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

Optimal. Leaf size=130 \[ \frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}+\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{\frac {1}{a x}+1}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]

[Out]

10*(1-1/a/x)^(1/4)/a/(1+1/a/x)^(1/4)+(1-1/a/x)^(5/4)*x/(1+1/a/x)^(1/4)+5*arctan((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4
))/a-5*arctanh((1+1/a/x)^(1/4)/(1-1/a/x)^(1/4))/a

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Rubi [A]  time = 0.05, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6170, 94, 93, 298, 203, 206} \[ \frac {x \left (1-\frac {1}{a x}\right )^{5/4}}{\sqrt [4]{\frac {1}{a x}+1}}+\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{\frac {1}{a x}+1}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(1 - 1/(a*x))^(1/4))/(a*(1 + 1/(a*x))^(1/4)) + ((1 - 1/(a*x))^(5/4)*x)/(1 + 1/(a*x))^(1/4) + (5*ArcTan[(1
+ 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/a - (5*ArcTanh[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/a

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 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 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]

Rubi steps

\begin {align*} \int e^{-\frac {5}{2} \coth ^{-1}(a x)} \, dx &=-\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 )\\ &=\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \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 )}{2 a}\\ &=\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \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 )}{2 a}\\ &=\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {10 \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 )}{a}\\ &=\frac {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}-\frac {5 \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 {5 \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 {10 \sqrt [4]{1-\frac {1}{a x}}}{a \sqrt [4]{1+\frac {1}{a x}}}+\frac {\left (1-\frac {1}{a x}\right )^{5/4} x}{\sqrt [4]{1+\frac {1}{a x}}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 31, normalized size = 0.24 \[ \frac {8 e^{-\frac {1}{2} \coth ^{-1}(a x)} \, _2F_1\left (-\frac {1}{4},2;\frac {3}{4};e^{2 \coth ^{-1}(a x)}\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(8*Hypergeometric2F1[-1/4, 2, 3/4, E^(2*ArcCoth[a*x])])/(a*E^(ArcCoth[a*x]/2))

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fricas [A]  time = 0.47, size = 86, normalized size = 0.66 \[ \frac {2 \, {\left (a x + 9\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a*x + 9)*((a*x - 1)/(a*x + 1))^(1/4) - 10*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 5*log(((a*x - 1)/(a*x
+ 1))^(1/4) + 1) + 5*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a

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giac [A]  time = 0.16, size = 129, normalized size = 0.99 \[ -\frac {1}{2} \, a {\left (\frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {5 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{2}} - \frac {16 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2}} + \frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2} {\left (\frac {a x - 1}{a x + 1} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(10*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^2 + 5*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^2 - 5*log(abs(((
a*x - 1)/(a*x + 1))^(1/4) - 1))/a^2 - 16*((a*x - 1)/(a*x + 1))^(1/4)/a^2 + 4*((a*x - 1)/(a*x + 1))^(1/4)/(a^2*
((a*x - 1)/(a*x + 1) - 1)))

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maple [F]  time = 0.37, size = 0, normalized size = 0.00 \[ \int \left (\frac {a x -1}{a x +1}\right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 132, normalized size = 1.02 \[ -\frac {1}{2} \, a {\left (\frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} + \frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{2}} - \frac {16 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*(4*((a*x - 1)/(a*x + 1))^(1/4)/((a*x - 1)*a^2/(a*x + 1) - a^2) + 10*arctan(((a*x - 1)/(a*x + 1))^(1/4))
/a^2 + 5*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^2 - 5*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^2 - 16*((a*x - 1)
/(a*x + 1))^(1/4)/a^2)

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mupad [B]  time = 1.19, size = 103, normalized size = 0.79 \[ \frac {2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}+\frac {8\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}}{a}-\frac {5\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a}+\frac {\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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