3.1068 \(\int \frac {e^{2 \tanh ^{-1}(a x)} x^4}{(c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=86 \[ \frac {11}{16 a^5 c^3 (1-a x)}-\frac {1}{16 a^5 c^3 (a x+1)}-\frac {3}{8 a^5 c^3 (1-a x)^2}+\frac {1}{12 a^5 c^3 (1-a x)^3}-\frac {\tanh ^{-1}(a x)}{4 a^5 c^3} \]

[Out]

1/12/a^5/c^3/(-a*x+1)^3-3/8/a^5/c^3/(-a*x+1)^2+11/16/a^5/c^3/(-a*x+1)-1/16/a^5/c^3/(a*x+1)-1/4*arctanh(a*x)/a^
5/c^3

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Rubi [A]  time = 0.14, 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} \[ \frac {11}{16 a^5 c^3 (1-a x)}-\frac {1}{16 a^5 c^3 (a x+1)}-\frac {3}{8 a^5 c^3 (1-a x)^2}+\frac {1}{12 a^5 c^3 (1-a x)^3}-\frac {\tanh ^{-1}(a x)}{4 a^5 c^3} \]

Antiderivative was successfully verified.

[In]

Int[(E^(2*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2)^3,x]

[Out]

1/(12*a^5*c^3*(1 - a*x)^3) - 3/(8*a^5*c^3*(1 - a*x)^2) + 11/(16*a^5*c^3*(1 - a*x)) - 1/(16*a^5*c^3*(1 + a*x))
- ArcTanh[a*x]/(4*a^5*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^4}{\left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {x^4}{(1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac {\int \left (\frac {1}{4 a^4 (-1+a x)^4}+\frac {3}{4 a^4 (-1+a x)^3}+\frac {11}{16 a^4 (-1+a x)^2}+\frac {1}{16 a^4 (1+a x)^2}+\frac {1}{4 a^4 \left (-1+a^2 x^2\right )}\right ) \, dx}{c^3}\\ &=\frac {1}{12 a^5 c^3 (1-a x)^3}-\frac {3}{8 a^5 c^3 (1-a x)^2}+\frac {11}{16 a^5 c^3 (1-a x)}-\frac {1}{16 a^5 c^3 (1+a x)}+\frac {\int \frac {1}{-1+a^2 x^2} \, dx}{4 a^4 c^3}\\ &=\frac {1}{12 a^5 c^3 (1-a x)^3}-\frac {3}{8 a^5 c^3 (1-a x)^2}+\frac {11}{16 a^5 c^3 (1-a x)}-\frac {1}{16 a^5 c^3 (1+a x)}-\frac {\tanh ^{-1}(a x)}{4 a^5 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 64, normalized size = 0.74 \[ \frac {-9 a^3 x^3+6 a^2 x^2+5 a x-3 (a x-1)^3 (a x+1) \tanh ^{-1}(a x)-4}{12 a^5 c^3 (a x-1)^3 (a x+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*ArcTanh[a*x])*x^4)/(c - a^2*c*x^2)^3,x]

[Out]

(-4 + 5*a*x + 6*a^2*x^2 - 9*a^3*x^3 - 3*(-1 + a*x)^3*(1 + a*x)*ArcTanh[a*x])/(12*a^5*c^3*(-1 + a*x)^3*(1 + a*x
))

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fricas [A]  time = 0.55, size = 123, normalized size = 1.43 \[ -\frac {18 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 10 \, a x + 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) + 8}{24 \, {\left (a^{9} c^{3} x^{4} - 2 \, a^{8} c^{3} x^{3} + 2 \, a^{6} c^{3} x - a^{5} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="fricas")

[Out]

-1/24*(18*a^3*x^3 - 12*a^2*x^2 - 10*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) + 8)/(a^9*c^3*x^4 - 2*a^8*c^3*x^3 + 2*a^6*c^3*x - a^5*c^3)

