3.11.73 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x (c-a^2 c x^2)^3} \, dx\) [1073]

Optimal. Leaf size=93 \[ \frac {1}{12 c^3 (1-a x)^3}+\frac {1}{4 c^3 (1-a x)^2}+\frac {11}{16 c^3 (1-a x)}+\frac {1}{16 c^3 (1+a x)}+\frac {\log (x)}{c^3}-\frac {13 \log (1-a x)}{16 c^3}-\frac {3 \log (1+a x)}{16 c^3} \]

[Out]

1/12/c^3/(-a*x+1)^3+1/4/c^3/(-a*x+1)^2+11/16/c^3/(-a*x+1)+1/16/c^3/(a*x+1)+ln(x)/c^3-13/16*ln(-a*x+1)/c^3-3/16
*ln(a*x+1)/c^3

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Rubi [A]
time = 0.08, antiderivative size = 93, 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 = {6285, 90} \begin {gather*} \frac {11}{16 c^3 (1-a x)}+\frac {1}{16 c^3 (a x+1)}+\frac {1}{4 c^3 (1-a x)^2}+\frac {1}{12 c^3 (1-a x)^3}-\frac {13 \log (1-a x)}{16 c^3}-\frac {3 \log (a x+1)}{16 c^3}+\frac {\log (x)}{c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(12*c^3*(1 - a*x)^3) + 1/(4*c^3*(1 - a*x)^2) + 11/(16*c^3*(1 - a*x)) + 1/(16*c^3*(1 + a*x)) + Log[x]/c^3 - (
13*Log[1 - a*x])/(16*c^3) - (3*Log[1 + a*x])/(16*c^3)

Rule 90

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 6285

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 \left (c-a^2 c x^2\right )^3} \, dx &=\frac {\int \frac {1}{x (1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac {\int \left (\frac {1}{x}+\frac {a}{4 (-1+a x)^4}-\frac {a}{2 (-1+a x)^3}+\frac {11 a}{16 (-1+a x)^2}-\frac {13 a}{16 (-1+a x)}-\frac {a}{16 (1+a x)^2}-\frac {3 a}{16 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac {1}{12 c^3 (1-a x)^3}+\frac {1}{4 c^3 (1-a x)^2}+\frac {11}{16 c^3 (1-a x)}+\frac {1}{16 c^3 (1+a x)}+\frac {\log (x)}{c^3}-\frac {13 \log (1-a x)}{16 c^3}-\frac {3 \log (1+a x)}{16 c^3}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 66, normalized size = 0.71 \begin {gather*} \frac {\frac {33}{1-a x}-\frac {4}{(-1+a x)^3}+\frac {12}{(-1+a x)^2}+\frac {3}{1+a x}+48 \log (x)-39 \log (1-a x)-9 \log (1+a x)}{48 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(33/(1 - a*x) - 4/(-1 + a*x)^3 + 12/(-1 + a*x)^2 + 3/(1 + a*x) + 48*Log[x] - 39*Log[1 - a*x] - 9*Log[1 + a*x])
/(48*c^3)

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Maple [A]
time = 0.07, size = 60, normalized size = 0.65

method	result	size
default	\(\frac {\frac {1}{16 a x +16}-\frac {3 \ln \left (a x +1\right )}{16}+\ln \left (x \right )-\frac {1}{12 \left (a x -1\right )^{3}}+\frac {1}{4 \left (a x -1\right )^{2}}-\frac {11}{16 \left (a x -1\right )}-\frac {13 \ln \left (a x -1\right )}{16}}{c^{3}}\)	\(60\)
risch	\(\frac {-\frac {5}{8} a^{3} x^{3}+\frac {3}{4} a^{2} x^{2}+\frac {19}{24} a x -\frac {13}{12}}{c^{3} \left (a x -1\right )^{2} \left (a^{2} x^{2}-1\right )}+\frac {\ln \left (-x \right )}{c^{3}}-\frac {13 \ln \left (-a x +1\right )}{16 c^{3}}-\frac {3 \ln \left (a x +1\right )}{16 c^{3}}\)	\(77\)
norman	\(\frac {-\frac {11 a x}{8 c}+\frac {5 a^{3} x^{3}}{3 c}-\frac {5 a^{5} x^{5}}{8 c}-\frac {2 a^{2} x^{2}}{c}-\frac {13 a^{6} x^{6}}{12 c}+\frac {11 a^{4} x^{4}}{4 c}}{\left (a^{2} x^{2}-1\right )^{3} c^{2}}+\frac {\ln \left (x \right )}{c^{3}}-\frac {13 \ln \left (a x -1\right )}{16 c^{3}}-\frac {3 \ln \left (a x +1\right )}{16 c^{3}}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^3*(1/16/(a*x+1)-3/16*ln(a*x+1)+ln(x)-1/12/(a*x-1)^3+1/4/(a*x-1)^2-11/16/(a*x-1)-13/16*ln(a*x-1))

