\(\int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx\) [80]

Optimal result

Integrand size = 12, antiderivative size = 176 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {25 \sqrt [4]{1+\frac {1}{a x}}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {25 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \]

[Out]

-25/2*(1+1/a/x)^(1/4)/a^2/(1-1/a/x)^(1/4)+5/4*(1+1/a/x)^(5/4)*x/a/(1-1/a/x)^(1/4)+1/2*(1+1/a/x)^(9/4)*x^2/(1-1
/a/x)^(1/4)+25/4*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^2+25/4*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6306, 98, 96, 95, 218, 212, 209} \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {25 \arctan \left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \text {arctanh}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}-\frac {25 \sqrt [4]{\frac {1}{a x}+1}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {x^2 \left (\frac {1}{a x}+1\right )^{9/4}}{2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 x \left (\frac {1}{a x}+1\right )^{5/4}}{4 a \sqrt [4]{1-\frac {1}{a x}}} \]

[In]

Int[E^((5*ArcCoth[a*x])/2)*x,x]

[Out]

(-25*(1 + 1/(a*x))^(1/4))/(2*a^2*(1 - 1/(a*x))^(1/4)) + (5*(1 + 1/(a*x))^(5/4)*x)/(4*a*(1 - 1/(a*x))^(1/4)) +
((1 + 1/(a*x))^(9/4)*x^2)/(2*(1 - 1/(a*x))^(1/4)) + (25*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(4*a^
2) + (25*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/(4*a^2)

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 6306

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]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^{5/4}}{x^3 \left (1-\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}-\frac {5 \text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^{5/4}}{x^2 \left (1-\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{4 a} \\ & = \frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}-\frac {25 \text {Subst}\left (\int \frac {\sqrt [4]{1+\frac {x}{a}}}{x \left (1-\frac {x}{a}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{8 a^2} \\ & = -\frac {25 \sqrt [4]{1+\frac {1}{a x}}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}-\frac {25 \text {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 )}{8 a^2} \\ & = -\frac {25 \sqrt [4]{1+\frac {1}{a x}}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}-\frac {25 \text {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 )}{2 a^2} \\ & = -\frac {25 \sqrt [4]{1+\frac {1}{a x}}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {25 \text {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 {25 \text {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 {25 \sqrt [4]{1+\frac {1}{a x}}}{2 a^2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \left (1+\frac {1}{a x}\right )^{5/4} x}{4 a \sqrt [4]{1-\frac {1}{a x}}}+\frac {\left (1+\frac {1}{a x}\right )^{9/4} x^2}{2 \sqrt [4]{1-\frac {1}{a x}}}+\frac {25 \arctan \left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \text {arctanh}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.45 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {-\frac {2 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (25-45 e^{2 \coth ^{-1}(a x)}+16 e^{4 \coth ^{-1}(a x)}\right )}{\left (-1+e^{2 \coth ^{-1}(a x)}\right )^2}+25 \arctan \left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+25 \text {arctanh}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \]

[In]

Integrate[E^((5*ArcCoth[a*x])/2)*x,x]

[Out]

((-2*E^(ArcCoth[a*x]/2)*(25 - 45*E^(2*ArcCoth[a*x]) + 16*E^(4*ArcCoth[a*x])))/(-1 + E^(2*ArcCoth[a*x]))^2 + 25
*ArcTan[E^(ArcCoth[a*x]/2)] + 25*ArcTanh[E^(ArcCoth[a*x]/2)])/(4*a^2)

Maple [F]

\[\int \frac {x}{\left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}}d x\]

[In]

int(1/((a*x-1)/(a*x+1))^(5/4)*x,x)

[Out]

int(1/((a*x-1)/(a*x+1))^(5/4)*x,x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.73 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {50 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 25 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 25 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (2 \, a^{3} x^{3} + 11 \, a^{2} x^{2} - 34 \, a x - 43\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{8 \, {\left (a^{3} x - a^{2}\right )}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x,x, algorithm="fricas")

[Out]

-1/8*(50*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 25*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) + 2
5*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) - 1) - 2*(2*a^3*x^3 + 11*a^2*x^2 - 34*a*x - 43)*((a*x - 1)/(a*x +
1))^(3/4))/(a^3*x - a^2)

Sympy [F]

\[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\int \frac {x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}}}\, dx \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(5/4)*x,x)

[Out]

Integral(x/((a*x - 1)/(a*x + 1))**(5/4), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.94 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {1}{8} \, a {\left (\frac {4 \, {\left (\frac {45 \, {\left (a x - 1\right )}}{a x + 1} - \frac {25 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 16\right )}}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{4}} - 2 \, a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} + a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} - \frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x,x, algorithm="maxima")

[Out]

1/8*a*(4*(45*(a*x - 1)/(a*x + 1) - 25*(a*x - 1)^2/(a*x + 1)^2 - 16)/(a^3*((a*x - 1)/(a*x + 1))^(9/4) - 2*a^3*(
(a*x - 1)/(a*x + 1))^(5/4) + a^3*((a*x - 1)/(a*x + 1))^(1/4)) - 50*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 + 2
5*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^3 - 25*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=-\frac {1}{8} \, a {\left (\frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {25 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} + \frac {64}{a^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {4 \, {\left (\frac {9 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a x + 1} - 13 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(5/4)*x,x, algorithm="giac")

[Out]

-1/8*a*(50*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^3 - 25*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^3 + 25*log(abs(
((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^3 + 64/(a^3*((a*x - 1)/(a*x + 1))^(1/4)) + 4*(9*(a*x - 1)*((a*x - 1)/(a*x
+ 1))^(3/4)/(a*x + 1) - 13*((a*x - 1)/(a*x + 1))^(3/4))/(a^3*((a*x - 1)/(a*x + 1) - 1)^2))

Mupad [B] (verification not implemented)

Time = 4.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.79 \[ \int e^{\frac {5}{2} \coth ^{-1}(a x)} x \, dx=\frac {25\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {25\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {\frac {25\,{\left (a\,x-1\right )}^2}{2\,{\left (a\,x+1\right )}^2}-\frac {45\,\left (a\,x-1\right )}{2\,\left (a\,x+1\right )}+8}{a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-2\,a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}+a^2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/4}} \]

[In]

int(x/((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

(25*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/(4*a^2) - (25*atan(((a*x - 1)/(a*x + 1))^(1/4)))/(4*a^2) - ((25*(a*x -
1)^2)/(2*(a*x + 1)^2) - (45*(a*x - 1))/(2*(a*x + 1)) + 8)/(a^2*((a*x - 1)/(a*x + 1))^(1/4) - 2*a^2*((a*x - 1)
/(a*x + 1))^(5/4) + a^2*((a*x - 1)/(a*x + 1))^(9/4))