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

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

[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)

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Rubi [A]  time = 0.0676022, antiderivative size = 176, normalized size of antiderivative = 1., 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, 212, 206, 203} \[ -\frac{25 \sqrt [4]{\frac{1}{a x}+1}}{2 a^2 \sqrt [4]{1-\frac{1}{a x}}}+\frac{25 \tan ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{25 \tanh ^{-1}\left (\frac{\sqrt [4]{\frac{1}{a x}+1}}{\sqrt [4]{1-\frac{1}{a x}}}\right )}{4 a^2}+\frac{x^2 \left (\frac{1}{a x}+1\right )^{9/4}}{2 \sqrt [4]{1-\frac{1}{a x}}}+\frac{5 x \left (\frac{1}{a x}+1\right )^{5/4}}{4 a \sqrt [4]{1-\frac{1}{a x}}} \]

Antiderivative was successfully verified.

[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 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]

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 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 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 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 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 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])

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 \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 \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 )}{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*}

Mathematica [A]  time = 0.212247, size = 80, normalized size = 0.45 \[ \frac{-\frac{2 e^{\frac{1}{2} \coth ^{-1}(a x)} \left (-45 e^{2 \coth ^{-1}(a x)}+16 e^{4 \coth ^{-1}(a x)}+25\right )}{\left (e^{2 \coth ^{-1}(a x)}-1\right )^2}+25 \tan ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )+25 \tanh ^{-1}\left (e^{\frac{1}{2} \coth ^{-1}(a x)}\right )}{4 a^2} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 0.325, size = 0, normalized size = 0. \begin{align*} \int{x \left ({\frac{ax-1}{ax+1}} \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]  time = 1.48703, size = 224, normalized size = 1.27 \begin{align*} \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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.67888, size = 332, normalized size = 1.89 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [A]  time = 1.16932, size = 217, normalized size = 1.23 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))