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

Optimal. Leaf size=18 \[ c^2 (-x)-\frac {c^2}{a^2 x} \]

[Out]

-c^2/a^2/x-c^2*x

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Rubi [A]  time = 0.09, antiderivative size = 18, 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 = {6131, 6129, 73, 14} \[ c^2 (-x)-\frac {c^2}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2/(a^2*x)) - c^2*x

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 6129

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 6131

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

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Mathematica [A]  time = 0.10, size = 18, normalized size = 1.00 \[ c^2 (-x)-\frac {c^2}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2/(a^2*x)) - c^2*x

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fricas [A]  time = 0.56, size = 22, normalized size = 1.22 \[ -\frac {a^{2} c^{2} x^{2} + c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*c^2*x^2 + c^2)/(a^2*x)

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giac [A]  time = 0.20, size = 18, normalized size = 1.00 \[ -c^{2} x - \frac {c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2*x - c^2/(a^2*x)

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maple [A]  time = 0.03, size = 20, normalized size = 1.11 \[ \frac {c^{2} \left (-a^{2} x -\frac {1}{x}\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2/a^2*(-a^2*x-1/x)

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maxima [A]  time = 0.35, size = 18, normalized size = 1.00 \[ -c^{2} x - \frac {c^{2}}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c^2*x - c^2/(a^2*x)

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mupad [B]  time = 0.04, size = 20, normalized size = 1.11 \[ -\frac {c^2\,\left (a^2\,x^2+1\right )}{a^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c^2*(a^2*x^2 + 1))/(a^2*x)

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sympy [A]  time = 0.10, size = 17, normalized size = 0.94 \[ \frac {- a^{2} c^{2} x - \frac {c^{2}}{x}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a**2*c**2*x - c**2/x)/a**2

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