3.947 \(\int \frac {e^{\tanh ^{-1}(a x)}}{x^3 (1-a^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=89 \[ \frac {a^2}{1-a x}+\frac {a^2}{8 (a x+1)}+\frac {a^2}{8 (1-a x)^2}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2} \]

[Out]

-1/2/x^2-a/x+1/8*a^2/(-a*x+1)^2+a^2/(-a*x+1)+1/8*a^2/(a*x+1)+3*a^2*ln(x)-39/16*a^2*ln(-a*x+1)-9/16*a^2*ln(a*x+
1)

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6150, 88} \[ \frac {a^2}{1-a x}+\frac {a^2}{8 (a x+1)}+\frac {a^2}{8 (1-a x)^2}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (a x+1)-\frac {a}{x}-\frac {1}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*x^2) - a/x + a^2/(8*(1 - a*x)^2) + a^2/(1 - a*x) + a^2/(8*(1 + a*x)) + 3*a^2*Log[x] - (39*a^2*Log[1 - a*
x])/16 - (9*a^2*Log[1 + a*x])/16

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \frac {e^{\tanh ^{-1}(a x)}}{x^3 \left (1-a^2 x^2\right )^{5/2}} \, dx &=\int \frac {1}{x^3 (1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (\frac {1}{x^3}+\frac {a}{x^2}+\frac {3 a^2}{x}-\frac {a^3}{4 (-1+a x)^3}+\frac {a^3}{(-1+a x)^2}-\frac {39 a^3}{16 (-1+a x)}-\frac {a^3}{8 (1+a x)^2}-\frac {9 a^3}{16 (1+a x)}\right ) \, dx\\ &=-\frac {1}{2 x^2}-\frac {a}{x}+\frac {a^2}{8 (1-a x)^2}+\frac {a^2}{1-a x}+\frac {a^2}{8 (1+a x)}+3 a^2 \log (x)-\frac {39}{16} a^2 \log (1-a x)-\frac {9}{16} a^2 \log (1+a x)\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 83, normalized size = 0.93 \[ \frac {1}{16} \left (\frac {16 a^2}{1-a x}+\frac {2 a^2}{a x+1}+\frac {2 a^2}{(a x-1)^2}+48 a^2 \log (x)-39 a^2 \log (1-a x)-9 a^2 \log (a x+1)-\frac {16 a}{x}-\frac {8}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8/x^2 - (16*a)/x + (16*a^2)/(1 - a*x) + (2*a^2)/(-1 + a*x)^2 + (2*a^2)/(1 + a*x) + 48*a^2*Log[x] - 39*a^2*Lo
g[1 - a*x] - 9*a^2*Log[1 + a*x])/16

________________________________________________________________________________________

fricas [B]  time = 0.69, size = 172, normalized size = 1.93 \[ -\frac {30 \, a^{4} x^{4} - 6 \, a^{3} x^{3} - 44 \, a^{2} x^{2} + 8 \, a x + 9 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 39 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 48 \, {\left (a^{5} x^{5} - a^{4} x^{4} - a^{3} x^{3} + a^{2} x^{2}\right )} \log \relax (x) + 8}{16 \, {\left (a^{3} x^{5} - a^{2} x^{4} - a x^{3} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(30*a^4*x^4 - 6*a^3*x^3 - 44*a^2*x^2 + 8*a*x + 9*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log(a*x + 1) +
39*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log(a*x - 1) - 48*(a^5*x^5 - a^4*x^4 - a^3*x^3 + a^2*x^2)*log(x) +
8)/(a^3*x^5 - a^2*x^4 - a*x^3 + x^2)

