3.1074 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^2 (c-a^2 c x^2)^3} \, dx\)

Optimal. Leaf size=109 \[ \frac {23 a}{16 c^3 (1-a x)}-\frac {a}{16 c^3 (a x+1)}+\frac {3 a}{8 c^3 (1-a x)^2}+\frac {a}{12 c^3 (1-a x)^3}+\frac {2 a \log (x)}{c^3}-\frac {9 a \log (1-a x)}{4 c^3}+\frac {a \log (a x+1)}{4 c^3}-\frac {1}{c^3 x} \]

[Out]

-1/c^3/x+1/12*a/c^3/(-a*x+1)^3+3/8*a/c^3/(-a*x+1)^2+23/16*a/c^3/(-a*x+1)-1/16*a/c^3/(a*x+1)+2*a*ln(x)/c^3-9/4*
a*ln(-a*x+1)/c^3+1/4*a*ln(a*x+1)/c^3

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Rubi [A]  time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {6150, 88} \[ \frac {23 a}{16 c^3 (1-a x)}-\frac {a}{16 c^3 (a x+1)}+\frac {3 a}{8 c^3 (1-a x)^2}+\frac {a}{12 c^3 (1-a x)^3}+\frac {2 a \log (x)}{c^3}-\frac {9 a \log (1-a x)}{4 c^3}+\frac {a \log (a x+1)}{4 c^3}-\frac {1}{c^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(c^3*x)) + a/(12*c^3*(1 - a*x)^3) + (3*a)/(8*c^3*(1 - a*x)^2) + (23*a)/(16*c^3*(1 - a*x)) - a/(16*c^3*(1 +
a*x)) + (2*a*Log[x])/c^3 - (9*a*Log[1 - a*x])/(4*c^3) + (a*Log[1 + a*x])/(4*c^3)

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])

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)}}{x^2 \left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {1}{x^2 (1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac {\int \left (\frac {1}{x^2}+\frac {2 a}{x}+\frac {a^2}{4 (-1+a x)^4}-\frac {3 a^2}{4 (-1+a x)^3}+\frac {23 a^2}{16 (-1+a x)^2}-\frac {9 a^2}{4 (-1+a x)}+\frac {a^2}{16 (1+a x)^2}+\frac {a^2}{4 (1+a x)}\right ) \, dx}{c^3}\\ &=-\frac {1}{c^3 x}+\frac {a}{12 c^3 (1-a x)^3}+\frac {3 a}{8 c^3 (1-a x)^2}+\frac {23 a}{16 c^3 (1-a x)}-\frac {a}{16 c^3 (1+a x)}+\frac {2 a \log (x)}{c^3}-\frac {9 a \log (1-a x)}{4 c^3}+\frac {a \log (1+a x)}{4 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 78, normalized size = 0.72 \[ \frac {\frac {69 a}{1-a x}-\frac {3 a}{a x+1}+\frac {18 a}{(a x-1)^2}-\frac {4 a}{(a x-1)^3}+96 a \log (x)-108 a \log (1-a x)+12 a \log (a x+1)-\frac {48}{x}}{48 c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-48/x + (69*a)/(1 - a*x) - (4*a)/(-1 + a*x)^3 + (18*a)/(-1 + a*x)^2 - (3*a)/(1 + a*x) + 96*a*Log[x] - 108*a*L
og[1 - a*x] + 12*a*Log[1 + a*x])/(48*c^3)

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fricas [A]  time = 1.09, size = 175, normalized size = 1.61 \[ -\frac {30 \, a^{4} x^{4} - 48 \, a^{3} x^{3} - 14 \, a^{2} x^{2} + 46 \, a x - 3 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + 2 \, a^{2} x^{2} - a x\right )} \log \left (a x + 1\right ) + 27 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + 2 \, a^{2} x^{2} - a x\right )} \log \left (a x - 1\right ) - 24 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + 2 \, a^{2} x^{2} - a x\right )} \log \relax (x) - 12}{12 \, {\left (a^{4} c^{3} x^{5} - 2 \, a^{3} c^{3} x^{4} + 2 \, a c^{3} x^{2} - c^{3} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(30*a^4*x^4 - 48*a^3*x^3 - 14*a^2*x^2 + 46*a*x - 3*(a^5*x^5 - 2*a^4*x^4 + 2*a^2*x^2 - a*x)*log(a*x + 1)
+ 27*(a^5*x^5 - 2*a^4*x^4 + 2*a^2*x^2 - a*x)*log(a*x - 1) - 24*(a^5*x^5 - 2*a^4*x^4 + 2*a^2*x^2 - a*x)*log(x)
- 12)/(a^4*c^3*x^5 - 2*a^3*c^3*x^4 + 2*a*c^3*x^2 - c^3*x)

