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

Optimal. Leaf size=56 \[ -\frac {1}{2 a^4 (1-a x)}+\frac {1}{8 a^4 (a x+1)}+\frac {1}{8 a^4 (1-a x)^2}+\frac {3 \tanh ^{-1}(a x)}{8 a^4} \]

[Out]

1/8/a^4/(-a*x+1)^2-1/2/a^4/(-a*x+1)+1/8/a^4/(a*x+1)+3/8*arctanh(a*x)/a^4

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Rubi [A]  time = 0.12, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6150, 88, 207} \[ -\frac {1}{2 a^4 (1-a x)}+\frac {1}{8 a^4 (a x+1)}+\frac {1}{8 a^4 (1-a x)^2}+\frac {3 \tanh ^{-1}(a x)}{8 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(8*a^4*(1 - a*x)^2) - 1/(2*a^4*(1 - a*x)) + 1/(8*a^4*(1 + a*x)) + (3*ArcTanh[a*x])/(8*a^4)

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 207

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

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 {x^3}{(1-a x)^3 (1+a x)^2} \, dx\\ &=\int \left (-\frac {1}{4 a^3 (-1+a x)^3}-\frac {1}{2 a^3 (-1+a x)^2}-\frac {1}{8 a^3 (1+a x)^2}-\frac {3}{8 a^3 \left (-1+a^2 x^2\right )}\right ) \, dx\\ &=\frac {1}{8 a^4 (1-a x)^2}-\frac {1}{2 a^4 (1-a x)}+\frac {1}{8 a^4 (1+a x)}-\frac {3 \int \frac {1}{-1+a^2 x^2} \, dx}{8 a^3}\\ &=\frac {1}{8 a^4 (1-a x)^2}-\frac {1}{2 a^4 (1-a x)}+\frac {1}{8 a^4 (1+a x)}+\frac {3 \tanh ^{-1}(a x)}{8 a^4}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 53, normalized size = 0.95 \[ \frac {5 a^2 x^2-a x+3 (a x-1)^2 (a x+1) \tanh ^{-1}(a x)-2}{8 a^4 (a x-1)^2 (a x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2 - a*x + 5*a^2*x^2 + 3*(-1 + a*x)^2*(1 + a*x)*ArcTanh[a*x])/(8*a^4*(-1 + a*x)^2*(1 + a*x))

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fricas [B]  time = 0.74, size = 101, normalized size = 1.80 \[ \frac {10 \, a^{2} x^{2} - 2 \, a x + 3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) - 3 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 4}{16 \, {\left (a^{7} x^{3} - a^{6} x^{2} - a^{5} x + a^{4}\right )}} \]

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*(10*a^2*x^2 - 2*a*x + 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) - 3*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(
a*x - 1) - 4)/(a^7*x^3 - a^6*x^2 - a^5*x + a^4)

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giac [A]  time = 0.19, size = 58, normalized size = 1.04 \[ \frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{4}} - \frac {3 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{4}} + \frac {5 \, a^{2} x^{2} - a x - 2}{8 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{2} a^{4}} \]

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]

3/16*log(abs(a*x + 1))/a^4 - 3/16*log(abs(a*x - 1))/a^4 + 1/8*(5*a^2*x^2 - a*x - 2)/((a*x + 1)*(a*x - 1)^2*a^4
)

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maple [A]  time = 0.04, size = 60, normalized size = 1.07 \[ \frac {1}{8 a^{4} \left (a x -1\right )^{2}}+\frac {1}{2 a^{4} \left (a x -1\right )}-\frac {3 \ln \left (a x -1\right )}{16 a^{4}}+\frac {1}{8 a^{4} \left (a x +1\right )}+\frac {3 \ln \left (a x +1\right )}{16 a^{4}} \]

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]

1/8/a^4/(a*x-1)^2+1/2/a^4/(a*x-1)-3/16/a^4*ln(a*x-1)+1/8/a^4/(a*x+1)+3/16*ln(a*x+1)/a^4

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maxima [A]  time = 0.31, size = 66, normalized size = 1.18 \[ \frac {5 \, a^{2} x^{2} - a x - 2}{8 \, {\left (a^{7} x^{3} - a^{6} x^{2} - a^{5} x + a^{4}\right )}} + \frac {3 \, \log \left (a x + 1\right )}{16 \, a^{4}} - \frac {3 \, \log \left (a x - 1\right )}{16 \, a^{4}} \]

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]

1/8*(5*a^2*x^2 - a*x - 2)/(a^7*x^3 - a^6*x^2 - a^5*x + a^4) + 3/16*log(a*x + 1)/a^4 - 3/16*log(a*x - 1)/a^4

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mupad [B]  time = 0.06, size = 53, normalized size = 0.95 \[ \frac {\frac {x}{8\,a^3}+\frac {1}{4\,a^4}-\frac {5\,x^2}{8\,a^2}}{-a^3\,x^3+a^2\,x^2+a\,x-1}+\frac {3\,\mathrm {atanh}\left (a\,x\right )}{8\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x/(8*a^3) + 1/(4*a^4) - (5*x^2)/(8*a^2))/(a*x + a^2*x^2 - a^3*x^3 - 1) + (3*atanh(a*x))/(8*a^4)

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sympy [A]  time = 0.31, size = 66, normalized size = 1.18 \[ - \frac {- 5 a^{2} x^{2} + a x + 2}{8 a^{7} x^{3} - 8 a^{6} x^{2} - 8 a^{5} x + 8 a^{4}} - \frac {\frac {3 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {3 \log {\left (x + \frac {1}{a} \right )}}{16}}{a^{4}} \]

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]

-(-5*a**2*x**2 + a*x + 2)/(8*a**7*x**3 - 8*a**6*x**2 - 8*a**5*x + 8*a**4) - (3*log(x - 1/a)/16 - 3*log(x + 1/a
)/16)/a**4

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