3.1066 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^4 (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=110 \[ \frac {9 a^3}{4 c^2 (1-a x)}+\frac {a^3}{4 c^2 (1-a x)^2}+\frac {6 a^3 \log (x)}{c^2}-\frac {49 a^3 \log (1-a x)}{8 c^2}+\frac {a^3 \log (a x+1)}{8 c^2}-\frac {4 a^2}{c^2 x}-\frac {a}{c^2 x^2}-\frac {1}{3 c^2 x^3} \]

[Out]

-1/3/c^2/x^3-a/c^2/x^2-4*a^2/c^2/x+1/4*a^3/c^2/(-a*x+1)^2+9/4*a^3/c^2/(-a*x+1)+6*a^3*ln(x)/c^2-49/8*a^3*ln(-a*
x+1)/c^2+1/8*a^3*ln(a*x+1)/c^2

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Rubi [A]  time = 0.13, antiderivative size = 110, 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} \[ \frac {9 a^3}{4 c^2 (1-a x)}+\frac {a^3}{4 c^2 (1-a x)^2}-\frac {4 a^2}{c^2 x}+\frac {6 a^3 \log (x)}{c^2}-\frac {49 a^3 \log (1-a x)}{8 c^2}+\frac {a^3 \log (a x+1)}{8 c^2}-\frac {a}{c^2 x^2}-\frac {1}{3 c^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(3*c^2*x^3) - a/(c^2*x^2) - (4*a^2)/(c^2*x) + a^3/(4*c^2*(1 - a*x)^2) + (9*a^3)/(4*c^2*(1 - a*x)) + (6*a^3*
Log[x])/c^2 - (49*a^3*Log[1 - a*x])/(8*c^2) + (a^3*Log[1 + a*x])/(8*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])

Rubi steps

\begin {align*} \int \frac {e^{2 \tanh ^{-1}(a x)}}{x^4 \left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {1}{x^4 (1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac {\int \left (\frac {1}{x^4}+\frac {2 a}{x^3}+\frac {4 a^2}{x^2}+\frac {6 a^3}{x}-\frac {a^4}{2 (-1+a x)^3}+\frac {9 a^4}{4 (-1+a x)^2}-\frac {49 a^4}{8 (-1+a x)}+\frac {a^4}{8 (1+a x)}\right ) \, dx}{c^2}\\ &=-\frac {1}{3 c^2 x^3}-\frac {a}{c^2 x^2}-\frac {4 a^2}{c^2 x}+\frac {a^3}{4 c^2 (1-a x)^2}+\frac {9 a^3}{4 c^2 (1-a x)}+\frac {6 a^3 \log (x)}{c^2}-\frac {49 a^3 \log (1-a x)}{8 c^2}+\frac {a^3 \log (1+a x)}{8 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 87, normalized size = 0.79 \[ \frac {\frac {9 a^3}{4-4 a x}+\frac {a^3}{4 (a x-1)^2}+6 a^3 \log (x)-\frac {49}{8} a^3 \log (1-a x)+\frac {1}{8} a^3 \log (a x+1)-\frac {4 a^2}{x}-\frac {a}{x^2}-\frac {1}{3 x^3}}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1/3*1/x^3 - a/x^2 - (4*a^2)/x + (9*a^3)/(4 - 4*a*x) + a^3/(4*(-1 + a*x)^2) + 6*a^3*Log[x] - (49*a^3*Log[1 -
a*x])/8 + (a^3*Log[1 + a*x])/8)/c^2

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fricas [A]  time = 0.63, size = 150, normalized size = 1.36 \[ -\frac {150 \, a^{4} x^{4} - 228 \, a^{3} x^{3} + 56 \, a^{2} x^{2} + 8 \, a x - 3 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x + 1\right ) + 147 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \left (a x - 1\right ) - 144 \, {\left (a^{5} x^{5} - 2 \, a^{4} x^{4} + a^{3} x^{3}\right )} \log \relax (x) + 8}{24 \, {\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="fricas")

[Out]

-1/24*(150*a^4*x^4 - 228*a^3*x^3 + 56*a^2*x^2 + 8*a*x - 3*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(a*x + 1) + 147*(
a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(a*x - 1) - 144*(a^5*x^5 - 2*a^4*x^4 + a^3*x^3)*log(x) + 8)/(a^2*c^2*x^5 - 2
*a*c^2*x^4 + c^2*x^3)

