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

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

[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 = 0.04, 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, 212, 206, 203} \[ \frac {x \left (\frac {1}{a x}+1\right )^{5/4}}{\sqrt [4]{1-\frac {1}{a x}}}-\frac {10 \sqrt [4]{\frac {1}{a x}+1}}{a \sqrt [4]{1-\frac {1}{a x}}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{a x}+1}}{\sqrt [4]{1-\frac {1}{a x}}}\right )}{a} \]

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 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 {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 \sqrt [4]{1-\frac {x}{a}} \left (1+\frac {x}{a}\right )^{3/4}} \, 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 {10 \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}\\ &=-\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 [A]  time = 0.14, size = 67, normalized size = 0.52 \[ \frac {-\frac {2 e^{\frac {1}{2} \coth ^{-1}(a x)} \left (4 e^{2 \coth ^{-1}(a x)}-5\right )}{e^{2 \coth ^{-1}(a x)}-1}+5 \tan ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )+5 \tanh ^{-1}\left (e^{\frac {1}{2} \coth ^{-1}(a x)}\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 117, normalized size = 0.90 \[ -\frac {10 \, {\left (a x - 1\right )} \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right ) - 5 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right ) + 5 \, {\left (a x - 1\right )} \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right ) - 2 \, {\left (a^{2} x^{2} - 8 \, a x - 9\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{4}}}{2 \, {\left (a^{2} x - a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(10*(a*x - 1)*arctan(((a*x - 1)/(a*x + 1))^(1/4)) - 5*(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) + 1) + 5*
(a*x - 1)*log(((a*x - 1)/(a*x + 1))^(1/4) - 1) - 2*(a^2*x^2 - 8*a*x - 9)*((a*x - 1)/(a*x + 1))^(3/4))/(a^2*x -
 a)

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giac [A]  time = 0.20, size = 141, normalized size = 1.08 \[ -\frac {1}{2} \, a {\left (\frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} - \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} + \frac {5 \, \log \left ({\left | \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1 \right |}\right )}{a^{2}} + \frac {4 \, {\left (\frac {5 \, {\left (a x - 1\right )}}{a x + 1} - 4\right )}}{a^{2} {\left (\frac {{\left (a x - 1\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}}{a x + 1} - \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((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 + 4*(5*(a*x - 1)/(a*x + 1) - 4)/(a^2*((a*x - 1)*((a*x - 1)/(a*x + 1))^(1/4
)/(a*x + 1) - ((a*x - 1)/(a*x + 1))^(1/4))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.40, size = 131, normalized size = 1.01 \[ -\frac {1}{2} \, a {\left (\frac {4 \, {\left (\frac {5 \, {\left (a x - 1\right )}}{a x + 1} - 4\right )}}{a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{4}} - a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}} + \frac {10 \, \arctan \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}}\right )}{a^{2}} - \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} + 1\right )}{a^{2}} + \frac {5 \, \log \left (\left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{4}} - 1\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.06, size = 98, normalized size = 0.75 \[ \frac {\frac {10\,\left (a\,x-1\right )}{a\,x+1}-8}{a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}-a\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/4}}-\frac {5\,\mathrm {atan}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a}+\frac {5\,\mathrm {atanh}\left ({\left (\frac {a\,x-1}{a\,x+1}\right )}^{1/4}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x - 1)/(a*x + 1))^(5/4),x)

[Out]

((10*(a*x - 1))/(a*x + 1) - 8)/(a*((a*x - 1)/(a*x + 1))^(1/4) - a*((a*x - 1)/(a*x + 1))^(5/4)) - (5*atan(((a*x
 - 1)/(a*x + 1))^(1/4)))/a + (5*atanh(((a*x - 1)/(a*x + 1))^(1/4)))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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