3.5.81 \(\int \frac {e^{4 \tanh ^{-1}(a x)}}{(c-\frac {c}{a x})^2} \, dx\) [481]

Optimal. Leaf size=71 \[ \frac {x}{c^2}+\frac {4}{3 a c^2 (1-a x)^3}-\frac {6}{a c^2 (1-a x)^2}+\frac {13}{a c^2 (1-a x)}+\frac {6 \log (1-a x)}{a c^2} \]

[Out]

x/c^2+4/3/a/c^2/(-a*x+1)^3-6/a/c^2/(-a*x+1)^2+13/a/c^2/(-a*x+1)+6*ln(-a*x+1)/a/c^2

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Rubi [A]
time = 0.09, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6266, 6264, 90} \begin {gather*} \frac {13}{a c^2 (1-a x)}-\frac {6}{a c^2 (1-a x)^2}+\frac {4}{3 a c^2 (1-a x)^3}+\frac {6 \log (1-a x)}{a c^2}+\frac {x}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x/c^2 + 4/(3*a*c^2*(1 - a*x)^3) - 6/(a*c^2*(1 - a*x)^2) + 13/(a*c^2*(1 - a*x)) + (6*Log[1 - a*x])/(a*c^2)

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 6264

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> Dist[c^p, Int[u*(1 + d*(x/c))^
p*((1 + a*x)^(n/2)/(1 - a*x)^(n/2)), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^2, 0] && (IntegerQ
[p] || GtQ[c, 0])

Rule 6266

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_))^(p_.), x_Symbol] :> Dist[d^p, Int[u*(1 + c*(x/d))^
p*(E^(n*ArcTanh[a*x])/x^p), x], x] /; FreeQ[{a, c, d, n}, x] && EqQ[c^2 - a^2*d^2, 0] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {e^{4 \tanh ^{-1}(a x)}}{\left (c-\frac {c}{a x}\right )^2} \, dx &=\frac {a^2 \int \frac {e^{4 \tanh ^{-1}(a x)} x^2}{(1-a x)^2} \, dx}{c^2}\\ &=\frac {a^2 \int \frac {x^2 (1+a x)^2}{(1-a x)^4} \, dx}{c^2}\\ &=\frac {a^2 \int \left (\frac {1}{a^2}+\frac {4}{a^2 (-1+a x)^4}+\frac {12}{a^2 (-1+a x)^3}+\frac {13}{a^2 (-1+a x)^2}+\frac {6}{a^2 (-1+a x)}\right ) \, dx}{c^2}\\ &=\frac {x}{c^2}+\frac {4}{3 a c^2 (1-a x)^3}-\frac {6}{a c^2 (1-a x)^2}+\frac {13}{a c^2 (1-a x)}+\frac {6 \log (1-a x)}{a c^2}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 63, normalized size = 0.89 \begin {gather*} \frac {-25+57 a x-30 a^2 x^2-9 a^3 x^3+3 a^4 x^4+18 (-1+a x)^3 \log (1-a x)}{3 a c^2 (-1+a x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-25 + 57*a*x - 30*a^2*x^2 - 9*a^3*x^3 + 3*a^4*x^4 + 18*(-1 + a*x)^3*Log[1 - a*x])/(3*a*c^2*(-1 + a*x)^3)

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Maple [A]
time = 0.77, size = 61, normalized size = 0.86

