3.1069 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x^3}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=86 \[ \frac {3}{16 a^4 c^3 (1-a x)}+\frac {1}{16 a^4 c^3 (a x+1)}-\frac {1}{4 a^4 c^3 (1-a x)^2}+\frac {1}{12 a^4 c^3 (1-a x)^3}+\frac {\tanh ^{-1}(a x)}{8 a^4 c^3} \]

[Out]

1/12/a^4/c^3/(-a*x+1)^3-1/4/a^4/c^3/(-a*x+1)^2+3/16/a^4/c^3/(-a*x+1)+1/16/a^4/c^3/(a*x+1)+1/8*arctanh(a*x)/a^4
/c^3

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Rubi [A]  time = 0.12, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6150, 88, 207} \[ \frac {3}{16 a^4 c^3 (1-a x)}+\frac {1}{16 a^4 c^3 (a x+1)}-\frac {1}{4 a^4 c^3 (1-a x)^2}+\frac {1}{12 a^4 c^3 (1-a x)^3}+\frac {\tanh ^{-1}(a x)}{8 a^4 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(12*a^4*c^3*(1 - a*x)^3) - 1/(4*a^4*c^3*(1 - a*x)^2) + 3/(16*a^4*c^3*(1 - a*x)) + 1/(16*a^4*c^3*(1 + a*x)) +
ArcTanh[a*x]/(8*a^4*c^3)

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^{2 \tanh ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {x^3}{(1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac {\int \left (\frac {1}{4 a^3 (-1+a x)^4}+\frac {1}{2 a^3 (-1+a x)^3}+\frac {3}{16 a^3 (-1+a x)^2}-\frac {1}{16 a^3 (1+a x)^2}-\frac {1}{8 a^3 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3}\\ &=\frac {1}{12 a^4 c^3 (1-a x)^3}-\frac {1}{4 a^4 c^3 (1-a x)^2}+\frac {3}{16 a^4 c^3 (1-a x)}+\frac {1}{16 a^4 c^3 (1+a x)}-\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{8 a^3 c^3}\\ &=\frac {1}{12 a^4 c^3 (1-a x)^3}-\frac {1}{4 a^4 c^3 (1-a x)^2}+\frac {3}{16 a^4 c^3 (1-a x)}+\frac {1}{16 a^4 c^3 (1+a x)}+\frac {\tanh ^{-1}(a x)}{8 a^4 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.74 \[ \frac {-3 a^3 x^3-6 a^2 x^2+7 a x+3 (a x-1)^3 (a x+1) \tanh ^{-1}(a x)-2}{24 a^4 c^3 (a x-1)^3 (a x+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2 + 7*a*x - 6*a^2*x^2 - 3*a^3*x^3 + 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[a*x])/(24*a^4*c^3*(-1 + a*x)^3*(1 + a*x
))

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fricas [A]  time = 0.61, size = 123, normalized size = 1.43 \[ -\frac {6 \, a^{3} x^{3} + 12 \, a^{2} x^{2} - 14 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 4}{48 \, {\left (a^{8} c^{3} x^{4} - 2 \, a^{7} c^{3} x^{3} + 2 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(6*a^3*x^3 + 12*a^2*x^2 - 14*a*x - 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 3*(a^4*x^4 - 2*a^3
*x^3 + 2*a*x - 1)*log(a*x - 1) + 4)/(a^8*c^3*x^4 - 2*a^7*c^3*x^3 + 2*a^5*c^3*x - a^4*c^3)

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giac [A]  time = 0.19, size = 75, normalized size = 0.87 \[ \frac {\log \left ({\left | a x + 1 \right |}\right )}{16 \, a^{4} c^{3}} - \frac {\log \left ({\left | a x - 1 \right |}\right )}{16 \, a^{4} c^{3}} - \frac {3 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 7 \, a x + 2}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a^{4} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*log(abs(a*x + 1))/(a^4*c^3) - 1/16*log(abs(a*x - 1))/(a^4*c^3) - 1/24*(3*a^3*x^3 + 6*a^2*x^2 - 7*a*x + 2)
/((a*x + 1)*(a*x - 1)^3*a^4*c^3)

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maple [A]  time = 0.04, size = 90, normalized size = 1.05 \[ -\frac {1}{12 c^{3} a^{4} \left (a x -1\right )^{3}}-\frac {1}{4 c^{3} a^{4} \left (a x -1\right )^{2}}-\frac {3}{16 c^{3} a^{4} \left (a x -1\right )}-\frac {\ln \left (a x -1\right )}{16 c^{3} a^{4}}+\frac {1}{16 a^{4} c^{3} \left (a x +1\right )}+\frac {\ln \left (a x +1\right )}{16 c^{3} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/c^3/a^4/(a*x-1)^3-1/4/c^3/a^4/(a*x-1)^2-3/16/c^3/a^4/(a*x-1)-1/16/c^3/a^4*ln(a*x-1)+1/16/a^4/c^3/(a*x+1)
+1/16/c^3/a^4*ln(a*x+1)

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maxima [A]  time = 0.34, size = 94, normalized size = 1.09 \[ -\frac {3 \, a^{3} x^{3} + 6 \, a^{2} x^{2} - 7 \, a x + 2}{24 \, {\left (a^{8} c^{3} x^{4} - 2 \, a^{7} c^{3} x^{3} + 2 \, a^{5} c^{3} x - a^{4} c^{3}\right )}} + \frac {\log \left (a x + 1\right )}{16 \, a^{4} c^{3}} - \frac {\log \left (a x - 1\right )}{16 \, a^{4} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*a^3*x^3 + 6*a^2*x^2 - 7*a*x + 2)/(a^8*c^3*x^4 - 2*a^7*c^3*x^3 + 2*a^5*c^3*x - a^4*c^3) + 1/16*log(a*x
+ 1)/(a^4*c^3) - 1/16*log(a*x - 1)/(a^4*c^3)

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mupad [B]  time = 0.93, size = 77, normalized size = 0.90 \[ \frac {\frac {1}{12\,a^4}-\frac {7\,x}{24\,a^3}+\frac {x^3}{8\,a}+\frac {x^2}{4\,a^2}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}+\frac {\mathrm {atanh}\left (a\,x\right )}{8\,a^4\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/(12*a^4) - (7*x)/(24*a^3) + x^3/(8*a) + x^2/(4*a^2))/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x) + atan
h(a*x)/(8*a^4*c^3)

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sympy [A]  time = 0.43, size = 88, normalized size = 1.02 \[ \frac {- 3 a^{3} x^{3} - 6 a^{2} x^{2} + 7 a x - 2}{24 a^{8} c^{3} x^{4} - 48 a^{7} c^{3} x^{3} + 48 a^{5} c^{3} x - 24 a^{4} c^{3}} + \frac {- \frac {\log {\left (x - \frac {1}{a} \right )}}{16} + \frac {\log {\left (x + \frac {1}{a} \right )}}{16}}{a^{4} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*a**3*x**3 - 6*a**2*x**2 + 7*a*x - 2)/(24*a**8*c**3*x**4 - 48*a**7*c**3*x**3 + 48*a**5*c**3*x - 24*a**4*c**
3) + (-log(x - 1/a)/16 + log(x + 1/a)/16)/(a**4*c**3)

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