3.1065 \(\int \frac {e^{2 \tanh ^{-1}(a x)}}{x^3 (c-a^2 c x^2)^2} \, dx\)

Optimal. Leaf size=99 \[ \frac {7 a^2}{4 c^2 (1-a x)}+\frac {a^2}{4 c^2 (1-a x)^2}+\frac {4 a^2 \log (x)}{c^2}-\frac {31 a^2 \log (1-a x)}{8 c^2}-\frac {a^2 \log (a x+1)}{8 c^2}-\frac {2 a}{c^2 x}-\frac {1}{2 c^2 x^2} \]

[Out]

-1/2/c^2/x^2-2*a/c^2/x+1/4*a^2/c^2/(-a*x+1)^2+7/4*a^2/c^2/(-a*x+1)+4*a^2*ln(x)/c^2-31/8*a^2*ln(-a*x+1)/c^2-1/8
*a^2*ln(a*x+1)/c^2

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Rubi [A]  time = 0.12, antiderivative size = 99, 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} \[ \frac {7 a^2}{4 c^2 (1-a x)}+\frac {a^2}{4 c^2 (1-a x)^2}+\frac {4 a^2 \log (x)}{c^2}-\frac {31 a^2 \log (1-a x)}{8 c^2}-\frac {a^2 \log (a x+1)}{8 c^2}-\frac {2 a}{c^2 x}-\frac {1}{2 c^2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*c^2*x^2) - (2*a)/(c^2*x) + a^2/(4*c^2*(1 - a*x)^2) + (7*a^2)/(4*c^2*(1 - a*x)) + (4*a^2*Log[x])/c^2 - (3
1*a^2*Log[1 - a*x])/(8*c^2) - (a^2*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^3 \left (c-a^2 c x^2\right )^2} \, dx &=\frac {\int \frac {1}{x^3 (1-a x)^3 (1+a x)} \, dx}{c^2}\\ &=\frac {\int \left (\frac {1}{x^3}+\frac {2 a}{x^2}+\frac {4 a^2}{x}-\frac {a^3}{2 (-1+a x)^3}+\frac {7 a^3}{4 (-1+a x)^2}-\frac {31 a^3}{8 (-1+a x)}-\frac {a^3}{8 (1+a x)}\right ) \, dx}{c^2}\\ &=-\frac {1}{2 c^2 x^2}-\frac {2 a}{c^2 x}+\frac {a^2}{4 c^2 (1-a x)^2}+\frac {7 a^2}{4 c^2 (1-a x)}+\frac {4 a^2 \log (x)}{c^2}-\frac {31 a^2 \log (1-a x)}{8 c^2}-\frac {a^2 \log (1+a x)}{8 c^2}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 72, normalized size = 0.73 \[ -\frac {\frac {14 a^2}{a x-1}-\frac {2 a^2}{(a x-1)^2}-32 a^2 \log (x)+31 a^2 \log (1-a x)+a^2 \log (a x+1)+\frac {16 a}{x}+\frac {4}{x^2}}{8 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(4/x^2 + (16*a)/x - (2*a^2)/(-1 + a*x)^2 + (14*a^2)/(-1 + a*x) - 32*a^2*Log[x] + 31*a^2*Log[1 - a*x] + a^
2*Log[1 + a*x])/c^2

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fricas [A]  time = 0.67, size = 141, normalized size = 1.42 \[ -\frac {30 \, a^{3} x^{3} - 44 \, a^{2} x^{2} + 8 \, a x + {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x + 1\right ) + 31 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \left (a x - 1\right ) - 32 \, {\left (a^{4} x^{4} - 2 \, a^{3} x^{3} + a^{2} x^{2}\right )} \log \relax (x) + 4}{8 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(30*a^3*x^3 - 44*a^2*x^2 + 8*a*x + (a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(a*x + 1) + 31*(a^4*x^4 - 2*a^3*x^3
+ a^2*x^2)*log(a*x - 1) - 32*(a^4*x^4 - 2*a^3*x^3 + a^2*x^2)*log(x) + 4)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2
)

