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

Optimal. Leaf size=63 \[ \frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+\frac {4 c^3 \log (x)}{a}+c^3 x \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 63, 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, 75} \[ \frac {5 c^3}{3 a^4 x^3}+\frac {c^3}{a^5 x^4}+\frac {c^3}{5 a^6 x^5}-\frac {5 c^3}{a^2 x}+\frac {4 c^3 \log (x)}{a}+c^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 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])

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

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Mathematica [A]  time = 0.02, size = 63, normalized size = 1.00 \[ \frac {c^3}{5 a^6 x^5}+\frac {c^3}{a^5 x^4}+\frac {5 c^3}{3 a^4 x^3}-\frac {5 c^3}{a^2 x}+\frac {4 c^3 \log (x)}{a}+c^3 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 67, normalized size = 1.06 \[ \frac {15 \, a^{6} c^{3} x^{6} + 60 \, a^{5} c^{3} x^{5} \log \relax (x) - 75 \, a^{4} c^{3} x^{4} + 25 \, a^{2} c^{3} x^{2} + 15 \, a c^{3} x + 3 \, c^{3}}{15 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a^6*c^3*x^6 + 60*a^5*c^3*x^5*log(x) - 75*a^4*c^3*x^4 + 25*a^2*c^3*x^2 + 15*a*c^3*x + 3*c^3)/(a^6*x^5)

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giac [A]  time = 0.14, size = 60, normalized size = 0.95 \[ c^{3} x + \frac {4 \, c^{3} \log \left ({\left | x \right |}\right )}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*x + 4*c^3*log(abs(x))/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

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maple [A]  time = 0.04, size = 60, normalized size = 0.95 \[ \frac {c^{3}}{5 a^{6} x^{5}}+\frac {c^{3}}{a^{5} x^{4}}+\frac {5 c^{3}}{3 a^{4} x^{3}}-\frac {5 c^{3}}{a^{2} x}+c^{3} x +\frac {4 c^{3} \ln \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 59, normalized size = 0.94 \[ c^{3} x + \frac {4 \, c^{3} \log \relax (x)}{a} - \frac {75 \, a^{4} c^{3} x^{4} - 25 \, a^{2} c^{3} x^{2} - 15 \, a c^{3} x - 3 \, c^{3}}{15 \, a^{6} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*x + 4*c^3*log(x)/a - 1/15*(75*a^4*c^3*x^4 - 25*a^2*c^3*x^2 - 15*a*c^3*x - 3*c^3)/(a^6*x^5)

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mupad [B]  time = 0.06, size = 48, normalized size = 0.76 \[ \frac {c^3\,\left (a\,x+\frac {5\,a^2\,x^2}{3}-5\,a^4\,x^4+a^6\,x^6+4\,a^5\,x^5\,\ln \relax (x)+\frac {1}{5}\right )}{a^6\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.32, size = 65, normalized size = 1.03 \[ \frac {a^{6} c^{3} x + 4 a^{5} c^{3} \log {\relax (x )} + \frac {- 75 a^{4} c^{3} x^{4} + 25 a^{2} c^{3} x^{2} + 15 a c^{3} x + 3 c^{3}}{15 x^{5}}}{a^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a**6*c**3*x + 4*a**5*c**3*log(x) + (-75*a**4*c**3*x**4 + 25*a**2*c**3*x**2 + 15*a*c**3*x + 3*c**3)/(15*x**5))
/a**6

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