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

Optimal. Leaf size=110 \[ \frac {39}{16 a c^3 (1-a x)}-\frac {1}{16 a c^3 (a x+1)}-\frac {5}{8 a c^3 (1-a x)^2}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {9 \log (1-a x)}{4 a c^3}-\frac {\log (a x+1)}{4 a c^3}+\frac {x}{c^3} \]

[Out]

x/c^3+1/12/a/c^3/(-a*x+1)^3-5/8/a/c^3/(-a*x+1)^2+39/16/a/c^3/(-a*x+1)-1/16/a/c^3/(a*x+1)+9/4*ln(-a*x+1)/a/c^3-
1/4*ln(a*x+1)/a/c^3

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Rubi [A]  time = 0.20, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6167, 6157, 6150, 88} \[ \frac {39}{16 a c^3 (1-a x)}-\frac {1}{16 a c^3 (a x+1)}-\frac {5}{8 a c^3 (1-a x)^2}+\frac {1}{12 a c^3 (1-a x)^3}+\frac {9 \log (1-a x)}{4 a c^3}-\frac {\log (a x+1)}{4 a c^3}+\frac {x}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/c^3 + 1/(12*a*c^3*(1 - a*x)^3) - 5/(8*a*c^3*(1 - a*x)^2) + 39/(16*a*c^3*(1 - a*x)) - 1/(16*a*c^3*(1 + a*x))
+ (9*Log[1 - a*x])/(4*a*c^3) - Log[1 + a*x]/(4*a*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])

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]

Rule 6167

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps

\begin {align*} \int \frac {e^{2 \coth ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx &=-\int \frac {e^{2 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a^2 x^2}\right )^3} \, dx\\ &=\frac {a^6 \int \frac {e^{2 \tanh ^{-1}(a x)} x^6}{\left (1-a^2 x^2\right )^3} \, dx}{c^3}\\ &=\frac {a^6 \int \frac {x^6}{(1-a x)^4 (1+a x)^2} \, dx}{c^3}\\ &=\frac {a^6 \int \left (\frac {1}{a^6}+\frac {1}{4 a^6 (-1+a x)^4}+\frac {5}{4 a^6 (-1+a x)^3}+\frac {39}{16 a^6 (-1+a x)^2}+\frac {9}{4 a^6 (-1+a x)}+\frac {1}{16 a^6 (1+a x)^2}-\frac {1}{4 a^6 (1+a x)}\right ) \, dx}{c^3}\\ &=\frac {x}{c^3}+\frac {1}{12 a c^3 (1-a x)^3}-\frac {5}{8 a c^3 (1-a x)^2}+\frac {39}{16 a c^3 (1-a x)}-\frac {1}{16 a c^3 (1+a x)}+\frac {9 \log (1-a x)}{4 a c^3}-\frac {\log (1+a x)}{4 a c^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 82, normalized size = 0.75 \[ \frac {\frac {2 \left (6 a^5 x^5-12 a^4 x^4-15 a^3 x^3+24 a^2 x^2+7 a x-11\right )}{(a x-1)^3 (a x+1)}+27 \log (1-a x)-3 \log (a x+1)}{12 a c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(-11 + 7*a*x + 24*a^2*x^2 - 15*a^3*x^3 - 12*a^4*x^4 + 6*a^5*x^5))/((-1 + a*x)^3*(1 + a*x)) + 27*Log[1 - a*
x] - 3*Log[1 + a*x])/(12*a*c^3)

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fricas [A]  time = 0.46, size = 137, normalized size = 1.25 \[ \frac {12 \, a^{5} x^{5} - 24 \, a^{4} x^{4} - 30 \, a^{3} x^{3} + 48 \, a^{2} x^{2} + 14 \, a x - 3 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x + 1\right ) + 27 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + 2 \, a x - 1\right )} \log \left (a x - 1\right ) - 22}{12 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*a^5*x^5 - 24*a^4*x^4 - 30*a^3*x^3 + 48*a^2*x^2 + 14*a*x - 3*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x
+ 1) + 27*(a^4*x^4 - 2*a^3*x^3 + 2*a*x - 1)*log(a*x - 1) - 22)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a
*c^3)

