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

Optimal. Leaf size=51 \[ -\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}+c^2 x \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 51, 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, 43} \[ -\frac {2 c^2}{a^3 x^2}-\frac {c^2}{3 a^4 x^3}-\frac {6 c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 )^2 \, dx &=\frac {c^2 \int \frac {e^{4 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^2}{x^4} \, dx}{a^4}\\ &=\frac {c^2 \int \frac {(1+a x)^4}{x^4} \, dx}{a^4}\\ &=\frac {c^2 \int \left (a^4+\frac {1}{x^4}+\frac {4 a}{x^3}+\frac {6 a^2}{x^2}+\frac {4 a^3}{x}\right ) \, dx}{a^4}\\ &=-\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+c^2 x+\frac {4 c^2 \log (x)}{a}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 51, normalized size = 1.00 \[ -\frac {c^2}{3 a^4 x^3}-\frac {2 c^2}{a^3 x^2}-\frac {6 c^2}{a^2 x}+\frac {4 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 56, normalized size = 1.10 \[ \frac {3 \, a^{4} c^{2} x^{4} + 12 \, a^{3} c^{2} x^{3} \log \relax (x) - 18 \, a^{2} c^{2} x^{2} - 6 \, a c^{2} x - c^{2}}{3 \, a^{4} x^{3}} \]

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)^2,x, algorithm="fricas")

[Out]

1/3*(3*a^4*c^2*x^4 + 12*a^3*c^2*x^3*log(x) - 18*a^2*c^2*x^2 - 6*a*c^2*x - c^2)/(a^4*x^3)

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giac [A]  time = 0.20, size = 47, normalized size = 0.92 \[ c^{2} x + \frac {4 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac {18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \]

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)^2,x, algorithm="giac")

[Out]

c^2*x + 4*c^2*log(abs(x))/a - 1/3*(18*a^2*c^2*x^2 + 6*a*c^2*x + c^2)/(a^4*x^3)

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maple [A]  time = 0.03, size = 50, normalized size = 0.98 \[ -\frac {c^{2}}{3 a^{4} x^{3}}-\frac {2 c^{2}}{x^{2} a^{3}}-\frac {6 c^{2}}{a^{2} x}+c^{2} x +\frac {4 c^{2} \ln \relax (x )}{a} \]

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)^2,x)

[Out]

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

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maxima [A]  time = 0.30, size = 46, normalized size = 0.90 \[ c^{2} x + \frac {4 \, c^{2} \log \relax (x)}{a} - \frac {18 \, a^{2} c^{2} x^{2} + 6 \, a c^{2} x + c^{2}}{3 \, a^{4} x^{3}} \]

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)^2,x, algorithm="maxima")

[Out]

c^2*x + 4*c^2*log(x)/a - 1/3*(18*a^2*c^2*x^2 + 6*a*c^2*x + c^2)/(a^4*x^3)

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mupad [B]  time = 0.85, size = 43, normalized size = 0.84 \[ -\frac {c^2\,\left (6\,a\,x+18\,a^2\,x^2-3\,a^4\,x^4-12\,a^3\,x^3\,\ln \relax (x)+1\right )}{3\,a^4\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^2*(6*a*x + 18*a^2*x^2 - 3*a^4*x^4 - 12*a^3*x^3*log(x) + 1))/(3*a^4*x^3)

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sympy [A]  time = 0.22, size = 53, normalized size = 1.04 \[ \frac {a^{4} c^{2} x + 4 a^{3} c^{2} \log {\relax (x )} + \frac {- 18 a^{2} c^{2} x^{2} - 6 a c^{2} x - c^{2}}{3 x^{3}}}{a^{4}} \]

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)**2,x)

[Out]

(a**4*c**2*x + 4*a**3*c**2*log(x) + (-18*a**2*c**2*x**2 - 6*a*c**2*x - c**2)/(3*x**3))/a**4

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