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Maxima [A]
time = 0.27, size = 89, normalized size = 0.96 \begin {gather*} -\frac {15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \, {\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\right )}} - \frac {3 \, \log \left (a x + 1\right )}{16 \, c^{3}} - \frac {13 \, \log \left (a x - 1\right )}{16 \, c^{3}} + \frac {\log \left (x\right )}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(15*a^3*x^3 - 18*a^2*x^2 - 19*a*x + 26)/(a^4*c^3*x^4 - 2*a^3*c^3*x^3 + 2*a*c^3*x - c^3) - 3/16*log(a*x +
1)/c^3 - 13/16*log(a*x - 1)/c^3 + log(x)/c^3

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Fricas [A]
time = 0.34, size = 143, normalized size = 1.54 \begin {gather*} -\frac {30 \, a^{3} x^{3} - 36 \, a^{2} x^{2} - 38 \, a x + 9 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 39 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 48 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (x\right ) + 52}{48 \, {\left (a^{4} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{3} + 2 \, a c^{3} x - c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(30*a^3*x^3 - 36*a^2*x^2 - 38*a*x + 9*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x + 1) + 39*(a^4*x^4 - 2*a
^3*x^3 + 2*a*x - 1)*log(a*x - 1) - 48*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(x) + 52)/(a^4*c^3*x^4 - 2*a^3*c^3*
x^3 + 2*a*c^3*x - c^3)

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Sympy [A]
time = 0.36, size = 87, normalized size = 0.94 \begin {gather*} \frac {- 15 a^{3} x^{3} + 18 a^{2} x^{2} + 19 a x - 26}{24 a^{4} c^{3} x^{4} - 48 a^{3} c^{3} x^{3} + 48 a c^{3} x - 24 c^{3}} + \frac {\log {\left (x \right )} - \frac {13 \log {\left (x - \frac {1}{a} \right )}}{16} - \frac {3 \log {\left (x + \frac {1}{a} \right )}}{16}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-15*a**3*x**3 + 18*a**2*x**2 + 19*a*x - 26)/(24*a**4*c**3*x**4 - 48*a**3*c**3*x**3 + 48*a*c**3*x - 24*c**3) +
(log(x) - 13*log(x - 1/a)/16 - 3*log(x + 1/a)/16)/c**3

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Giac [A]
time = 0.43, size = 73, normalized size = 0.78 \begin {gather*} -\frac {3 \, \log \left ({\left | a x + 1 \right |}\right )}{16 \, c^{3}} - \frac {13 \, \log \left ({\left | a x - 1 \right |}\right )}{16 \, c^{3}} + \frac {\log \left ({\left | x \right |}\right )}{c^{3}} - \frac {15 \, a^{3} x^{3} - 18 \, a^{2} x^{2} - 19 \, a x + 26}{24 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*log(abs(a*x + 1))/c^3 - 13/16*log(abs(a*x - 1))/c^3 + log(abs(x))/c^3 - 1/24*(15*a^3*x^3 - 18*a^2*x^2 -
19*a*x + 26)/((a*x + 1)*(a*x - 1)^3*c^3)

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Mupad [B]
time = 0.95, size = 88, normalized size = 0.95 \begin {gather*} \frac {\ln \left (x\right )}{c^3}-\frac {13\,\ln \left (a\,x-1\right )}{16\,c^3}-\frac {3\,\ln \left (a\,x+1\right )}{16\,c^3}-\frac {-\frac {5\,a^3\,x^3}{8}+\frac {3\,a^2\,x^2}{4}+\frac {19\,a\,x}{24}-\frac {13}{12}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x)/c^3 - (13*log(a*x - 1))/(16*c^3) - (3*log(a*x + 1))/(16*c^3) - ((19*a*x)/24 + (3*a^2*x^2)/4 - (5*a^3*x^
3)/8 - 13/12)/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x)

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