3.57 \(\int \frac{e^{-3 \coth ^{-1}(a x)}}{x^4} \, dx\)

Optimal. Leaf size=96 \[ \frac{1}{6} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (28 a-\frac{3}{x}\right )+\frac{1}{3} a \sqrt{1-\frac{1}{a^2 x^2}} \left (3 a-\frac{1}{x}\right )^2+\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{11}{2} a^3 \csc ^{-1}(a x) \]

[Out]

(a^2*Sqrt[1 - 1/(a^2*x^2)]*(28*a - 3/x))/6 + (a - x^(-1))^3/Sqrt[1 - 1/(a^2*x^2)] + (a*Sqrt[1 - 1/(a^2*x^2)]*(
3*a - x^(-1))^2)/3 + (11*a^3*ArcCsc[a*x])/2

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Rubi [A]  time = 0.766351, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6169, 1633, 1593, 12, 852, 1635, 1654, 780, 216} \[ \frac{1}{6} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (28 a-\frac{3}{x}\right )+\frac{1}{3} a \sqrt{1-\frac{1}{a^2 x^2}} \left (3 a-\frac{1}{x}\right )^2+\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{11}{2} a^3 \csc ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcCoth[a*x])*x^4),x]

[Out]

(a^2*Sqrt[1 - 1/(a^2*x^2)]*(28*a - 3/x))/6 + (a - x^(-1))^3/Sqrt[1 - 1/(a^2*x^2)] + (a*Sqrt[1 - 1/(a^2*x^2)]*(
3*a - x^(-1))^2)/3 + (11*a^3*ArcCsc[a*x])/2

Rule 6169

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(x_)^(m_.), x_Symbol] :> -Subst[Int[(1 + x/a)^((n + 1)/2)/(x^(m + 2)*(1 - x/
a)^((n - 1)/2)*Sqrt[1 - x^2/a^2]), x], x, 1/x] /; FreeQ[a, x] && IntegerQ[(n - 1)/2] && IntegerQ[m]

Rule 1633

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d*e, Int[(d + e*x)^(m - 1)*
PolynomialQuotient[Pq, a*e + c*d*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && EqQ[c*d^2 + a*e^2, 0] && EqQ[PolynomialRemainder[Pq, a*e + c*d*x, x], 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1635

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, -Simp[(d*f*(d + e*x)^m*(a + c*x^2)^(p + 1))/(2*
a*e*(p + 1)), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)*Q
+ f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p +
1/2, 0] && GtQ[m, 0]

Rule 1654

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + c*x^2)^(p + 1))/(c*e^(q - 1)*(m + q + 2*p + 1)), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(a*e^2*(m + q - 1) - c*d^2*(m + q + 2*p + 1) - 2*c*d*e*(
m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq
, x] && NeQ[c*d^2 + a*e^2, 0] &&  !(EqQ[d, 0] && True) &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[
p] || ILtQ[p + 1/2, 0]))

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

\begin{align*} \int \frac{e^{-3 \coth ^{-1}(a x)}}{x^4} \, dx &=-\operatorname{Subst}\left (\int \frac{x^2 \left (1-\frac{x}{a}\right )^2}{\left (1+\frac{x}{a}\right ) \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}} \left (a x^2-x^3\right )}{\left (1+\frac{x}{a}\right )^2} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(a-x) x^2 \sqrt{1-\frac{x^2}{a^2}}}{\left (1+\frac{x}{a}\right )^2} \, dx,x,\frac{1}{x}\right )}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{a^2 x^2 \left (1-\frac{x^2}{a^2}\right )^{3/2}}{\left (1+\frac{x}{a}\right )^3} \, dx,x,\frac{1}{x}\right )}{a^2}\\ &=-\operatorname{Subst}\left (\int \frac{x^2 \left (1-\frac{x^2}{a^2}\right )^{3/2}}{\left (1+\frac{x}{a}\right )^3} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int \frac{x^2 \left (1-\frac{x}{a}\right )^3}{\left (1-\frac{x^2}{a^2}\right )^{3/2}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\operatorname{Subst}\left (\int \frac{\left (1-\frac{x}{a}\right )^2 \left (3 a^2-a x\right )}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{1}{3} a \sqrt{1-\frac{1}{a^2 x^2}} \left (3 a-\frac{1}{x}\right )^2-\frac{1}{3} \operatorname{Subst}\left (\int \frac{\left (-5+\frac{3 x}{a}\right ) \left (3 a^2-a x\right )}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{6} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (28 a-\frac{3}{x}\right )+\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{1}{3} a \sqrt{1-\frac{1}{a^2 x^2}} \left (3 a-\frac{1}{x}\right )^2+\frac{1}{2} \left (11 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{6} a^2 \sqrt{1-\frac{1}{a^2 x^2}} \left (28 a-\frac{3}{x}\right )+\frac{\left (a-\frac{1}{x}\right )^3}{\sqrt{1-\frac{1}{a^2 x^2}}}+\frac{1}{3} a \sqrt{1-\frac{1}{a^2 x^2}} \left (3 a-\frac{1}{x}\right )^2+\frac{11}{2} a^3 \csc ^{-1}(a x)\\ \end{align*}

