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

Optimal. Leaf size=176 \[ -\frac {25 \sqrt [4]{1-\frac {1}{a x}}}{2 a^2 \sqrt [4]{\frac {1}{a x}+1}}-\frac {25 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}-\frac {5 x \left (1-\frac {1}{a x}\right )^{5/4}}{4 a \sqrt [4]{\frac {1}{a x}+1}} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {25 \sqrt [4]{1-\frac {1}{a x}}}{2 a^2 \sqrt [4]{\frac {1}{a x}+1}}-\frac {25 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {x^2 \left (1-\frac {1}{a x}\right )^{9/4}}{2 \sqrt [4]{\frac {1}{a x}+1}}-\frac {5 x \left (1-\frac {1}{a x}\right )^{5/4}}{4 a \sqrt [4]{\frac {1}{a x}+1}} \]

Antiderivative was successfully verified.

[In]

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

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 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]

Rubi steps

\begin {align*} \int e^{-\frac {5}{2} \coth ^{-1}(a x)} x \, dx &=-\operatorname {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 \operatorname {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 \operatorname {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 \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 {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 \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 {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 \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 {25 \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 {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 \tan ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}+\frac {25 \tanh ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{4 a^2}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 121, normalized size = 0.69 \[ -\frac {e^{-\frac {1}{2} \coth ^{-1}(a x)} \left (-90 e^{2 \coth ^{-1}(a x)}+50 e^{4 \coth ^{-1}(a x)}+25 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (e^{2 \coth ^{-1}(a x)}-1\right )^2 \tan ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-25 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (e^{2 \coth ^{-1}(a x)}-1\right )^2 \tanh ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+32\right )}{4 a^2 \left (e^{2 \coth ^{-1}(a x)}-1\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*(32 - 90*E^(2*ArcCoth[a*x]) + 50*E^(4*ArcCoth[a*x]) + 25*E^(ArcCoth[a*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2*A
rcTan[E^(ArcCoth[a*x]/2)] - 25*E^(ArcCoth[a*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2*ArcTanh[E^(ArcCoth[a*x]/2)])/(a^
2*E^(ArcCoth[a*x]/2)*(-1 + E^(2*ArcCoth[a*x]))^2)

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fricas [A]  time = 0.90, size = 95, normalized size = 0.54 \[ \frac {2 \, {\left (2 \, a^{2} x^{2} - 9 \, a x - 43\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) + 25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) - 25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{8 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(2*a^2*x^2 - 9*a*x - 43)*((a*x - 1)/(a*x + 1))^(1/4) + 50*arctan(((a*x - 1)/(a*x + 1))^(1/4)) + 25*log(
((a*x - 1)/(a*x + 1))^(1/4) + 1) - 25*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2

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giac [A]  time = 0.19, size = 161, normalized size = 0.91 \[ \frac {1}{8} \, a {\left (\frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} - \frac {25 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{3}} - \frac {64 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{3}} + \frac {4 \, {\left (\frac {13 \, {\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{a^{3} {\left (\frac {a x - 1}{a x + 1} - 1\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*((a*x-1)/(a*x+1))^(5/4),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*x - 1)/(a*x + 1))^(1/4)/a^3 + 4*(13*(a*x - 1)*((a*x - 1)/(a*x +
1))^(1/4)/(a*x + 1) - 9*((a*x - 1)/(a*x + 1))^(1/4))/(a^3*((a*x - 1)/(a*x + 1) - 1)^2))

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maple [F]  time = 0.36, size = 0, normalized size = 0.00 \[ \int x \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 172, normalized size = 0.98 \[ -\frac {1}{8} \, a {\left (\frac {4 \, {\left (13 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - 9 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{3}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{3}}{{\left (a x + 1\right )}^{2}} - a^{3}} - \frac {50 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{3}} - \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{3}} + \frac {25 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{3}} + \frac {64 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a*(4*(13*((a*x - 1)/(a*x + 1))^(5/4) - 9*((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) - 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(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^3 + 64*((a*x - 1)/(a*x + 1))^(1/4)/a^3)

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mupad [B]  time = 1.20, size = 145, normalized size = 0.82 \[ \frac {25\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{4\,a^2}-\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a^2}-\frac {\frac {9\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{2}-\frac {13\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}{2}}{a^2+\frac {a^2\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {2\,a^2\,\left (a\,x-1\right )}{a\,x+1}}-\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,25{}\mathrm {i}}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

(25*atan(((a*x - 1)/(a*x + 1))^(1/4)))/(4*a^2) - ((9*((a*x - 1)/(a*x + 1))^(1/4))/2 - (13*((a*x - 1)/(a*x + 1)
)^(5/4))/2)/(a^2 + (a^2*(a*x - 1)^2)/(a*x + 1)^2 - (2*a^2*(a*x - 1))/(a*x + 1)) - (8*((a*x - 1)/(a*x + 1))^(1/
4))/a^2 - (atan(((a*x - 1)/(a*x + 1))^(1/4)*1i)*25i)/(4*a^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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