3.56 \(\int \frac {e^{-3 \tanh ^{-1}(a x)}}{x^2} \, dx\)

Optimal. Leaf size=62 \[ -\frac {4 a \sqrt {1-a^2 x^2}}{a x+1}-\frac {\sqrt {1-a^2 x^2}}{x}+3 a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.69, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 = {6124, 6742, 264, 266, 63, 208, 651} \[ -\frac {4 a \sqrt {1-a^2 x^2}}{a x+1}-\frac {\sqrt {1-a^2 x^2}}{x}+3 a \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(E^(3*ArcTanh[a*x])*x^2),x]

[Out]

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

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 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

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 651

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

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 57, normalized size = 0.92 \[ \sqrt {1-a^2 x^2} \left (-\frac {4 a}{a x+1}-\frac {1}{x}\right )+3 a \log \left (\sqrt {1-a^2 x^2}+1\right )-3 a \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(E^(3*ArcTanh[a*x])*x^2),x]

[Out]

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

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fricas [A]  time = 0.43, size = 75, normalized size = 1.21 \[ -\frac {4 \, a^{2} x^{2} + 4 \, a x + 3 \, {\left (a^{2} x^{2} + a x\right )} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + \sqrt {-a^{2} x^{2} + 1} {\left (5 \, a x + 1\right )}}{a x^{2} + x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*a^2*x^2 + 4*a*x + 3*(a^2*x^2 + a*x)*log((sqrt(-a^2*x^2 + 1) - 1)/x) + sqrt(-a^2*x^2 + 1)*(5*a*x + 1))/(a*x
^2 + x)

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giac [B]  time = 0.20, size = 150, normalized size = 2.42 \[ \frac {3 \, a^{2} \log \left (\frac {{\left | -2 \, \sqrt {-a^{2} x^{2} + 1} {\left | a \right |} - 2 \, a \right |}}{2 \, a^{2} {\left | x \right |}}\right )}{{\left | a \right |}} + \frac {{\left (a^{2} + \frac {17 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )}}{x}\right )} a^{2} x}{2 \, {\left (\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a\right )} {\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{a^{2} x} + 1\right )} {\left | a \right |}} - \frac {\sqrt {-a^{2} x^{2} + 1} {\left | a \right |} + a}{2 \, x {\left | a \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^2*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) + 1/2*(a^2 + 17*(sqrt(-a^2*x^2 + 1)
*abs(a) + a)/x)*a^2*x/((sqrt(-a^2*x^2 + 1)*abs(a) + a)*((sqrt(-a^2*x^2 + 1)*abs(a) + a)/(a^2*x) + 1)*abs(a)) -
1/2*(sqrt(-a^2*x^2 + 1)*abs(a) + a)/(x*abs(a))

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maple [B]  time = 0.05, size = 261, normalized size = 4.21 \[ -a \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}-3 a \sqrt {-a^{2} x^{2}+1}+3 a \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {5}{2}}}{x}-a^{2} x \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}-\frac {3 a^{2} x \sqrt {-a^{2} x^{2}+1}}{2}-\frac {3 a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}-\frac {\left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {5}{2}}}{a^{2} \left (x +\frac {1}{a}\right )^{3}}+a \left (-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )\right )^{\frac {3}{2}}+\frac {3 a^{2} \sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}\, x}{2}+\frac {3 a^{2} \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{2 \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(-a^2*x^2+1)^(3/2)-3*a*(-a^2*x^2+1)^(1/2)+3*a*arctanh(1/(-a^2*x^2+1)^(1/2))-1/x*(-a^2*x^2+1)^(5/2)-a^2*x*(-
a^2*x^2+1)^(3/2)-3/2*a^2*x*(-a^2*x^2+1)^(1/2)-3/2*a^2/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))-1/a
^2/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+a*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(3/2)+3/2*a^2*(-a^2*(x+1/a)^2+2
*a*(x+1/a))^(1/2)*x+3/2*a^2/(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 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{{\left (a x + 1\right )}^{3} x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.80, size = 81, normalized size = 1.31 \[ 3\,a\,\mathrm {atanh}\left (\sqrt {1-a^2\,x^2}\right )-\frac {\sqrt {1-a^2\,x^2}}{x}+\frac {4\,a^2\,\sqrt {1-a^2\,x^2}}{\left (x\,\sqrt {-a^2}+\frac {\sqrt {-a^2}}{a}\right )\,\sqrt {-a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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