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

Optimal. Leaf size=116 \[ -\frac{4 a^3 \sqrt{1-a^2 x^2}}{a x+1}-\frac{14 a^2 \sqrt{1-a^2 x^2}}{3 x}+\frac{3 a \sqrt{1-a^2 x^2}}{2 x^2}-\frac{\sqrt{1-a^2 x^2}}{3 x^3}+\frac{11}{2} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

[Out]

-Sqrt[1 - a^2*x^2]/(3*x^3) + (3*a*Sqrt[1 - a^2*x^2])/(2*x^2) - (14*a^2*Sqrt[1 - a^2*x^2])/(3*x) - (4*a^3*Sqrt[
1 - a^2*x^2])/(1 + a*x) + (11*a^3*ArcTanh[Sqrt[1 - a^2*x^2]])/2

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Rubi [A]  time = 0.747871, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6124, 6742, 271, 264, 266, 51, 63, 208, 651} \[ -\frac{4 a^3 \sqrt{1-a^2 x^2}}{a x+1}-\frac{14 a^2 \sqrt{1-a^2 x^2}}{3 x}+\frac{3 a \sqrt{1-a^2 x^2}}{2 x^2}-\frac{\sqrt{1-a^2 x^2}}{3 x^3}+\frac{11}{2} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[1 - a^2*x^2]/(3*x^3) + (3*a*Sqrt[1 - a^2*x^2])/(2*x^2) - (14*a^2*Sqrt[1 - a^2*x^2])/(3*x) - (4*a^3*Sqrt[
1 - a^2*x^2])/(1 + a*x) + (11*a^3*ArcTanh[Sqrt[1 - a^2*x^2]])/2

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]]

Rule 271

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 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]

Rubi steps

\begin{align*} \int \frac{e^{-3 \tanh ^{-1}(a x)}}{x^4} \, dx &=\int \frac{(1-a x)^2}{x^4 (1+a x) \sqrt{1-a^2 x^2}} \, dx\\ &=\int \left (\frac{1}{x^4 \sqrt{1-a^2 x^2}}-\frac{3 a}{x^3 \sqrt{1-a^2 x^2}}+\frac{4 a^2}{x^2 \sqrt{1-a^2 x^2}}-\frac{4 a^3}{x \sqrt{1-a^2 x^2}}+\frac{4 a^4}{(1+a x) \sqrt{1-a^2 x^2}}\right ) \, dx\\ &=-\left ((3 a) \int \frac{1}{x^3 \sqrt{1-a^2 x^2}} \, dx\right )+\left (4 a^2\right ) \int \frac{1}{x^2 \sqrt{1-a^2 x^2}} \, dx-\left (4 a^3\right ) \int \frac{1}{x \sqrt{1-a^2 x^2}} \, dx+\left (4 a^4\right ) \int \frac{1}{(1+a x) \sqrt{1-a^2 x^2}} \, dx+\int \frac{1}{x^4 \sqrt{1-a^2 x^2}} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2}}{3 x^3}-\frac{4 a^2 \sqrt{1-a^2 x^2}}{x}-\frac{4 a^3 \sqrt{1-a^2 x^2}}{1+a x}-\frac{1}{2} (3 a) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1-a^2 x}} \, dx,x,x^2\right )+\frac{1}{3} \left (2 a^2\right ) \int \frac{1}{x^2 \sqrt{1-a^2 x^2}} \, dx-\left (2 a^3\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}}{3 x^3}+\frac{3 a \sqrt{1-a^2 x^2}}{2 x^2}-\frac{14 a^2 \sqrt{1-a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1-a^2 x^2}}{1+a x}+(4 a) \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{1}{4} \left (3 a^3\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}}{3 x^3}+\frac{3 a \sqrt{1-a^2 x^2}}{2 x^2}-\frac{14 a^2 \sqrt{1-a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1-a^2 x^2}}{1+a x}+4 a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )+\frac{1}{2} (3 a) \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}}{3 x^3}+\frac{3 a \sqrt{1-a^2 x^2}}{2 x^2}-\frac{14 a^2 \sqrt{1-a^2 x^2}}{3 x}-\frac{4 a^3 \sqrt{1-a^2 x^2}}{1+a x}+\frac{11}{2} a^3 \tanh ^{-1}\left (\sqrt{1-a^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.075407, size = 82, normalized size = 0.71 \[ \frac{1}{6} \left (-\frac{\sqrt{1-a^2 x^2} \left (52 a^3 x^3+19 a^2 x^2-7 a x+2\right )}{x^3 (a x+1)}+33 a^3 \log \left (\sqrt{1-a^2 x^2}+1\right )-33 a^3 \log (x)\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.057, size = 338, normalized size = 2.9 \begin{align*} -{\frac{16\,{a}^{2}}{3\,x} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{16\,{a}^{4}x}{3} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-8\,{a}^{4}x\sqrt{-{a}^{2}{x}^{2}+1}-8\,{\frac{{a}^{4}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+2\,{\frac{a \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{5/2}}{ \left ( x+{a}^{-1} \right ) ^{2}}}+{\frac{16\,{a}^{3}}{3} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{3}{2}}}}+8\,{a}^{4}\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }x+8\,{\frac{{a}^{4}}{\sqrt{{a}^{2}}}\arctan \left ({\frac{\sqrt{{a}^{2}}x}{\sqrt{-{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) }}} \right ) }-{\frac{11\,{a}^{3}}{6} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{11\,{a}^{3}}{2}\sqrt{-{a}^{2}{x}^{2}+1}}+{\frac{11\,{a}^{3}}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{a}^{2}{x}^{2}+1}}} \right ) }+{\frac{3\,a}{2\,{x}^{2}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{1}{3\,{x}^{3}} \left ( -{a}^{2}{x}^{2}+1 \right ) ^{{\frac{5}{2}}}}-{\frac{1}{ \left ( x+{a}^{-1} \right ) ^{3}} \left ( -{a}^{2} \left ( x+{a}^{-1} \right ) ^{2}+2\,a \left ( x+{a}^{-1} \right ) \right ) ^{{\frac{5}{2}}}} \end{align*}

