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

Optimal. Leaf size=91 \[ -\frac {9}{2} a^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {9}{2} a^2 \csc ^{-1}(a x)-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right )} \]

[Out]

-a^5*(1-1/a^2/x^2)^(5/2)/(a-1/x)^3-3/2*a^3*(1-1/a^2/x^2)^(3/2)/(a-1/x)+9/2*a^2*arccsc(a*x)-9/2*a^2*(1-1/a^2/x^
2)^(1/2)

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Rubi [A]  time = 0.45, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {6169, 1633, 1593, 12, 793, 665, 216} \[ -\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right )}-\frac {9}{2} a^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {9}{2} a^2 \csc ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^2*Sqrt[1 - 1/(a^2*x^2)])/2 - (a^5*(1 - 1/(a^2*x^2))^(5/2))/(a - x^(-1))^3 - (3*a^3*(1 - 1/(a^2*x^2))^(3/
2))/(2*(a - x^(-1))) + (9*a^2*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 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(a + c*x^2)^p)/(
e*(m + 2*p + 1)), x] - Dist[(2*c*d*p)/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1), x], x] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[
m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 793

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d*g - e*f)*(
d + e*x)^m*(a + c*x^2)^(p + 1))/(2*c*d*(m + p + 1)), x] + Dist[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d
)*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[c*d^2
+ a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) &&
NeQ[m + p + 1, 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 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 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^3} \, dx &=-\operatorname {Subst}\left (\int \frac {x \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 {\left (-a x-x^2\right ) \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-x) x \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 \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 \left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^3} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}+(3 a) \operatorname {Subst}\left (\int \frac {\left (1-\frac {x^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right )}+\frac {1}{2} (9 a) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{1-\frac {x}{a}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {9}{2} a^2 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right )}+\frac {1}{2} (9 a) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {9}{2} a^2 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{\left (a-\frac {1}{x}\right )^3}-\frac {3 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{2 \left (a-\frac {1}{x}\right )}+\frac {9}{2} a^2 \csc ^{-1}(a x)\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 56, normalized size = 0.62 \[ \frac {1}{2} a \left (\frac {\sqrt {1-\frac {1}{a^2 x^2}} \left (-14 a^2 x^2+5 a x+1\right )}{x (a x-1)}+9 a \sin ^{-1}\left (\frac {1}{a x}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*((Sqrt[1 - 1/(a^2*x^2)]*(1 + 5*a*x - 14*a^2*x^2))/(x*(-1 + a*x)) + 9*a*ArcSin[1/(a*x)]))/2

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fricas [A]  time = 0.38, size = 88, normalized size = 0.97 \[ -\frac {18 \, {\left (a^{3} x^{3} - a^{2} x^{2}\right )} \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + {\left (14 \, a^{3} x^{3} + 9 \, a^{2} x^{2} - 6 \, a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, {\left (a x^{3} - x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(18*(a^3*x^3 - a^2*x^2)*arctan(sqrt((a*x - 1)/(a*x + 1))) + (14*a^3*x^3 + 9*a^2*x^2 - 6*a*x - 1)*sqrt((a*
x - 1)/(a*x + 1)))/(a*x^3 - x^2)

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giac [A]  time = 0.14, size = 108, normalized size = 1.19 \[ -{\left (9 \, a \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + \frac {4 \, a}{\sqrt {\frac {a x - 1}{a x + 1}}} + \frac {\frac {5 \, {\left (a x - 1\right )} a \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1} + 7 \, a \sqrt {\frac {a x - 1}{a x + 1}}}{{\left (\frac {a x - 1}{a x + 1} + 1\right )}^{2}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(9*a*arctan(sqrt((a*x - 1)/(a*x + 1))) + 4*a/sqrt((a*x - 1)/(a*x + 1)) + (5*(a*x - 1)*a*sqrt((a*x - 1)/(a*x +
1))/(a*x + 1) + 7*a*sqrt((a*x - 1)/(a*x + 1)))/((a*x - 1)/(a*x + 1) + 1)^2)*a

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maple [B]  time = 0.06, size = 641, normalized size = 7.04 \[ \frac {-6 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{5} a^{5}+6 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{3} a^{3}+21 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{4} a^{4}+9 a^{4} x^{4} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+6 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}-6 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{4} a^{4}-6 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{4} a^{5}-11 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x^{2} a^{2}-24 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{3} a^{3}-18 a^{3} x^{3} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )-12 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}-4 \sqrt {a^{2}}\, \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} x^{2} a^{2}+12 \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, x^{3} a^{3}+12 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{3} a^{4}+4 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, x a +9 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{2} a^{2}+9 a^{2} x^{2} \sqrt {a^{2}}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )+6 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}-6 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}-6 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+\left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}}{2 \sqrt {a^{2}}\, x^{2} \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x +1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-6*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^5*a^5+6*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^3*a^3+21*(a^2*x^2-1)^(1/2)*(a^
2)^(1/2)*x^4*a^4+9*a^4*x^4*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+6*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/
(a^2)^(1/2))*x^4*a^5-6*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^4*a^4-6*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(
1/2))/(a^2)^(1/2))*x^4*a^5-11*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x^2*a^2-24*(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3-1
8*a^3*x^3*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))-12*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3
*a^4-4*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(3/2)*x^2*a^2+12*(a^2)^(1/2)*((a*x-1)*(a*x+1))^(1/2)*x^3*a^3+12*ln((a^2*x
+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^3*a^4+4*(a^2*x^2-1)^(3/2)*(a^2)^(1/2)*x*a+9*(a^2*x^2-1)^(
1/2)*(a^2)^(1/2)*x^2*a^2+9*a^2*x^2*(a^2)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))+6*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)
^(1/2))/(a^2)^(1/2))*x^2*a^3-6*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x^2*a^2-6*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)
*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+(a^2*x^2-1)^(3/2)*(a^2)^(1/2))/(a^2)^(1/2)/x^2/((a*x-1)*(a*x+1))^(1/2)/(a*x
+1)/((a*x-1)/(a*x+1))^(3/2)

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maxima [A]  time = 0.41, size = 110, normalized size = 1.21 \[ -{\left (9 \, a \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) + \frac {\frac {15 \, {\left (a x - 1\right )} a}{a x + 1} + \frac {9 \, {\left (a x - 1\right )}^{2} a}{{\left (a x + 1\right )}^{2}} + 4 \, a}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 2 \, \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + \sqrt {\frac {a x - 1}{a x + 1}}}\right )} a \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(9*a*arctan(sqrt((a*x - 1)/(a*x + 1))) + (15*(a*x - 1)*a/(a*x + 1) + 9*(a*x - 1)^2*a/(a*x + 1)^2 + 4*a)/(((a*
x - 1)/(a*x + 1))^(5/2) + 2*((a*x - 1)/(a*x + 1))^(3/2) + sqrt((a*x - 1)/(a*x + 1))))*a

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mupad [B]  time = 0.08, size = 83, normalized size = 0.91 \[ \frac {1}{2\,x^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}}-\frac {7\,a^2}{\sqrt {\frac {a\,x-1}{a\,x+1}}}-9\,a^2\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )+\frac {5\,a}{2\,x\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(2*x^2*((a*x - 1)/(a*x + 1))^(1/2)) - (7*a^2)/((a*x - 1)/(a*x + 1))^(1/2) - 9*a^2*atan(((a*x - 1)/(a*x + 1))
^(1/2)) + (5*a)/(2*x*((a*x - 1)/(a*x + 1))^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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