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

Optimal. Leaf size=110 \[ \frac {1}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {(1-a x)^{5/4} (a x+1)^{3/4}}{2 x^2}+\frac {a \sqrt [4]{1-a x} (a x+1)^{3/4}}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6126, 96, 94, 93, 298, 203, 206} \[ \frac {1}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {1}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {(1-a x)^{5/4} (a x+1)^{3/4}}{2 x^2}+\frac {a \sqrt [4]{1-a x} (a x+1)^{3/4}}{4 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(1 - a*x)^(1/4)*(1 + a*x)^(3/4))/(4*x) - ((1 - a*x)^(5/4)*(1 + a*x)^(3/4))/(2*x^2) + (a^2*ArcTan[(1 + a*x)^
(1/4)/(1 - a*x)^(1/4)])/4 - (a^2*ArcTanh[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/4

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 6126

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 69, normalized size = 0.63 \[ \frac {\sqrt [4]{1-a x} \left (-2 a^2 x^2 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {1-a x}{a x+1}\right )+3 a^2 x^2+a x-2\right )}{4 x^2 \sqrt [4]{a x+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 - a*x)^(1/4)*(-2 + a*x + 3*a^2*x^2 - 2*a^2*x^2*Hypergeometric2F1[1/4, 1, 5/4, (1 - a*x)/(1 + a*x)]))/(4*x^
2*(1 + a*x)^(1/4))

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fricas [A]  time = 0.52, size = 148, normalized size = 1.35 \[ \frac {2 \, a^{2} x^{2} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + a^{2} x^{2} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right ) + 2 \, \sqrt {-a^{2} x^{2} + 1} {\left (3 \, a x - 2\right )} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*a^2*x^2*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - a^2*x^2*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) +
1) + a^2*x^2*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1) + 2*sqrt(-a^2*x^2 + 1)*(3*a*x - 2)*sqrt(-sqrt(-a^2*
x^2 + 1)/(a*x - 1)))/x^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}}\, x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt {\frac {a x + 1}{\sqrt {-a^{2} x^{2} + 1}}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,\sqrt {\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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