3.5.98 \(\int \frac {e^{-2 \tanh ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx\) [498]

Optimal. Leaf size=20 \[ -\frac {x}{c}+\frac {\log (1+a x)}{a c} \]

[Out]

-x/c+ln(a*x+1)/a/c

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Rubi [A]
time = 0.06, antiderivative size = 20, 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 = {6266, 6264, 45} \begin {gather*} \frac {\log (a x+1)}{a c}-\frac {x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))),x]

[Out]

-(x/c) + Log[1 + a*x]/(a*c)

Rule 45

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 6264

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 6266

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

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Mathematica [A]
time = 0.01, size = 18, normalized size = 0.90 \begin {gather*} \frac {-a x+\log (1+a x)}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(E^(2*ArcTanh[a*x])*(c - c/(a*x))),x]

[Out]

(-(a*x) + Log[1 + a*x])/(a*c)

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Maple [A]
time = 0.48, size = 23, normalized size = 1.15

method	result	size
risch	\(-\frac {x}{c}+\frac {\ln \left (a x +1\right )}{a c}\)	\(21\)
default	\(\frac {a \left (-\frac {x}{a}+\frac {\ln \left (a x +1\right )}{a^{2}}\right )}{c}\)	\(23\)
norman	\(\frac {-\frac {x}{c}-\frac {a \,x^{2}}{c}}{a x +1}+\frac {\ln \left (a x +1\right )}{a c}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 1.00 \begin {gather*} -\frac {x}{c} + \frac {\log \left (a x + 1\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/c + log(a*x + 1)/(a*c)

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Fricas [A]
time = 0.36, size = 20, normalized size = 1.00 \begin {gather*} -\frac {a x - \log \left (a x + 1\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a*x - log(a*x + 1))/(a*c)

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Sympy [A]
time = 0.05, size = 19, normalized size = 0.95 \begin {gather*} - a \left (\frac {x}{a c} - \frac {\log {\left (a x + 1 \right )}}{a^{2} c}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*(x/(a*c) - log(a*x + 1)/(a**2*c))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (20) = 40\).
time = 0.41, size = 41, normalized size = 2.05 \begin {gather*} -\frac {a x + 1}{a c} - \frac {\log \left (\frac {{\left | a x + 1 \right |}}{{\left (a x + 1\right )}^{2} {\left | a \right |}}\right )}{a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 18, normalized size = 0.90 \begin {gather*} \frac {\ln \left (a\,x+1\right )-a\,x}{a\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(a*x + 1) - a*x)/(a*c)

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