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giac [A]  time = 0.19, size = 88, normalized size = 0.81 \[ \frac {a \log \left ({\left | a x + 1 \right |}\right )}{4 \, c^{3}} - \frac {9 \, a \log \left ({\left | a x - 1 \right |}\right )}{4 \, c^{3}} + \frac {2 \, a \log \left ({\left | x \right |}\right )}{c^{3}} - \frac {15 \, a^{4} x^{4} - 24 \, a^{3} x^{3} - 7 \, a^{2} x^{2} + 23 \, a x - 6}{6 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} c^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*log(abs(a*x + 1))/c^3 - 9/4*a*log(abs(a*x - 1))/c^3 + 2*a*log(abs(x))/c^3 - 1/6*(15*a^4*x^4 - 24*a^3*x^3
- 7*a^2*x^2 + 23*a*x - 6)/((a*x + 1)*(a*x - 1)^3*c^3*x)

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maple [A]  time = 0.04, size = 94, normalized size = 0.86 \[ -\frac {1}{c^{3} x}+\frac {2 a \ln \relax (x )}{c^{3}}-\frac {a}{12 c^{3} \left (a x -1\right )^{3}}+\frac {3 a}{8 c^{3} \left (a x -1\right )^{2}}-\frac {23 a}{16 c^{3} \left (a x -1\right )}-\frac {9 a \ln \left (a x -1\right )}{4 c^{3}}-\frac {a}{16 c^{3} \left (a x +1\right )}+\frac {a \ln \left (a x +1\right )}{4 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/c^3/x+2*a*ln(x)/c^3-1/12/c^3*a/(a*x-1)^3+3/8/c^3*a/(a*x-1)^2-23/16/c^3*a/(a*x-1)-9/4/c^3*a*ln(a*x-1)-1/16*a
/c^3/(a*x+1)+1/4*a*ln(a*x+1)/c^3

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maxima [A]  time = 0.31, size = 104, normalized size = 0.95 \[ -\frac {15 \, a^{4} x^{4} - 24 \, a^{3} x^{3} - 7 \, a^{2} x^{2} + 23 \, a x - 6}{6 \, {\left (a^{4} c^{3} x^{5} - 2 \, a^{3} c^{3} x^{4} + 2 \, a c^{3} x^{2} - c^{3} x\right )}} + \frac {a \log \left (a x + 1\right )}{4 \, c^{3}} - \frac {9 \, a \log \left (a x - 1\right )}{4 \, c^{3}} + \frac {2 \, a \log \relax (x)}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*a^4*x^4 - 24*a^3*x^3 - 7*a^2*x^2 + 23*a*x - 6)/(a^4*c^3*x^5 - 2*a^3*c^3*x^4 + 2*a*c^3*x^2 - c^3*x) +
1/4*a*log(a*x + 1)/c^3 - 9/4*a*log(a*x - 1)/c^3 + 2*a*log(x)/c^3

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mupad [B]  time = 0.99, size = 104, normalized size = 0.95 \[ \frac {2\,a\,\ln \relax (x)}{c^3}-\frac {-\frac {5\,a^4\,x^4}{2}+4\,a^3\,x^3+\frac {7\,a^2\,x^2}{6}-\frac {23\,a\,x}{6}+1}{-a^4\,c^3\,x^5+2\,a^3\,c^3\,x^4-2\,a\,c^3\,x^2+c^3\,x}-\frac {9\,a\,\ln \left (a\,x-1\right )}{4\,c^3}+\frac {a\,\ln \left (a\,x+1\right )}{4\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*log(x))/c^3 - ((7*a^2*x^2)/6 - (23*a*x)/6 + 4*a^3*x^3 - (5*a^4*x^4)/2 + 1)/(c^3*x - 2*a*c^3*x^2 + 2*a^3*c
^3*x^4 - a^4*c^3*x^5) - (9*a*log(a*x - 1))/(4*c^3) + (a*log(a*x + 1))/(4*c^3)

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sympy [A]  time = 0.74, size = 104, normalized size = 0.95 \[ \frac {- 15 a^{4} x^{4} + 24 a^{3} x^{3} + 7 a^{2} x^{2} - 23 a x + 6}{6 a^{4} c^{3} x^{5} - 12 a^{3} c^{3} x^{4} + 12 a c^{3} x^{2} - 6 c^{3} x} + \frac {2 a \log {\relax (x )} - \frac {9 a \log {\left (x - \frac {1}{a} \right )}}{4} + \frac {a \log {\left (x + \frac {1}{a} \right )}}{4}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-15*a**4*x**4 + 24*a**3*x**3 + 7*a**2*x**2 - 23*a*x + 6)/(6*a**4*c**3*x**5 - 12*a**3*c**3*x**4 + 12*a*c**3*x*
*2 - 6*c**3*x) + (2*a*log(x) - 9*a*log(x - 1/a)/4 + a*log(x + 1/a)/4)/c**3

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