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^4,x)

[Out]

-16/3*a^2/x*(-a^2*x^2+1)^(5/2)-16/3*a^4*x*(-a^2*x^2+1)^(3/2)-8*a^4*x*(-a^2*x^2+1)^(1/2)-8*a^4/(a^2)^(1/2)*arct
an((a^2)^(1/2)*x/(-a^2*x^2+1)^(1/2))+2*a/(x+1/a)^2*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(5/2)+16/3*a^3*(-a^2*(x+1/a)^2
+2*a*(x+1/a))^(3/2)+8*a^4*(-a^2*(x+1/a)^2+2*a*(x+1/a))^(1/2)*x+8*a^4/(a^2)^(1/2)*arctan((a^2)^(1/2)*x/(-a^2*(x
+1/a)^2+2*a*(x+1/a))^(1/2))-11/6*a^3*(-a^2*x^2+1)^(3/2)-11/2*a^3*(-a^2*x^2+1)^(1/2)+11/2*a^3*arctanh(1/(-a^2*x
^2+1)^(1/2))+3/2*a/x^2*(-a^2*x^2+1)^(5/2)-1/3/x^3*(-a^2*x^2+1)^(5/2)-1/(x+1/a)^3*(-a^2*(x+1/a)^2+2*a*(x+1/a))^
(5/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-a^{2} x^{2} + 1\right )}^{\frac{3}{2}}}{{\left (a x + 1\right )}^{3} x^{4}}\,{d x} \end{align*}

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^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.86297, size = 219, normalized size = 1.89 \begin{align*} -\frac{24 \, a^{4} x^{4} + 24 \, a^{3} x^{3} + 33 \,{\left (a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (\frac{\sqrt{-a^{2} x^{2} + 1} - 1}{x}\right ) +{\left (52 \, a^{3} x^{3} + 19 \, a^{2} x^{2} - 7 \, a x + 2\right )} \sqrt{-a^{2} x^{2} + 1}}{6 \,{\left (a x^{4} + x^{3}\right )}} \end{align*}

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^4,x, algorithm="fricas")

[Out]

-1/6*(24*a^4*x^4 + 24*a^3*x^3 + 33*(a^4*x^4 + a^3*x^3)*log((sqrt(-a^2*x^2 + 1) - 1)/x) + (52*a^3*x^3 + 19*a^2*
x^2 - 7*a*x + 2)*sqrt(-a^2*x^2 + 1))/(a*x^4 + x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac{3}{2}}}{x^{4} \left (a x + 1\right )^{3}}\, dx \end{align*}

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**4,x)

[Out]

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

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Giac [B]  time = 1.21526, size = 358, normalized size = 3.09 \begin{align*} \frac{{\left (a^{4} - \frac{8 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{2}}{x} + \frac{48 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2}}{x^{2}} + \frac{249 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{a^{2} x^{3}}\right )} a^{6} x^{3}}{24 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}{\left (\frac{\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a}{a^{2} x} + 1\right )}{\left | a \right |}} + \frac{11 \, a^{4} \log \left (\frac{{\left | -2 \, \sqrt{-a^{2} x^{2} + 1}{\left | a \right |} - 2 \, a \right |}}{2 \, a^{2}{\left | x \right |}}\right )}{2 \,{\left | a \right |}} - \frac{\frac{57 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )} a^{4}}{x} - \frac{9 \,{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{2} a^{2}}{x^{2}} + \frac{{\left (\sqrt{-a^{2} x^{2} + 1}{\left | a \right |} + a\right )}^{3}}{x^{3}}}{24 \, a^{2}{\left | a \right |}} \end{align*}

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^4,x, algorithm="giac")

[Out]

1/24*(a^4 - 8*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^2/x + 48*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2/x^2 + 249*(sqrt(-a^
2*x^2 + 1)*abs(a) + a)^3/(a^2*x^3))*a^6*x^3/((sqrt(-a^2*x^2 + 1)*abs(a) + a)^3*((sqrt(-a^2*x^2 + 1)*abs(a) + a
)/(a^2*x) + 1)*abs(a)) + 11/2*a^4*log(1/2*abs(-2*sqrt(-a^2*x^2 + 1)*abs(a) - 2*a)/(a^2*abs(x)))/abs(a) - 1/24*
(57*(sqrt(-a^2*x^2 + 1)*abs(a) + a)*a^4/x - 9*(sqrt(-a^2*x^2 + 1)*abs(a) + a)^2*a^2/x^2 + (sqrt(-a^2*x^2 + 1)*
abs(a) + a)^3/x^3)/(a^2*abs(a))