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giac [A]  time = 0.16, size = 87, normalized size = 0.79 \[ \frac {a^{3} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {49 \, a^{3} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {6 \, a^{3} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \, {\left (a x - 1\right )}^{2} c^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="giac")

[Out]

1/8*a^3*log(abs(a*x + 1))/c^2 - 49/8*a^3*log(abs(a*x - 1))/c^2 + 6*a^3*log(abs(x))/c^2 - 1/12*(75*a^4*x^4 - 11
4*a^3*x^3 + 28*a^2*x^2 + 4*a*x + 4)/((a*x - 1)^2*c^2*x^3)

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maple [A]  time = 0.04, size = 98, normalized size = 0.89 \[ -\frac {1}{3 c^{2} x^{3}}-\frac {a}{c^{2} x^{2}}-\frac {4 a^{2}}{c^{2} x}+\frac {6 a^{3} \ln \relax (x )}{c^{2}}+\frac {a^{3}}{4 c^{2} \left (a x -1\right )^{2}}-\frac {9 a^{3}}{4 c^{2} \left (a x -1\right )}-\frac {49 a^{3} \ln \left (a x -1\right )}{8 c^{2}}+\frac {a^{3} \ln \left (a x +1\right )}{8 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/c^2/x^3-a/c^2/x^2-4*a^2/c^2/x+6*a^3*ln(x)/c^2+1/4/c^2*a^3/(a*x-1)^2-9/4/c^2*a^3/(a*x-1)-49/8/c^2*a^3*ln(a
*x-1)+1/8*a^3*ln(a*x+1)/c^2

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maxima [A]  time = 0.31, size = 100, normalized size = 0.91 \[ \frac {a^{3} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac {49 \, a^{3} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {6 \, a^{3} \log \relax (x)}{c^{2}} - \frac {75 \, a^{4} x^{4} - 114 \, a^{3} x^{3} + 28 \, a^{2} x^{2} + 4 \, a x + 4}{12 \, {\left (a^{2} c^{2} x^{5} - 2 \, a c^{2} x^{4} + c^{2} x^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)^2/(-a^2*x^2+1)/x^4/(-a^2*c*x^2+c)^2,x, algorithm="maxima")

[Out]

1/8*a^3*log(a*x + 1)/c^2 - 49/8*a^3*log(a*x - 1)/c^2 + 6*a^3*log(x)/c^2 - 1/12*(75*a^4*x^4 - 114*a^3*x^3 + 28*
a^2*x^2 + 4*a*x + 4)/(a^2*c^2*x^5 - 2*a*c^2*x^4 + c^2*x^3)

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mupad [B]  time = 0.99, size = 100, normalized size = 0.91 \[ \frac {6\,a^3\,\ln \relax (x)}{c^2}-\frac {\frac {25\,a^4\,x^4}{4}-\frac {19\,a^3\,x^3}{2}+\frac {7\,a^2\,x^2}{3}+\frac {a\,x}{3}+\frac {1}{3}}{a^2\,c^2\,x^5-2\,a\,c^2\,x^4+c^2\,x^3}-\frac {49\,a^3\,\ln \left (a\,x-1\right )}{8\,c^2}+\frac {a^3\,\ln \left (a\,x+1\right )}{8\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(6*a^3*log(x))/c^2 - ((a*x)/3 + (7*a^2*x^2)/3 - (19*a^3*x^3)/2 + (25*a^4*x^4)/4 + 1/3)/(c^2*x^3 - 2*a*c^2*x^4
+ a^2*c^2*x^5) - (49*a^3*log(a*x - 1))/(8*c^2) + (a^3*log(a*x + 1))/(8*c^2)

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sympy [A]  time = 0.65, size = 100, normalized size = 0.91 \[ - \frac {75 a^{4} x^{4} - 114 a^{3} x^{3} + 28 a^{2} x^{2} + 4 a x + 4}{12 a^{2} c^{2} x^{5} - 24 a c^{2} x^{4} + 12 c^{2} x^{3}} - \frac {- 6 a^{3} \log {\relax (x )} + \frac {49 a^{3} \log {\left (x - \frac {1}{a} \right )}}{8} - \frac {a^{3} \log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(75*a**4*x**4 - 114*a**3*x**3 + 28*a**2*x**2 + 4*a*x + 4)/(12*a**2*c**2*x**5 - 24*a*c**2*x**4 + 12*c**2*x**3)
- (-6*a**3*log(x) + 49*a**3*log(x - 1/a)/8 - a**3*log(x + 1/a)/8)/c**2

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