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

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

[Out]

11/2*a^3*arccsc(a*x)+(a-1/x)^3/(1-1/a^2/x^2)^(1/2)+1/6*a^2*(28*a-3/x)*(1-1/a^2/x^2)^(1/2)+1/3*a*(3*a-1/x)^2*(1
-1/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.77, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, 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 12

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

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]

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.11, size = 66, normalized size = 0.69 \[ \frac {1}{6} a \left (33 a^2 \sin ^{-1}\left (\frac {1}{a x}\right )+\frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (52 a^3 x^3+19 a^2 x^2-7 a x+2\right )}{x^2 (a x+1)}\right ) \]

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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fricas [A]  time = 0.50, size = 69, normalized size = 0.72 \[ -\frac {66 \, a^{3} x^{3} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - {\left (52 \, a^{3} x^{3} + 19 \, a^{2} x^{2} - 7 \, a x + 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{6 \, x^{3}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

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

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maple [B]  time = 0.06, size = 666, normalized size = 6.94 \[ -\frac {\left (-30 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{6} a^{6}+30 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{4} a^{4}-93 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{5} a^{5}-33 \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}\, x^{5} a^{5}+30 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{5} a^{6}+30 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{5} a^{5}-30 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{5} a^{6}+51 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{3} a^{3}-96 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{4} a^{4}-66 a^{4} x^{4} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+60 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}+12 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} x^{3} a^{3}+60 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{4} a^{4}-60 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}+14 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{2} a^{2}-33 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{3} a^{3}-33 a^{3} x^{3} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+30 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}+30 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{3} a^{3}-30 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}-5 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x a +2 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{6 \sqrt {a^{2}}\, x^{3} \left (a x -1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}} \]

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*x^2-1)^(1/2)*(a^2)^(1/2)*x^6*a^6+30*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^4*a^4-93*(a^2*x^2-1)^(1/2)*
(a^2)^(1/2)*x^5*a^5-33*arctan(1/(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+30*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^5*a^5-30*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*
(a^2)^(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^4-66*a^4*x^4*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+60*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/
2))*x^4*a^5+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*x-1)*(a*x+1))^(1/2)*(a^2)^(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^3-33*a^3*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+30*ln((a^2*x+(a^2*x^
2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4+30*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3-30*ln((a^2*x+((a*x
-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4-5*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*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 = 0.42, size = 157, normalized size = 1.64 \[ -\frac {1}{3} \, {\left (33 \, a^{2} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 12 \, a^{2} \sqrt {\frac {a x - 1}{a x + 1}} - \frac {39 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 52 \, a^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 21 \, a^{2} \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {3 \, {\left (a x - 1\right )}}{a x + 1} + \frac {3 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {{\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} + 1}\right )} a \]

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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mupad [B]  time = 1.23, size = 153, normalized size = 1.59 \[ \frac {7\,a^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}+\frac {52\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{3}+13\,a^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{\frac {3\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}+\frac {{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {3\,\left (a\,x-1\right )}{a\,x+1}+1}+4\,a^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}-11\,a^3\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*a^3*((a*x - 1)/(a*x + 1))^(1/2) + (52*a^3*((a*x - 1)/(a*x + 1))^(3/2))/3 + 13*a^3*((a*x - 1)/(a*x + 1))^(5/
2))/((3*(a*x - 1)^2)/(a*x + 1)^2 + (a*x - 1)^3/(a*x + 1)^3 + (3*(a*x - 1))/(a*x + 1) + 1) + 4*a^3*((a*x - 1)/(
a*x + 1))^(1/2) - 11*a^3*atan(((a*x - 1)/(a*x + 1))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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