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

Optimal. Leaf size=53 \[ \frac {5}{a c (1-a x)}-\frac {1}{a c (1-a x)^2}+\frac {4 \log (1-a x)}{a c}+\frac {x}{c} \]

[Out]

x/c-1/a/c/(-a*x+1)^2+5/a/c/(-a*x+1)+4*ln(-a*x+1)/a/c

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Rubi [A]  time = 0.13, antiderivative size = 53, 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 = {6157, 6150, 77} \[ \frac {5}{a c (1-a x)}-\frac {1}{a c (1-a x)^2}+\frac {4 \log (1-a x)}{a c}+\frac {x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/c - 1/(a*c*(1 - a*x)^2) + 5/(a*c*(1 - a*x)) + (4*Log[1 - a*x])/(a*c)

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 53, normalized size = 1.00 \[ \frac {5}{a c (1-a x)}-\frac {1}{a c (1-a x)^2}+\frac {4 \log (1-a x)}{a c}+\frac {x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/c - 1/(a*c*(1 - a*x)^2) + 5/(a*c*(1 - a*x)) + (4*Log[1 - a*x])/(a*c)

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fricas [A]  time = 1.00, size = 64, normalized size = 1.21 \[ \frac {a^{3} x^{3} - 2 \, a^{2} x^{2} - 4 \, a x + 4 \, {\left (a^{2} x^{2} - 2 \, a x + 1\right )} \log \left (a x - 1\right ) + 4}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} \]

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^2/x^2),x, algorithm="fricas")

[Out]

(a^3*x^3 - 2*a^2*x^2 - 4*a*x + 4*(a^2*x^2 - 2*a*x + 1)*log(a*x - 1) + 4)/(a^3*c*x^2 - 2*a^2*c*x + a*c)

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giac [A]  time = 0.17, size = 42, normalized size = 0.79 \[ \frac {x}{c} + \frac {4 \, \log \left ({\left | a x - 1 \right |}\right )}{a c} - \frac {5 \, a x - 4}{{\left (a x - 1\right )}^{2} a c} \]

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^2/x^2),x, algorithm="giac")

[Out]

x/c + 4*log(abs(a*x - 1))/(a*c) - (5*a*x - 4)/((a*x - 1)^2*a*c)

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maple [A]  time = 0.03, size = 51, normalized size = 0.96 \[ \frac {x}{c}-\frac {5}{c a \left (a x -1\right )}+\frac {4 \ln \left (a x -1\right )}{a c}-\frac {1}{a c \left (a x -1\right )^{2}} \]

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^2/x^2),x)

[Out]

x/c-5/c/a/(a*x-1)+4/a/c*ln(a*x-1)-1/a/c/(a*x-1)^2

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maxima [A]  time = 0.31, size = 49, normalized size = 0.92 \[ -\frac {5 \, a x - 4}{a^{3} c x^{2} - 2 \, a^{2} c x + a c} + \frac {x}{c} + \frac {4 \, \log \left (a x - 1\right )}{a c} \]

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^2/x^2),x, algorithm="maxima")

[Out]

-(5*a*x - 4)/(a^3*c*x^2 - 2*a^2*c*x + a*c) + x/c + 4*log(a*x - 1)/(a*c)

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mupad [B]  time = 0.07, size = 48, normalized size = 0.91 \[ \frac {x}{c}-\frac {5\,x-\frac {4}{a}}{c\,a^2\,x^2-2\,c\,a\,x+c}+\frac {4\,\ln \left (a\,x-1\right )}{a\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/c - (5*x - 4/a)/(c + a^2*c*x^2 - 2*a*c*x) + (4*log(a*x - 1))/(a*c)

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sympy [A]  time = 0.24, size = 41, normalized size = 0.77 \[ \frac {- 5 a x + 4}{a^{3} c x^{2} - 2 a^{2} c x + a c} + \frac {x}{c} + \frac {4 \log {\left (a x - 1 \right )}}{a c} \]

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**2/x**2),x)

[Out]

(-5*a*x + 4)/(a**3*c*x**2 - 2*a**2*c*x + a*c) + x/c + 4*log(a*x - 1)/(a*c)

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