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

Optimal. Leaf size=76 \[ -\frac {7}{4 a c^2 (1-a x)}+\frac {1}{4 a c^2 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a c^2}+\frac {\log (a x+1)}{8 a c^2}-\frac {x}{c^2} \]

[Out]

-x/c^2+1/4/a/c^2/(-a*x+1)^2-7/4/a/c^2/(-a*x+1)-17/8*ln(-a*x+1)/a/c^2+1/8*ln(a*x+1)/a/c^2

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Rubi [A]  time = 0.15, antiderivative size = 76, 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} \[ -\frac {7}{4 a c^2 (1-a x)}+\frac {1}{4 a c^2 (1-a x)^2}-\frac {17 \log (1-a x)}{8 a c^2}+\frac {\log (a x+1)}{8 a c^2}-\frac {x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/c^2) + 1/(4*a*c^2*(1 - a*x)^2) - 7/(4*a*c^2*(1 - a*x)) - (17*Log[1 - a*x])/(8*a*c^2) + Log[1 + a*x]/(8*a*c
^2)

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 \frac {e^{2 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^2} \, dx &=\frac {a^4 \int \frac {e^{2 \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)^3 (1+a x)} \, dx}{c^2}\\ &=\frac {a^4 \int \left (-\frac {1}{a^4}-\frac {1}{2 a^4 (-1+a x)^3}-\frac {7}{4 a^4 (-1+a x)^2}-\frac {17}{8 a^4 (-1+a x)}+\frac {1}{8 a^4 (1+a x)}\right ) \, dx}{c^2}\\ &=-\frac {x}{c^2}+\frac {1}{4 a c^2 (1-a x)^2}-\frac {7}{4 a c^2 (1-a x)}-\frac {17 \log (1-a x)}{8 a c^2}+\frac {\log (1+a x)}{8 a c^2}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.91 \[ \frac {-8 a^3 x^3+16 a^2 x^2+6 a x-17 (a x-1)^2 \log (1-a x)+(a x-1)^2 \log (a x+1)-12}{8 a c^2 (a x-1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-12 + 6*a*x + 16*a^2*x^2 - 8*a^3*x^3 - 17*(-1 + a*x)^2*Log[1 - a*x] + (-1 + a*x)^2*Log[1 + a*x])/(8*a*c^2*(-1
+ a*x)^2)

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fricas [A]  time = 0.49, size = 93, normalized size = 1.22 \[ -\frac {8 \, a^{3} x^{3} - 16 \, a^{2} x^{2} - 6 \, a x - {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x + 1\right ) + 17 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 12}{8 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} \]

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)^2,x, algorithm="fricas")

[Out]

-1/8*(8*a^3*x^3 - 16*a^2*x^2 - 6*a*x - (a^2*x^2 - 2*a*x + 1)*log(a*x + 1) + 17*(a^2*x^2 - 2*a*x + 1)*log(a*x -
1) + 12)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2)

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giac [A]  time = 0.18, size = 58, normalized size = 0.76 \[ -\frac {x}{c^{2}} + \frac {\log \left ({\left | a x + 1 \right |}\right )}{8 \, a c^{2}} - \frac {17 \, \log \left ({\left | a x - 1 \right |}\right )}{8 \, a c^{2}} + \frac {7 \, a x - 6}{4 \, {\left (a x - 1\right )}^{2} a c^{2}} \]

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)^2,x, algorithm="giac")

[Out]

-x/c^2 + 1/8*log(abs(a*x + 1))/(a*c^2) - 17/8*log(abs(a*x - 1))/(a*c^2) + 1/4*(7*a*x - 6)/((a*x - 1)^2*a*c^2)

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maple [A]  time = 0.04, size = 66, normalized size = 0.87 \[ -\frac {x}{c^{2}}+\frac {1}{4 a \,c^{2} \left (a x -1\right )^{2}}+\frac {7}{4 a \,c^{2} \left (a x -1\right )}-\frac {17 \ln \left (a x -1\right )}{8 a \,c^{2}}+\frac {\ln \left (a x +1\right )}{8 a \,c^{2}} \]

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)^2,x)

[Out]

-x/c^2+1/4/a/c^2/(a*x-1)^2+7/4/a/c^2/(a*x-1)-17/8/a/c^2*ln(a*x-1)+1/8*ln(a*x+1)/a/c^2

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maxima [A]  time = 0.31, size = 70, normalized size = 0.92 \[ \frac {7 \, a x - 6}{4 \, {\left (a^{3} c^{2} x^{2} - 2 \, a^{2} c^{2} x + a c^{2}\right )}} - \frac {x}{c^{2}} + \frac {\log \left (a x + 1\right )}{8 \, a c^{2}} - \frac {17 \, \log \left (a x - 1\right )}{8 \, a c^{2}} \]

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)^2,x, algorithm="maxima")

[Out]

1/4*(7*a*x - 6)/(a^3*c^2*x^2 - 2*a^2*c^2*x + a*c^2) - x/c^2 + 1/8*log(a*x + 1)/(a*c^2) - 17/8*log(a*x - 1)/(a*
c^2)

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mupad [B]  time = 0.10, size = 68, normalized size = 0.89 \[ \frac {\frac {7\,x}{4}-\frac {3}{2\,a}}{a^2\,c^2\,x^2-2\,a\,c^2\,x+c^2}-\frac {x}{c^2}-\frac {17\,\ln \left (a\,x-1\right )}{8\,a\,c^2}+\frac {\ln \left (a\,x+1\right )}{8\,a\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((7*x)/4 - 3/(2*a))/(c^2 + a^2*c^2*x^2 - 2*a*c^2*x) - x/c^2 - (17*log(a*x - 1))/(8*a*c^2) + log(a*x + 1)/(8*a*
c^2)

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sympy [A]  time = 0.41, size = 75, normalized size = 0.99 \[ - a^{4} \left (\frac {- 7 a x + 6}{4 a^{7} c^{2} x^{2} - 8 a^{6} c^{2} x + 4 a^{5} c^{2}} + \frac {x}{a^{4} c^{2}} + \frac {\frac {17 \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {\log {\left (x + \frac {1}{a} \right )}}{8}}{a^{5} c^{2}}\right ) \]

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)**2,x)

[Out]

-a**4*((-7*a*x + 6)/(4*a**7*c**2*x**2 - 8*a**6*c**2*x + 4*a**5*c**2) + x/(a**4*c**2) + (17*log(x - 1/a)/8 - lo
g(x + 1/a)/8)/(a**5*c**2))

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