2} \sqrt [4]{1-\frac{1}{a x}} \left (1+\frac{1}{a x}\right )^{7/4} x^2-\frac{9 \tan ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{9 \tanh ^{-1}\left (\frac{\sqrt [4]{1+\frac{1}{a x}}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.171909, size = 70, normalized size = 0.49 $\frac{\frac{2 e^{\frac{3}{2} \coth ^{-1}(a x)} \left (7 e^{2 \coth ^{-1}(a x)}-3\right )}{\left (e^{2 \coth ^{-1}(a x)}-1\right )^2}-9 \tan ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )+9 \tanh ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )}{4 a^2}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^((3*ArcCoth[a*x])/2)*x,x]

[Out]

((2*E^((3*ArcCoth[a*x])/2)*(-3 + 7*E^(2*ArcCoth[a*x])))/(-1 + E^(2*ArcCoth[a*x]))^2 - 9*ArcTan[E^(ArcCoth[a*x]
/2)] + 9*ArcTanh[E^(ArcCoth[a*x]/2)])/(4*a^2)

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Maple [F]  time = 0.13, size = 0, normalized size = 0. \begin{align*} \int{x \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(3/4)*x,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(3/4)*x,x)

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Maxima [A]  time = 1.50407, size = 205, normalized size = 1.44 \begin{align*} \frac{1}{8} \, a{\left (\frac{4 \,{\left (3 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{4}} - 7 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}}{\frac{2 \,{\left (a x - 1\right )} a^{3}}{a x + 1} - \frac{{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} + \frac{18 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{3}} + \frac{9 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{3}} - \frac{9 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{a^{3}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x,x, algorithm="maxima")

[Out]

1/8*a*(4*(3*((a*x - 1)/(a*x + 1))^(5/4) - 7*((a*x - 1)/(a*x + 1))^(1/4))/(2*(a*x - 1)*a^3/(a*x + 1) - (a*x - 1
)^2*a^3/(a*x + 1)^2 - a^3) + 18*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 + 9*log(((a*x - 1)/(a*x + 1))^(1/4) +
1)/a^3 - 9*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3)

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Fricas [A]  time = 1.76629, size = 254, normalized size = 1.79 \begin{align*} \frac{2 \,{\left (2 \, a^{2} x^{2} + 7 \, a x + 5\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 18 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right ) + 9 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right ) - 9 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1\right )}{8 \, a^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x,x, algorithm="fricas")

[Out]

1/8*(2*(2*a^2*x^2 + 7*a*x + 5)*((a*x - 1)/(a*x + 1))^(1/4) + 18*arctan(((a*x - 1)/(a*x + 1))^(1/4)) + 9*log(((
a*x - 1)/(a*x + 1))^(1/4) + 1) - 9*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{4}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/4)*x,x)

[Out]

Integral(x/((a*x - 1)/(a*x + 1))**(3/4), x)

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Giac [A]  time = 1.21384, size = 190, normalized size = 1.34 \begin{align*} \frac{1}{8} \, a{\left (\frac{18 \, \arctan \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}{a^{3}} + \frac{9 \, \log \left (\left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} + 1\right )}{a^{3}} - \frac{9 \, \log \left ({\left | \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}} - 1 \right |}\right )}{a^{3}} - \frac{4 \,{\left (\frac{3 \,{\left (a x - 1\right )} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}}{a x + 1} - 7 \, \left (\frac{a x - 1}{a x + 1}\right )^{\frac{1}{4}}\right )}}{a^{3}{\left (\frac{a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/4)*x,x, algorithm="giac")

[Out]

1/8*a*(18*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 + 9*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^3 - 9*log(abs(((a
*x - 1)/(a*x + 1))^(1/4) - 1))/a^3 - 4*(3*(a*x - 1)*((a*x - 1)/(a*x + 1))^(1/4)/(a*x + 1) - 7*((a*x - 1)/(a*x
+ 1))^(1/4))/(a^3*((a*x - 1)/(a*x + 1) - 1)^2))