________________________________________________________________________________________

giac [A]  time = 0.40, size = 82, normalized size = 0.92 \[ -\frac {9}{16} \, a^{2} \log \left ({\left | a x + 1 \right |}\right ) - \frac {39}{16} \, a^{2} \log \left ({\left | a x - 1 \right |}\right ) + 3 \, a^{2} \log \left ({\left | x \right |}\right ) - \frac {15 \, a^{4} x^{4} - 3 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 4}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/16*a^2*log(abs(a*x + 1)) - 39/16*a^2*log(abs(a*x - 1)) + 3*a^2*log(abs(x)) - 1/8*(15*a^4*x^4 - 3*a^3*x^3 -
22*a^2*x^2 + 4*a*x + 4)/((a*x + 1)*(a*x - 1)^2*x^2)

________________________________________________________________________________________

maple [A]  time = 0.04, size = 78, normalized size = 0.88 \[ -\frac {1}{2 x^{2}}-\frac {a}{x}+3 a^{2} \ln \relax (x )-\frac {a^{2}}{a x -1}+\frac {a^{2}}{8 \left (a x -1\right )^{2}}-\frac {39 a^{2} \ln \left (a x -1\right )}{16}+\frac {a^{2}}{8 a x +8}-\frac {9 a^{2} \ln \left (a x +1\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2-a/x+3*a^2*ln(x)-a^2/(a*x-1)+1/8*a^2/(a*x-1)^2-39/16*a^2*ln(a*x-1)+1/8*a^2/(a*x+1)-9/16*a^2*ln(a*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.33, size = 89, normalized size = 1.00 \[ -\frac {9}{16} \, a^{2} \log \left (a x + 1\right ) - \frac {39}{16} \, a^{2} \log \left (a x - 1\right ) + 3 \, a^{2} \log \relax (x) - \frac {15 \, a^{4} x^{4} - 3 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 4}{8 \, {\left (a^{3} x^{5} - a^{2} x^{4} - a x^{3} + x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-9/16*a^2*log(a*x + 1) - 39/16*a^2*log(a*x - 1) + 3*a^2*log(x) - 1/8*(15*a^4*x^4 - 3*a^3*x^3 - 22*a^2*x^2 + 4*
a*x + 4)/(a^3*x^5 - a^2*x^4 - a*x^3 + x^2)

________________________________________________________________________________________

mupad [B]  time = 0.96, size = 89, normalized size = 1.00 \[ 3\,a^2\,\ln \relax (x)-\frac {39\,a^2\,\ln \left (a\,x-1\right )}{16}-\frac {9\,a^2\,\ln \left (a\,x+1\right )}{16}+\frac {\frac {15\,a^4\,x^4}{8}-\frac {3\,a^3\,x^3}{8}-\frac {11\,a^2\,x^2}{4}+\frac {a\,x}{2}+\frac {1}{2}}{-a^3\,x^5+a^2\,x^4+a\,x^3-x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a^2*log(x) - (39*a^2*log(a*x - 1))/16 - (9*a^2*log(a*x + 1))/16 + ((a*x)/2 - (11*a^2*x^2)/4 - (3*a^3*x^3)/8
+ (15*a^4*x^4)/8 + 1/2)/(a*x^3 - x^2 + a^2*x^4 - a^3*x^5)

________________________________________________________________________________________

sympy [A]  time = 0.56, size = 95, normalized size = 1.07 \[ 3 a^{2} \log {\relax (x )} - \frac {39 a^{2} \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {9 a^{2} \log {\left (x + \frac {1}{a} \right )}}{16} - \frac {15 a^{4} x^{4} - 3 a^{3} x^{3} - 22 a^{2} x^{2} + 4 a x + 4}{8 a^{3} x^{5} - 8 a^{2} x^{4} - 8 a x^{3} + 8 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**2*log(x) - 39*a**2*log(x - 1/a)/16 - 9*a**2*log(x + 1/a)/16 - (15*a**4*x**4 - 3*a**3*x**3 - 22*a**2*x**2
+ 4*a*x + 4)/(8*a**3*x**5 - 8*a**2*x**4 - 8*a*x**3 + 8*x**2)

________________________________________________________________________________________