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giac [A]  time = 1.09, size = 79, normalized size = 0.80 \[ -\frac {a^{2} \log \left ({\left | a x + 1 \right |}\right )}{8 \, c^{2}} - \frac {31 \, a^{2} \log \left ({\left | a x - 1 \right |}\right )}{8 \, c^{2}} + \frac {4 \, a^{2} \log \left ({\left | x \right |}\right )}{c^{2}} - \frac {15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \, {\left (a x - 1\right )}^{2} c^{2} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^2*log(abs(a*x + 1))/c^2 - 31/8*a^2*log(abs(a*x - 1))/c^2 + 4*a^2*log(abs(x))/c^2 - 1/4*(15*a^3*x^3 - 22
*a^2*x^2 + 4*a*x + 2)/((a*x - 1)^2*c^2*x^2)

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maple [A]  time = 0.04, size = 87, normalized size = 0.88 \[ -\frac {1}{2 c^{2} x^{2}}-\frac {2 a}{c^{2} x}+\frac {4 a^{2} \ln \relax (x )}{c^{2}}+\frac {a^{2}}{4 c^{2} \left (a x -1\right )^{2}}-\frac {7 a^{2}}{4 c^{2} \left (a x -1\right )}-\frac {31 a^{2} \ln \left (a x -1\right )}{8 c^{2}}-\frac {a^{2} \ln \left (a x +1\right )}{8 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/c^2/x^2-2*a/c^2/x+4*a^2*ln(x)/c^2+1/4/c^2*a^2/(a*x-1)^2-7/4/c^2*a^2/(a*x-1)-31/8/c^2*a^2*ln(a*x-1)-1/8*a^
2*ln(a*x+1)/c^2

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maxima [A]  time = 0.31, size = 92, normalized size = 0.93 \[ -\frac {a^{2} \log \left (a x + 1\right )}{8 \, c^{2}} - \frac {31 \, a^{2} \log \left (a x - 1\right )}{8 \, c^{2}} + \frac {4 \, a^{2} \log \relax (x)}{c^{2}} - \frac {15 \, a^{3} x^{3} - 22 \, a^{2} x^{2} + 4 \, a x + 2}{4 \, {\left (a^{2} c^{2} x^{4} - 2 \, a c^{2} x^{3} + c^{2} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^2*log(a*x + 1)/c^2 - 31/8*a^2*log(a*x - 1)/c^2 + 4*a^2*log(x)/c^2 - 1/4*(15*a^3*x^3 - 22*a^2*x^2 + 4*a*
x + 2)/(a^2*c^2*x^4 - 2*a*c^2*x^3 + c^2*x^2)

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mupad [B]  time = 0.12, size = 91, normalized size = 0.92 \[ \frac {4\,a^2\,\ln \relax (x)}{c^2}-\frac {\frac {15\,a^3\,x^3}{4}-\frac {11\,a^2\,x^2}{2}+a\,x+\frac {1}{2}}{a^2\,c^2\,x^4-2\,a\,c^2\,x^3+c^2\,x^2}-\frac {31\,a^2\,\ln \left (a\,x-1\right )}{8\,c^2}-\frac {a^2\,\ln \left (a\,x+1\right )}{8\,c^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*a^2*log(x))/c^2 - (a*x - (11*a^2*x^2)/2 + (15*a^3*x^3)/4 + 1/2)/(c^2*x^2 - 2*a*c^2*x^3 + a^2*c^2*x^4) - (31
*a^2*log(a*x - 1))/(8*c^2) - (a^2*log(a*x + 1))/(8*c^2)

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sympy [A]  time = 0.61, size = 92, normalized size = 0.93 \[ - \frac {15 a^{3} x^{3} - 22 a^{2} x^{2} + 4 a x + 2}{4 a^{2} c^{2} x^{4} - 8 a c^{2} x^{3} + 4 c^{2} x^{2}} - \frac {- 4 a^{2} \log {\relax (x )} + \frac {31 a^{2} \log {\left (x - \frac {1}{a} \right )}}{8} + \frac {a^{2} \log {\left (x + \frac {1}{a} \right )}}{8}}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(15*a**3*x**3 - 22*a**2*x**2 + 4*a*x + 2)/(4*a**2*c**2*x**4 - 8*a*c**2*x**3 + 4*c**2*x**2) - (-4*a**2*log(x)
+ 31*a**2*log(x - 1/a)/8 + a**2*log(x + 1/a)/8)/c**2

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