Mathematica [A]  time = 0.101494, size = 66, normalized size = 0.69 \[ \frac{1}{6} a \left (\frac{\sqrt{1-\frac{1}{a^2 x^2}} \left (52 a^3 x^3+19 a^2 x^2-7 a x+2\right )}{x^2 (a x+1)}+33 a^2 \sin ^{-1}\left (\frac{1}{a x}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*x^4),x]

[Out]

(a*((Sqrt[1 - 1/(a^2*x^2)]*(2 - 7*a*x + 19*a^2*x^2 + 52*a^3*x^3))/(x^2*(1 + a*x)) + 33*a^2*ArcSin[1/(a*x)]))/6

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Maple [B]  time = 0.147, size = 666, normalized size = 6.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a*x-1)/(a*x+1))^(3/2)/x^4,x)

[Out]

-1/6*(-30*(a^2)^(1/2)*(a^2*x^2-1)^(1/2)*x^6*a^6+30*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*x^4*a^4-93*(a^2*x^2-1)^(1/2)*
(a^2)^(1/2)*x^5*a^5+30*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^5*a^6-33*(a^2)^(1/2)*arctan(1/(
a^2*x^2-1)^(1/2))*x^5*a^5+30*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^5*a^5-30*ln((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*
x+1))^(1/2))/(a^2)^(1/2))*x^5*a^6+51*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^3*a^3-96*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^
4*a^4+60*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^4*a^5-66*a^4*x^4*(a^2)^(1/2)*arctan(1/(a^2*x^
2-1)^(1/2))+12*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x^3*a^3+60*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^4*a^4-60*l
n((a^2*x+(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^4*a^5+14*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^2*a^2-33
*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3+30*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4-33*a^3
*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+30*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3-30*ln((a^2*x+(a^2)
^(1/2)*((a*x-1)*(a*x+1))^(1/2))/(a^2)^(1/2))*x^3*a^4-5*(a^2)^(1/2)*(a^2*x^2-1)^(3/2)*x*a+2*(a^2*x^2-1)^(3/2)*(
a^2)^(1/2))*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/x^3/(a*x-1)/((a*x-1)*(a*x+1))^(1/2)

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Maxima [A]  time = 1.52274, size = 212, normalized size = 2.21 \begin{align*} -\frac{1}{3} \,{\left (33 \, a^{2} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) - 12 \, a^{2} \sqrt{\frac{a x - 1}{a x + 1}} - \frac{39 \, a^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{5}{2}} + 52 \, a^{2} \left (\frac{a x - 1}{a x + 1}\right )^{\frac{3}{2}} + 21 \, a^{2} \sqrt{\frac{a x - 1}{a x + 1}}}{\frac{3 \,{\left (a x - 1\right )}}{a x + 1} + \frac{3 \,{\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac{{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + 1}\right )} a \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="maxima")

[Out]

-1/3*(33*a^2*arctan(sqrt((a*x - 1)/(a*x + 1))) - 12*a^2*sqrt((a*x - 1)/(a*x + 1)) - (39*a^2*((a*x - 1)/(a*x +
1))^(5/2) + 52*a^2*((a*x - 1)/(a*x + 1))^(3/2) + 21*a^2*sqrt((a*x - 1)/(a*x + 1)))/(3*(a*x - 1)/(a*x + 1) + 3*
(a*x - 1)^2/(a*x + 1)^2 + (a*x - 1)^3/(a*x + 1)^3 + 1))*a

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Fricas [A]  time = 1.61868, size = 166, normalized size = 1.73 \begin{align*} -\frac{66 \, a^{3} x^{3} \arctan \left (\sqrt{\frac{a x - 1}{a x + 1}}\right ) -{\left (52 \, a^{3} x^{3} + 19 \, a^{2} x^{2} - 7 \, a x + 2\right )} \sqrt{\frac{a x - 1}{a x + 1}}}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="fricas")

[Out]

-1/6*(66*a^3*x^3*arctan(sqrt((a*x - 1)/(a*x + 1))) - (52*a^3*x^3 + 19*a^2*x^2 - 7*a*x + 2)*sqrt((a*x - 1)/(a*x
+ 1)))/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/x**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/x^4,x, algorithm="giac")

[Out]

undef