3.2.17 \(\int \frac {e^{-\frac {5}{2} \tanh ^{-1}(a x)}}{x^3} \, dx\) [117]

Optimal. Leaf size=136 \[ \frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {25}{4} a^2 \text {ArcTan}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {25}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right ) \]

[Out]

25/2*a^2*(-a*x+1)^(1/4)/(a*x+1)^(1/4)+5/4*a*(-a*x+1)^(5/4)/x/(a*x+1)^(1/4)-1/2*(-a*x+1)^(9/4)/x^2/(a*x+1)^(1/4
)+25/4*a^2*arctan((a*x+1)^(1/4)/(-a*x+1)^(1/4))-25/4*a^2*arctanh((a*x+1)^(1/4)/(-a*x+1)^(1/4))

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Rubi [A]
time = 0.04, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6261, 98, 96, 95, 304, 209, 212} \begin {gather*} \frac {25}{4} a^2 \text {ArcTan}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )+\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{a x+1}}-\frac {25}{4} a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{a x+1}}{\sqrt [4]{1-a x}}\right )-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{a x+1}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{a x+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^((5*ArcTanh[a*x])/2)*x^3),x]

[Out]

(25*a^2*(1 - a*x)^(1/4))/(2*(1 + a*x)^(1/4)) + (5*a*(1 - a*x)^(5/4))/(4*x*(1 + a*x)^(1/4)) - (1 - a*x)^(9/4)/(
2*x^2*(1 + a*x)^(1/4)) + (25*a^2*ArcTan[(1 + a*x)^(1/4)/(1 - a*x)^(1/4)])/4 - (25*a^2*ArcTanh[(1 + a*x)^(1/4)/
(1 - a*x)^(1/4)])/4

Rule 95

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 96

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[m, 1] ||  !SumSimplerQ[p, 1]) && NeQ[m, -1]

Rule 98

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 209

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

Rule 212

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

Rule 304

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 6261

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 {5}{2} \tanh ^{-1}(a x)}}{x^3} \, dx &=\int \frac {(1-a x)^{5/4}}{x^3 (1+a x)^{5/4}} \, dx\\ &=-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}-\frac {1}{4} (5 a) \int \frac {(1-a x)^{5/4}}{x^2 (1+a x)^{5/4}} \, dx\\ &=\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {1}{8} \left (25 a^2\right ) \int \frac {\sqrt [4]{1-a x}}{x (1+a x)^{5/4}} \, dx\\ &=\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {1}{8} \left (25 a^2\right ) \int \frac {1}{x (1-a x)^{3/4} \sqrt [4]{1+a x}} \, dx\\ &=\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {1}{2} \left (25 a^2\right ) \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ &=\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}-\frac {1}{4} \left (25 a^2\right ) \text {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} \left (25 a^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )\\ &=\frac {25 a^2 \sqrt [4]{1-a x}}{2 \sqrt [4]{1+a x}}+\frac {5 a (1-a x)^{5/4}}{4 x \sqrt [4]{1+a x}}-\frac {(1-a x)^{9/4}}{2 x^2 \sqrt [4]{1+a x}}+\frac {25}{4} a^2 \tan ^{-1}\left (\frac {\sqrt [4]{1+a x}}{\sqrt [4]{1-a x}}\right )-\frac {25}{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] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 70, normalized size = 0.51 \begin {gather*} \frac {\sqrt [4]{1-a x} \left (-2+9 a x+43 a^2 x^2-50 a^2 x^2 \, _2F_1\left (\frac {1}{4},1;\frac {5}{4};\frac {1-a x}{1+a x}\right )\right )}{4 x^2 \sqrt [4]{1+a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^((5*ArcTanh[a*x])/2)*x^3),x]

[Out]

((1 - a*x)^(1/4)*(-2 + 9*a*x + 43*a^2*x^2 - 50*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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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (\frac {a x +1}{\sqrt {-a^{2} x^{2}+1}}\right )^{\frac {5}{2}} x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)

[Out]

int(1/((a*x+1)/(-a^2*x^2+1)^(1/2))^(5/2)/x^3,x)

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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))^(5/2)/x^3,x, algorithm="maxima")

[Out]

integrate(1/(x^3*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)

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Fricas [A]
time = 0.38, size = 192, normalized size = 1.41 \begin {gather*} \frac {2 \, {\left (43 \, a^{2} x^{2} + 9 \, a x - 2\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 50 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \arctan \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}}\right ) - 25 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} + 1\right ) + 25 \, {\left (a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (\sqrt {-\frac {\sqrt {-a^{2} x^{2} + 1}}{a x - 1}} - 1\right )}{8 \, {\left (a x^{3} + x^{2}\right )}} \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))^(5/2)/x^3,x, algorithm="fricas")

[Out]

1/8*(2*(43*a^2*x^2 + 9*a*x - 2)*sqrt(-a^2*x^2 + 1)*sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) + 50*(a^3*x^3 + a^2*x^2
)*arctan(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))) - 25*(a^3*x^3 + a^2*x^2)*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1))
+ 1) + 25*(a^3*x^3 + a^2*x^2)*log(sqrt(-sqrt(-a^2*x^2 + 1)/(a*x - 1)) - 1))/(a*x^3 + x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{3} \left (\frac {a x + 1}{\sqrt {- a^{2} x^{2} + 1}}\right )^{\frac {5}{2}}}\, 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))**(5/2)/x**3,x)

[Out]

Integral(1/(x**3*((a*x + 1)/sqrt(-a**2*x**2 + 1))**(5/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not 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))^(5/2)/x^3,x, algorithm="giac")

[Out]

integrate(1/(x^3*((a*x + 1)/sqrt(-a^2*x^2 + 1))^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^3\,{\left (\frac {a\,x+1}{\sqrt {1-a^2\,x^2}}\right )}^{5/2}} \,d x \end {gather*}

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))^(5/2)),x)

[Out]

int(1/(x^3*((a*x + 1)/(1 - a^2*x^2)^(1/2))^(5/2)), x)

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