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

Optimal. Leaf size=78 \[ \frac {c^3}{5 a^6 x^5}+\frac {c^3}{2 a^5 x^4}-\frac {c^3}{3 a^4 x^3}-\frac {2 c^3}{a^3 x^2}-\frac {c^3}{a^2 x}-\frac {2 c^3 \log (x)}{a}+c^3 (-x) \]

[Out]

1/5*c^3/a^6/x^5+1/2*c^3/a^5/x^4-1/3*c^3/a^4/x^3-2*c^3/a^3/x^2-c^3/a^2/x-c^3*x-2*c^3*ln(x)/a

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Rubi [A]  time = 0.12, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6157, 6150, 88} \[ -\frac {2 c^3}{a^3 x^2}-\frac {c^3}{3 a^4 x^3}+\frac {c^3}{2 a^5 x^4}+\frac {c^3}{5 a^6 x^5}-\frac {c^3}{a^2 x}-\frac {2 c^3 \log (x)}{a}+c^3 (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c^3/(5*a^6*x^5) + c^3/(2*a^5*x^4) - c^3/(3*a^4*x^3) - (2*c^3)/(a^3*x^2) - c^3/(a^2*x) - c^3*x - (2*c^3*Log[x])
/a

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])

Rule 6157

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

Rubi steps

\begin {align*} \int e^{2 \tanh ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right )^3 \, dx &=-\frac {c^3 \int \frac {e^{2 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^3}{x^6} \, dx}{a^6}\\ &=-\frac {c^3 \int \frac {(1-a x)^2 (1+a x)^4}{x^6} \, dx}{a^6}\\ &=-\frac {c^3 \int \left (a^6+\frac {1}{x^6}+\frac {2 a}{x^5}-\frac {a^2}{x^4}-\frac {4 a^3}{x^3}-\frac {a^4}{x^2}+\frac {2 a^5}{x}\right ) \, dx}{a^6}\\ &=\frac {c^3}{5 a^6 x^5}+\frac {c^3}{2 a^5 x^4}-\frac {c^3}{3 a^4 x^3}-\frac {2 c^3}{a^3 x^2}-\frac {c^3}{a^2 x}-c^3 x-\frac {2 c^3 \log (x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 78, normalized size = 1.00 \[ \frac {c^3}{5 a^6 x^5}+\frac {c^3}{2 a^5 x^4}-\frac {c^3}{3 a^4 x^3}-\frac {2 c^3}{a^3 x^2}-\frac {c^3}{a^2 x}-\frac {2 c^3 \log (x)}{a}+c^3 (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c^3/(5*a^6*x^5) + c^3/(2*a^5*x^4) - c^3/(3*a^4*x^3) - (2*c^3)/(a^3*x^2) - c^3/(a^2*x) - c^3*x - (2*c^3*Log[x])
/a

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fricas [A]  time = 0.73, size = 78, normalized size = 1.00 \[ -\frac {30 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \relax (x) + 30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(30*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) + 30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x
- 6*c^3)/(a^6*x^5)

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giac [A]  time = 0.19, size = 72, normalized size = 0.92 \[ -c^{3} x - \frac {2 \, c^{3} \log \left ({\left | x \right |}\right )}{a} - \frac {30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^3*x - 2*c^3*log(abs(x))/a - 1/30*(30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x - 6*c^3)/(a
^6*x^5)

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maple [A]  time = 0.03, size = 73, normalized size = 0.94 \[ \frac {c^{3}}{5 a^{6} x^{5}}+\frac {c^{3}}{2 a^{5} x^{4}}-\frac {c^{3}}{3 a^{4} x^{3}}-\frac {2 c^{3}}{x^{2} a^{3}}-\frac {c^{3}}{a^{2} x}-c^{3} x -\frac {2 c^{3} \ln \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*c^3/a^6/x^5+1/2*c^3/a^5/x^4-1/3*c^3/a^4/x^3-2*c^3/x^2/a^3-c^3/a^2/x-c^3*x-2*c^3*ln(x)/a

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maxima [A]  time = 0.31, size = 71, normalized size = 0.91 \[ -c^{3} x - \frac {2 \, c^{3} \log \relax (x)}{a} - \frac {30 \, a^{4} c^{3} x^{4} + 60 \, a^{3} c^{3} x^{3} + 10 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 6 \, c^{3}}{30 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^3*x - 2*c^3*log(x)/a - 1/30*(30*a^4*c^3*x^4 + 60*a^3*c^3*x^3 + 10*a^2*c^3*x^2 - 15*a*c^3*x - 6*c^3)/(a^6*x^
5)

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mupad [B]  time = 0.88, size = 57, normalized size = 0.73 \[ -\frac {c^3\,\left (\frac {a^2\,x^2}{3}-\frac {a\,x}{2}+2\,a^3\,x^3+a^4\,x^4+a^6\,x^6+2\,a^5\,x^5\,\ln \relax (x)-\frac {1}{5}\right )}{a^6\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.35, size = 78, normalized size = 1.00 \[ \frac {- a^{6} c^{3} x - 2 a^{5} c^{3} \log {\relax (x )} - \frac {30 a^{4} c^{3} x^{4} + 60 a^{3} c^{3} x^{3} + 10 a^{2} c^{3} x^{2} - 15 a c^{3} x - 6 c^{3}}{30 x^{5}}}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a**6*c**3*x - 2*a**5*c**3*log(x) - (30*a**4*c**3*x**4 + 60*a**3*c**3*x**3 + 10*a**2*c**3*x**2 - 15*a*c**3*x
- 6*c**3)/(30*x**5))/a**6

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