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giac [A]  time = 0.19, size = 75, normalized size = 0.87 \[ -\frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a^{5} c^{3}} + \frac {\log \left ({\left | a x - 1 \right |}\right )}{8 \, a^{5} c^{3}} - \frac {9 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 5 \, a x + 4}{12 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a^{5} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="giac")

[Out]

-1/8*log(abs(a*x + 1))/(a^5*c^3) + 1/8*log(abs(a*x - 1))/(a^5*c^3) - 1/12*(9*a^3*x^3 - 6*a^2*x^2 - 5*a*x + 4)/
((a*x + 1)*(a*x - 1)^3*a^5*c^3)

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maple [A]  time = 0.04, size = 90, normalized size = 1.05 \[ -\frac {1}{12 c^{3} a^{5} \left (a x -1\right )^{3}}-\frac {3}{8 c^{3} a^{5} \left (a x -1\right )^{2}}-\frac {11}{16 c^{3} a^{5} \left (a x -1\right )}+\frac {\ln \left (a x -1\right )}{8 c^{3} a^{5}}-\frac {1}{16 a^{5} c^{3} \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{8 c^{3} a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^3,x)

[Out]

-1/12/c^3/a^5/(a*x-1)^3-3/8/c^3/a^5/(a*x-1)^2-11/16/c^3/a^5/(a*x-1)+1/8/c^3/a^5*ln(a*x-1)-1/16/a^5/c^3/(a*x+1)
-1/8/c^3/a^5*ln(a*x+1)

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maxima [A]  time = 0.33, size = 94, normalized size = 1.09 \[ -\frac {9 \, a^{3} x^{3} - 6 \, a^{2} x^{2} - 5 \, a x + 4}{12 \, {\left (a^{9} c^{3} x^{4} - 2 \, a^{8} c^{3} x^{3} + 2 \, a^{6} c^{3} x - a^{5} c^{3}\right )}} - \frac {\log \left (a x + 1\right )}{8 \, a^{5} c^{3}} + \frac {\log \left (a x - 1\right )}{8 \, a^{5} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)*x^4/(-a^2*c*x^2+c)^3,x, algorithm="maxima")

[Out]

-1/12*(9*a^3*x^3 - 6*a^2*x^2 - 5*a*x + 4)/(a^9*c^3*x^4 - 2*a^8*c^3*x^3 + 2*a^6*c^3*x - a^5*c^3) - 1/8*log(a*x
+ 1)/(a^5*c^3) + 1/8*log(a*x - 1)/(a^5*c^3)

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mupad [B]  time = 0.07, size = 78, normalized size = 0.91 \[ -\frac {\frac {5\,x}{12\,a^4}-\frac {1}{3\,a^5}-\frac {3\,x^3}{4\,a^2}+\frac {x^2}{2\,a^3}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}-\frac {\mathrm {atanh}\left (a\,x\right )}{4\,a^5\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4*(a*x + 1)^2)/((c - a^2*c*x^2)^3*(a^2*x^2 - 1)),x)

[Out]

- ((5*x)/(12*a^4) - 1/(3*a^5) - (3*x^3)/(4*a^2) + x^2/(2*a^3))/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x)
- atanh(a*x)/(4*a^5*c^3)

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sympy [A]  time = 0.45, size = 88, normalized size = 1.02 \[ \frac {- 9 a^{3} x^{3} + 6 a^{2} x^{2} + 5 a x - 4}{12 a^{9} c^{3} x^{4} - 24 a^{8} c^{3} x^{3} + 24 a^{6} c^{3} x - 12 a^{5} c^{3}} + \frac {\frac {\log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)**2/(-a**2*x**2+1)*x**4/(-a**2*c*x**2+c)**3,x)

[Out]

(-9*a**3*x**3 + 6*a**2*x**2 + 5*a*x - 4)/(12*a**9*c**3*x**4 - 24*a**8*c**3*x**3 + 24*a**6*c**3*x - 12*a**5*c**
3) + (log(x - 1/a)/8 - log(x + 1/a)/8)/(a**5*c**3)

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