method	result	size
risch	\(\frac {x}{c^{2}}+\frac {-13 a \,c^{2} x^{2}+20 x \,c^{2}-\frac {25 c^{2}}{3 a}}{c^{4} \left (a x -1\right )^{3}}+\frac {6 \ln \left (a x -1\right )}{a \,c^{2}}\)	\(56\)
default	\(\frac {a^{2} \left (\frac {x}{a^{2}}-\frac {13}{a^{3} \left (a x -1\right )}-\frac {6}{\left (a x -1\right )^{2} a^{3}}-\frac {4}{3 \left (a x -1\right )^{3} a^{3}}+\frac {6 \ln \left (a x -1\right )}{a^{3}}\right )}{c^{2}}\)	\(61\)
norman	\(\frac {\frac {a^{4} x^{5}}{c}+\frac {9 a \,x^{2}}{c}-\frac {6 x}{c}+\frac {11 a^{2} x^{3}}{3 c}-\frac {31 a^{3} x^{4}}{3 c}}{c \left (a x +1\right ) \left (a x -1\right )^{3}}+\frac {6 \ln \left (a x -1\right )}{a \,c^{2}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2/c^2*(x/a^2-13/a^3/(a*x-1)-6/(a*x-1)^2/a^3-4/3/(a*x-1)^3/a^3+6/a^3*ln(a*x-1))

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Maxima [A]
time = 0.26, size = 75, normalized size = 1.06 \begin {gather*} -\frac {39 \, a^{2} x^{2} - 60 \, a x + 25}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} + \frac {x}{c^{2}} + \frac {6 \, \log \left (a x - 1\right )}{a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(39*a^2*x^2 - 60*a*x + 25)/(a^4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2) + x/c^2 + 6*log(a*x - 1)/(
a*c^2)

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Fricas [A]
time = 0.35, size = 100, normalized size = 1.41 \begin {gather*} \frac {3 \, a^{4} x^{4} - 9 \, a^{3} x^{3} - 30 \, a^{2} x^{2} + 57 \, a x + 18 \, {\left (a^{3} x^{3} - 3 \, a^{2} x^{2} + 3 \, a x - 1\right )} \log \left (a x - 1\right ) - 25}{3 \, {\left (a^{4} c^{2} x^{3} - 3 \, a^{3} c^{2} x^{2} + 3 \, a^{2} c^{2} x - a c^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^4*x^4 - 9*a^3*x^3 - 30*a^2*x^2 + 57*a*x + 18*(a^3*x^3 - 3*a^2*x^2 + 3*a*x - 1)*log(a*x - 1) - 25)/(a^
4*c^2*x^3 - 3*a^3*c^2*x^2 + 3*a^2*c^2*x - a*c^2)

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Sympy [A]
time = 0.20, size = 73, normalized size = 1.03 \begin {gather*} \frac {- 39 a^{2} x^{2} + 60 a x - 25}{3 a^{4} c^{2} x^{3} - 9 a^{3} c^{2} x^{2} + 9 a^{2} c^{2} x - 3 a c^{2}} + \frac {x}{c^{2}} + \frac {6 \log {\left (a x - 1 \right )}}{a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-39*a**2*x**2 + 60*a*x - 25)/(3*a**4*c**2*x**3 - 9*a**3*c**2*x**2 + 9*a**2*c**2*x - 3*a*c**2) + x/c**2 + 6*lo
g(a*x - 1)/(a*c**2)

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Giac [A]
time = 0.43, size = 50, normalized size = 0.70 \begin {gather*} \frac {x}{c^{2}} + \frac {6 \, \log \left ({\left | a x - 1 \right |}\right )}{a c^{2}} - \frac {39 \, a^{2} x^{2} - 60 \, a x + 25}{3 \, {\left (a x - 1\right )}^{3} a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c^2 + 6*log(abs(a*x - 1))/(a*c^2) - 1/3*(39*a^2*x^2 - 60*a*x + 25)/((a*x - 1)^3*a*c^2)

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Mupad [B]
time = 0.84, size = 71, normalized size = 1.00 \begin {gather*} \frac {13\,a\,x^2-20\,x+\frac {25}{3\,a}}{-a^3\,c^2\,x^3+3\,a^2\,c^2\,x^2-3\,a\,c^2\,x+c^2}+\frac {x}{c^2}+\frac {6\,\ln \left (a\,x-1\right )}{a\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(13*a*x^2 - 20*x + 25/(3*a))/(c^2 + 3*a^2*c^2*x^2 - a^3*c^2*x^3 - 3*a*c^2*x) + x/c^2 + (6*log(a*x - 1))/(a*c^2
)

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