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

Optimal. Leaf size=98 \[ x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {6170, 94, 93, 212, 206, 203} \[ x \left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{\frac {1}{a x}+1}-\frac {3 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(1 - 1/(a*x))^(3/4)*(1 + 1/(a*x))^(1/4)*x - (3*ArcTan[(1 + 1/(a*x))^(1/4)/(1 - 1/(a*x))^(1/4)])/a - (3*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 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 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 {3}{2} \coth ^{-1}(a x)} \, dx &=-\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 )\\ &=\left (1-\frac {1}{a x}\right )^{3/4} \sqrt [4]{1+\frac {1}{a x}} x+\frac {3 \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 {6 \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 {3 \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 {3 \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 {3 \tan ^{-1}\left (\frac {\sqrt [4]{1+\frac {1}{a x}}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}-\frac {3 \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 [A]  time = 0.12, size = 55, normalized size = 0.56 \[ \frac {\frac {2 e^{\frac {1}{2} \coth ^{-1}(a x)}}{e^{2 \coth ^{-1}(a x)}-1}-3 \tan ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )-3 \tanh ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 86, normalized size = 0.88 \[ \frac {2 \, {\left (a x + 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}} + 6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a*x + 1)*((a*x - 1)/(a*x + 1))^(3/4) + 6*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 3*log(((a*x - 1)/(a*x +
1))^(1/4) + 1) + 3*log(((a*x - 1)/(a*x + 1))^(1/4) - 1))/a

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giac [A]  time = 0.17, size = 109, normalized size = 1.11 \[ \frac {1}{2} \, a {\left (\frac {6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} - \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} + \frac {3 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{2}} - \frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{a^{2} {\left (\frac {a x - 1}{a x + 1} - 1\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(6*arctan(((a*x - 1)/(a*x + 1))^(1/4))/a^2 - 3*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^2 + 3*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)))

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 112, normalized size = 1.14 \[ -\frac {1}{2} \, a {\left (\frac {4 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} - \frac {6 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} + \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} - \frac {3 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/4),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) - 6*arctan(((a*x - 1)/(a*x + 1))^(1/4))/
a^2 + 3*log(((a*x - 1)/(a*x + 1))^(1/4) + 1)/a^2 - 3*log(((a*x - 1)/(a*x + 1))^(1/4) - 1)/a^2)

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mupad [B]  time = 1.18, size = 79, normalized size = 0.81 \[ \frac {2\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/4}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}}+\frac {3\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a}-\frac {3\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x - 1)/(a*x + 1))^(3/4),x)

[Out]

(2*((a*x - 1)/(a*x + 1))^(3/4))/(a - (a*(a*x - 1))/(a*x + 1)) + (3*atan(((a*x - 1)/(a*x + 1))^(1/4)))/a - (3*a
tanh(((a*x - 1)/(a*x + 1))^(1/4)))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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