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

Optimal. Leaf size=13 \[ -\frac {c \log (x)}{a}-c x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 13, 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 = {6131, 6129, 43} \[ -\frac {c \log (x)}{a}-c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.04, size = 13, normalized size = 1.00 \[ -\frac {c \log (x)}{a}-c x \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c*x) - (c*Log[x])/a

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fricas [A]  time = 0.51, size = 14, normalized size = 1.08 \[ -\frac {a c x + c \log \relax (x)}{a} \]

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),x, algorithm="fricas")

[Out]

-(a*c*x + c*log(x))/a

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giac [A]  time = 0.18, size = 14, normalized size = 1.08 \[ -c x - \frac {c \log \left ({\left | x \right |}\right )}{a} \]

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),x, algorithm="giac")

[Out]

-c*x - c*log(abs(x))/a

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maple [A]  time = 0.03, size = 14, normalized size = 1.08 \[ -c x -\frac {c \ln \relax (x )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 13, normalized size = 1.00 \[ -c x - \frac {c \log \relax (x)}{a} \]

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),x, algorithm="maxima")

[Out]

-c*x - c*log(x)/a

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mupad [B]  time = 0.03, size = 12, normalized size = 0.92 \[ -\frac {c\,\left (\ln \relax (x)+a\,x\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*(log(x) + a*x))/a

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sympy [A]  time = 0.09, size = 12, normalized size = 0.92 \[ \frac {- a c x - c \log {\relax (x )}}{a} \]

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),x)

[Out]

(-a*c*x - c*log(x))/a

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