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

Optimal. Leaf size=21 \[ \frac {c}{a^2 x}-\frac {2 c \log (x)}{a}+c (-x) \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6157, 6150, 43} \[ \frac {c}{a^2 x}-\frac {2 c \log (x)}{a}+c (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \[ \frac {c}{a^2 x}-\frac {2 c \log (x)}{a}+c (-x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

c/(a^2*x) - c*x - (2*c*Log[x])/a

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fricas [A]  time = 0.47, size = 27, normalized size = 1.29 \[ -\frac {a^{2} c x^{2} + 2 \, a c x \log \relax (x) - c}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 22, normalized size = 1.05 \[ -c x - \frac {2 \, c \log \left ({\left | x \right |}\right )}{a} + \frac {c}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*x - 2*c*log(abs(x))/a + c/(a^2*x)

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maple [A]  time = 0.03, size = 22, normalized size = 1.05 \[ \frac {c}{a^{2} x}-c x -\frac {2 c \ln \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 21, normalized size = 1.00 \[ -c x - \frac {2 \, c \log \relax (x)}{a} + \frac {c}{a^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*x - 2*c*log(x)/a + c/(a^2*x)

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mupad [B]  time = 0.84, size = 24, normalized size = 1.14 \[ -\frac {c\,\left (a^2\,x^2+2\,a\,x\,\ln \relax (x)-1\right )}{a^2\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.12, size = 20, normalized size = 0.95 \[ \frac {- a^{2} c x - 2 a c \log {\relax (x )} + \frac {c}{x}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a**2*c*x - 2*a*c*log(x) + c/x)/a**2

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