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

Optimal. Leaf size=71 \[ \frac {6}{a c^2 (1-a x)}-\frac {2}{a c^2 (1-a x)^2}+\frac {1}{3 a c^2 (1-a x)^3}+\frac {4 \log (1-a x)}{a c^2}+\frac {x}{c^2} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 71, 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} \[ \frac {6}{a c^2 (1-a x)}-\frac {2}{a c^2 (1-a x)^2}+\frac {1}{3 a c^2 (1-a x)^3}+\frac {4 \log (1-a x)}{a c^2}+\frac {x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 63, normalized size = 0.89 \[ \frac {3 a^4 x^4-9 a^3 x^3-9 a^2 x^2+27 a x+12 (a x-1)^3 \log (1-a x)-13}{3 a c^2 (a x-1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13 + 27*a*x - 9*a^2*x^2 - 9*a^3*x^3 + 3*a^4*x^4 + 12*(-1 + a*x)^3*Log[1 - a*x])/(3*a*c^2*(-1 + a*x)^3)

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fricas [A]  time = 0.55, size = 100, normalized size = 1.41 \[ \frac {3 \, a^{4} x^{4} - 9 \, a^{3} x^{3} - 9 \, a^{2} x^{2} + 27 \, a x + 12 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 13}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \]

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*x^4 - 9*a^3*x^3 - 9*a^2*x^2 + 27*a*x + 12*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 13)/(a^4
*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

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giac [A]  time = 0.17, size = 50, normalized size = 0.70 \[ \frac {x}{c^{2}} + \frac {4 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac {18 \, a^{2} x^{2} - 30 \, a x + 13}{3 \, {\left (a x - 1\right )}^{3} a c^{2}} \]

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]

x/c^2 + 4*log(abs(a*x - 1))/(a*c^2) - 1/3*(18*a^2*x^2 - 30*a*x + 13)/((a*x - 1)^3*a*c^2)

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maple [A]  time = 0.03, size = 66, normalized size = 0.93 \[ \frac {x}{c^{2}}-\frac {6}{a \,c^{2} \left (a x -1\right )}-\frac {1}{3 a \,c^{2} \left (a x -1\right )^{3}}+\frac {4 \ln \left (a x -1\right )}{a \,c^{2}}-\frac {2}{a \,c^{2} \left (a x -1\right )^{2}} \]

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]

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

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maxima [A]  time = 0.31, size = 75, normalized size = 1.06 \[ -\frac {18 \, a^{2} x^{2} - 30 \, a x + 13}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} + \frac {x}{c^{2}} + \frac {4 \, \log \left (a x - 1\right )}{a c^{2}} \]

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]

-1/3*(18*a^2*x^2 - 30*a*x + 13)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2) + x/c^2 + 4*log(a*x - 1)/(
a*c^2)

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mupad [B]  time = 0.07, size = 71, normalized size = 1.00 \[ \frac {6\,a\,x^2-10\,x+\frac {13}{3\,a}}{-a^3\,c^2\,x^3+3\,a^2\,c^2\,x^2-3\,a\,c^2\,x+c^2}+\frac {x}{c^2}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*a*x^2 - 10*x + 13/(3*a))/(c^2 + 3*a^2*c^2*x^2 - a^3*c^2*x^3 - 3*a*c^2*x) + x/c^2 + (4*log(a*x - 1))/(a*c^2)

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sympy [A]  time = 0.33, size = 83, normalized size = 1.17 \[ a^{4} \left (\frac {- 18 a^{2} x^{2} + 30 a x - 13}{3 a^{8} c^{2} x^{3} - 9 a^{7} c^{2} x^{2} + 9 a^{6} c^{2} x - 3 a^{5} c^{2}} + \frac {x}{a^{4} c^{2}} + \frac {4 \log {\left (a x - 1 \right )}}{a^{5} c^{2}}\right ) \]

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*((-18*a**2*x**2 + 30*a*x - 13)/(3*a**8*c**2*x**3 - 9*a**7*c**2*x**2 + 9*a**6*c**2*x - 3*a**5*c**2) + x/(a
**4*c**2) + 4*log(a*x - 1)/(a**5*c**2))

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