3.42 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{x^5} \, dx\)

Optimal. Leaf size=114 \[ -\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+\frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x} \]

[Out]

-3/8*a^4*arctanh((-a^2*x^2+1)^(1/2))-1/4*(-a^2*x^2+1)^(1/2)/x^4+1/3*a*(-a^2*x^2+1)^(1/2)/x^3-3/8*a^2*(-a^2*x^2
+1)^(1/2)/x^2+2/3*a^3*(-a^2*x^2+1)^(1/2)/x

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Rubi [A]  time = 0.10, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6124, 835, 807, 266, 63, 208} \[ \frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {\sqrt {1-a^2 x^2}}{4 x^4}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^ArcTanh[a*x]*x^5),x]

[Out]

-Sqrt[1 - a^2*x^2]/(4*x^4) + (a*Sqrt[1 - a^2*x^2])/(3*x^3) - (3*a^2*Sqrt[1 - a^2*x^2])/(8*x^2) + (2*a^3*Sqrt[1
- a^2*x^2])/(3*x) - (3*a^4*ArcTanh[Sqrt[1 - a^2*x^2]])/8

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 6124

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

Rubi steps

\begin {align*} \int \frac {e^{-\tanh ^{-1}(a x)}}{x^5} \, dx &=\int \frac {1-a x}{x^5 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}-\frac {1}{4} \int \frac {4 a-3 a^2 x}{x^4 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}+\frac {1}{12} \int \frac {9 a^2-8 a^3 x}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}-\frac {1}{24} \int \frac {16 a^3-9 a^4 x}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{8} \left (3 a^4\right ) \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{16} \left (3 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x}-\frac {1}{8} \left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {\sqrt {1-a^2 x^2}}{4 x^4}+\frac {a \sqrt {1-a^2 x^2}}{3 x^3}-\frac {3 a^2 \sqrt {1-a^2 x^2}}{8 x^2}+\frac {2 a^3 \sqrt {1-a^2 x^2}}{3 x}-\frac {3}{8} a^4 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 74, normalized size = 0.65 \[ \frac {1}{24} \left (9 a^4 \log (x)-9 a^4 \log \left (\sqrt {1-a^2 x^2}+1\right )+\frac {\sqrt {1-a^2 x^2} \left (16 a^3 x^3-9 a^2 x^2+8 a x-6\right )}{x^4}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^ArcTanh[a*x]*x^5),x]

[Out]

((Sqrt[1 - a^2*x^2]*(-6 + 8*a*x - 9*a^2*x^2 + 16*a^3*x^3))/x^4 + 9*a^4*Log[x] - 9*a^4*Log[1 + Sqrt[1 - a^2*x^2
]])/24

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fricas [A]  time = 0.51, size = 68, normalized size = 0.60 \[ \frac {9 \, a^{4} x^{4} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (16 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 8 \, a x - 6\right )} \sqrt {-a^{2} x^{2} + 1}}{24 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^5,x, algorithm="fricas")

[Out]

1/24*(9*a^4*x^4*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (16*a^3*x^3 - 9*a^2*x^2 + 8*a*x - 6)*sqrt(-a^2*x^2 + 1))/x^4

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giac [B]  time = 0.20, size = 273, normalized size = 2.39 \[ \frac {{\left (3 \, a^{5} - \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{3}}{x} + \frac {24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a}{x^{2}} - \frac {72 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3}}{a x^{3}}\right )} a^{8} x^{4}}{192 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} {\left | a \right |}} - \frac {3 \, a^{5} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{8 \, {\left | a \right |}} + \frac {\frac {72 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} a^{5} {\left | a \right |}}{x} - \frac {24 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{2} a^{3} {\left | a \right |}}{x^{2}} + \frac {8 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{3} a {\left | a \right |}}{x^{3}} - \frac {3 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}^{4} {\left | a \right |}}{a x^{4}}}{192 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^5,x, algorithm="giac")

[Out]

1/192*(3*a^5 - 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^3/x + 24*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a/x^2 - 72*(sqrt
(-a^2*x^2 + 1)*abs(a) + a)^3/(a*x^3))*a^8*x^4/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)) - 3/8*a^5*log(1/2*abs
(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/192*(72*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^5*abs(
a)/x - 24*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^3*abs(a)/x^2 + 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*a*abs(a)/x^3
- 3*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^4*abs(a)/(a*x^4))/a^4

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maple [B]  time = 0.05, size = 226, normalized size = 1.98 \[ \frac {3 a^{4} \sqrt {-a^{2} x^{2}+1}}{8}-\frac {3 a^{4} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{8}+\frac {a^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}+a^{5} x \sqrt {-a^{2} x^{2}+1}+\frac {a^{5} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{\sqrt {a^{2}}}+\frac {a \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 x^{3}}-\frac {5 a^{2} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{8 x^{2}}-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{4 x^{4}}-a^{4} \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}-\frac {a^{5} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1}}{{\left (a x + 1\right )} x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.03, size = 98, normalized size = 0.86 \[ \frac {a\,\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {\sqrt {1-a^2\,x^2}}{4\,x^4}-\frac {3\,a^2\,\sqrt {1-a^2\,x^2}}{8\,x^2}+\frac {2\,a^3\,\sqrt {1-a^2\,x^2}}{3\,x}+\frac {a^4\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^4*atan((1 - a^2*x^2)^(1/2)*1i)*3i)/8 - (1 - a^2*x^2)^(1/2)/(4*x^4) + (a*(1 - a^2*x^2)^(1/2))/(3*x^3) - (3*a
^2*(1 - a^2*x^2)^(1/2))/(8*x^2) + (2*a^3*(1 - a^2*x^2)^(1/2))/(3*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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