3.1.41 \(\int \frac {e^{-\tanh ^{-1}(a x)}}{x^4} \, dx\) [41]

Optimal. Leaf size=90 \[ -\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{2} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6259, 849, 821, 272, 65, 214} \begin {gather*} -\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {1}{2} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 214

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

Rule 272

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 821

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 849

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 6259

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^4} \, dx &=\int \frac {1-a x}{x^4 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 x^3}-\frac {1}{3} \int \frac {3 a-2 a^2 x}{x^3 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {a \sqrt {1-a^2 x^2}}{2 x^2}+\frac {1}{6} \int \frac {4 a^2-3 a^3 x}{x^2 \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}-\frac {1}{2} a^3 \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {\sqrt {1-a^2 x^2}}{3 x^3}+\frac {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}-\frac {1}{4} a^3 \text {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 {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{2} a \text {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 {a \sqrt {1-a^2 x^2}}{2 x^2}-\frac {2 a^2 \sqrt {1-a^2 x^2}}{3 x}+\frac {1}{2} a^3 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 66, normalized size = 0.73 \begin {gather*} \frac {1}{6} \left (\frac {\left (-2+3 a x-4 a^2 x^2\right ) \sqrt {1-a^2 x^2}}{x^3}-3 a^3 \log (x)+3 a^3 \log \left (1+\sqrt {1-a^2 x^2}\right )\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(241\) vs. \(2(74)=148\).
time = 0.80, size = 242, normalized size = 2.69

method	result	size
risch	\(\frac {4 a^{4} x^{4}-3 a^{3} x^{3}-2 a^{2} x^{2}+3 a x -2}{6 x^{3} \sqrt {-a^{2} x^{2}+1}}+\frac {a^{3} \arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )}{2}\)	\(67\)
default	\(-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{3 x^{3}}+a^{3} \left (\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}+\frac {a \arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} \left (x +\frac {1}{a}\right )^{2}+2 a \left (x +\frac {1}{a}\right )}}\right )}{\sqrt {a^{2}}}\right )-a \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{2 x^{2}}-\frac {a^{2} \left (\sqrt {-a^{2} x^{2}+1}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )}{2}\right )+a^{2} \left (-\frac {\left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{x}-2 a^{2} \left (\frac {x \sqrt {-a^{2} x^{2}+1}}{2}+\frac {\arctan \left (\frac {\sqrt {a^{2}}\, x}{\sqrt {-a^{2} x^{2}+1}}\right )}{2 \sqrt {a^{2}}}\right )\right )-a^{3} \left (\sqrt {-a^{2} x^{2}+1}-\arctanh \left (\frac {1}{\sqrt {-a^{2} x^{2}+1}}\right )\right )\)	\(242\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 60, normalized size = 0.67 \begin {gather*} -\frac {3 \, a^{3} x^{3} \log \left (\frac {\sqrt {-a^{2} x^{2} + 1} - 1}{x}\right ) + {\left (4 \, a^{2} x^{2} - 3 \, a x + 2\right )} \sqrt {-a^{2} x^{2} + 1}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}{x^{4} \left (a x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a*x+1)*(-a^2*x^2+1)^(1/2)/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [B]
time = 0.03, size = 78, normalized size = 0.87 \begin {gather*} \frac {a\,\sqrt {1-a^2\,x^2}}{2\,x^2}-\frac {\sqrt {1-a^2\,x^2}}{3\,x^3}-\frac {2\,a^2\,\sqrt {1-a^2\,x^2}}{3\,x}-\frac {a^3\,\mathrm {atan}\left (\sqrt {1-a^2\,x^2}\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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