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giac [A]  time = 0.13, size = 80, normalized size = 0.73 \[ \frac {x}{c^{3}} - \frac {\log \left ({\left | a x + 1 \right |}\right )}{4 \, a c^{3}} + \frac {9 \, \log \left ({\left | a x - 1 \right |}\right )}{4 \, a c^{3}} - \frac {15 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 13 \, a x + 11}{6 \, {\left (a x + 1\right )} {\left (a x - 1\right )}^{3} a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^3 - 1/4*log(abs(a*x + 1))/(a*c^3) + 9/4*log(abs(a*x - 1))/(a*c^3) - 1/6*(15*a^3*x^3 - 12*a^2*x^2 - 13*a*x
+ 11)/((a*x + 1)*(a*x - 1)^3*a*c^3)

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maple [A]  time = 0.05, size = 95, normalized size = 0.86 \[ \frac {x}{c^{3}}-\frac {1}{12 c^{3} a \left (a x -1\right )^{3}}-\frac {5}{8 c^{3} a \left (a x -1\right )^{2}}-\frac {39}{16 a \,c^{3} \left (a x -1\right )}+\frac {9 \ln \left (a x -1\right )}{4 c^{3} a}-\frac {1}{16 a \,c^{3} \left (a x +1\right )}-\frac {\ln \left (a x +1\right )}{4 a \,c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^3-1/12/c^3/a/(a*x-1)^3-5/8/c^3/a/(a*x-1)^2-39/16/a/c^3/(a*x-1)+9/4/c^3/a*ln(a*x-1)-1/16/a/c^3/(a*x+1)-1/4*
ln(a*x+1)/a/c^3

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maxima [A]  time = 0.31, size = 97, normalized size = 0.88 \[ -\frac {15 \, a^{3} x^{3} - 12 \, a^{2} x^{2} - 13 \, a x + 11}{6 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{4} c^{3} x^{3} + 2 \, a^{2} c^{3} x - a c^{3}\right )}} + \frac {x}{c^{3}} - \frac {\log \left (a x + 1\right )}{4 \, a c^{3}} + \frac {9 \, \log \left (a x - 1\right )}{4 \, a c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(15*a^3*x^3 - 12*a^2*x^2 - 13*a*x + 11)/(a^5*c^3*x^4 - 2*a^4*c^3*x^3 + 2*a^2*c^3*x - a*c^3) + x/c^3 - 1/4
*log(a*x + 1)/(a*c^3) + 9/4*log(a*x - 1)/(a*c^3)

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mupad [B]  time = 1.28, size = 94, normalized size = 0.85 \[ \frac {x}{c^3}-\frac {\frac {13\,x}{6}+2\,a\,x^2-\frac {11}{6\,a}-\frac {5\,a^2\,x^3}{2}}{-a^4\,c^3\,x^4+2\,a^3\,c^3\,x^3-2\,a\,c^3\,x+c^3}+\frac {9\,\ln \left (a\,x-1\right )}{4\,a\,c^3}-\frac {\ln \left (a\,x+1\right )}{4\,a\,c^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^3 - ((13*x)/6 + 2*a*x^2 - 11/(6*a) - (5*a^2*x^3)/2)/(c^3 + 2*a^3*c^3*x^3 - a^4*c^3*x^4 - 2*a*c^3*x) + (9*l
og(a*x - 1))/(4*a*c^3) - log(a*x + 1)/(4*a*c^3)

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sympy [A]  time = 0.59, size = 102, normalized size = 0.93 \[ a^{6} \left (\frac {- 15 a^{3} x^{3} + 12 a^{2} x^{2} + 13 a x - 11}{6 a^{11} c^{3} x^{4} - 12 a^{10} c^{3} x^{3} + 12 a^{8} c^{3} x - 6 a^{7} c^{3}} + \frac {x}{a^{6} c^{3}} + \frac {\frac {9 \log {\left (x - \frac {1}{a} \right )}}{4} - \frac {\log {\left (x + \frac {1}{a} \right )}}{4}}{a^{7} c^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**6*((-15*a**3*x**3 + 12*a**2*x**2 + 13*a*x - 11)/(6*a**11*c**3*x**4 - 12*a**10*c**3*x**3 + 12*a**8*c**3*x -
6*a**7*c**3) + x/(a**6*c**3) + (9*log(x - 1/a)/4 - log(x + 1/a)/4)/(a**7*c**3))

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