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

Optimal. Leaf size=27 \[ -\frac {c^2}{a^2 x}+\frac {2 c^2 \log (x)}{a}+c^2 x \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 27, 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 = {6131, 6129, 43} \[ -\frac {c^2}{a^2 x}+\frac {2 c^2 \log (x)}{a}+c^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

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Mathematica [A]  time = 0.10, size = 29, normalized size = 1.07 \[ -\frac {c^2}{a^2 x}+\frac {2 c^2 \log (a x)}{a}+c^2 x \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 32, normalized size = 1.19 \[ \frac {a^{2} c^{2} x^{2} + 2 \, a c^{2} x \log \relax (x) - c^{2}}{a^{2} x} \]

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]

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

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giac [A]  time = 0.20, size = 28, normalized size = 1.04 \[ c^{2} x + \frac {2 \, c^{2} \log \left ({\left | x \right |}\right )}{a} - \frac {c^{2}}{a^{2} x} \]

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]

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

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maple [A]  time = 0.03, size = 28, normalized size = 1.04 \[ -\frac {c^{2}}{a^{2} x}+c^{2} x +\frac {2 c^{2} \ln \relax (x )}{a} \]

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)

[Out]

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

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maxima [A]  time = 0.35, size = 27, normalized size = 1.00 \[ c^{2} x + \frac {2 \, c^{2} \log \relax (x)}{a} - \frac {c^{2}}{a^{2} x} \]

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]

c^2*x + 2*c^2*log(x)/a - c^2/(a^2*x)

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mupad [B]  time = 0.81, size = 25, normalized size = 0.93 \[ \frac {c^2\,\left (a^2\,x^2+2\,a\,x\,\ln \relax (x)-1\right )}{a^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.13, size = 26, normalized size = 0.96 \[ \frac {a^{2} c^{2} x + 2 a c^{2} \log {\relax (x )} - \frac {c^{2}}{x}}{a^{2}} \]

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]

(a**2*c**2*x + 2*a*c**2*log(x) - c**2/x)/a**2

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