3.9.55 \(\int e^{-2 \tanh ^{-1}(a+b x)} \, dx\) [855]

Optimal. Leaf size=16 \[ -x+\frac {2 \log (1+a+b x)}{b} \]

[Out]

-x+2*ln(b*x+a+1)/b

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6296, 45} \begin {gather*} \frac {2 \log (a+b x+1)}{b}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(-2*ArcTanh[a + b*x]),x]

[Out]

-x + (2*Log[1 + a + b*x])/b

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 6296

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.)), x_Symbol] :> Int[(1 + a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^(
n/2), x] /; FreeQ[{a, b, c, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} -x+\frac {2 \log (1+a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(-2*ArcTanh[a + b*x]),x]

[Out]

-x + (2*Log[1 + a + b*x])/b

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Maple [A]
time = 0.06, size = 17, normalized size = 1.06

method	result	size
default	\(-x +\frac {2 \ln \left (b x +a +1\right )}{b}\)	\(17\)
risch	\(-x +\frac {2 \ln \left (b x +a +1\right )}{b}\)	\(17\)
norman	\(\frac {\frac {a^{2}+2 a +1}{b}-b \,x^{2}}{b x +a +1}+\frac {2 \ln \left (b x +a +1\right )}{b}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x+2*ln(b*x+a+1)/b

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Maxima [A]
time = 0.26, size = 16, normalized size = 1.00 \begin {gather*} -x + \frac {2 \, \log \left (b x + a + 1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(b*x + a + 1)/b

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Fricas [A]
time = 0.34, size = 18, normalized size = 1.12 \begin {gather*} -\frac {b x - 2 \, \log \left (b x + a + 1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*x - 2*log(b*x + a + 1))/b

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Sympy [A]
time = 0.05, size = 12, normalized size = 0.75 \begin {gather*} - x + \frac {2 \log {\left (a + b x + 1 \right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x + 2*log(a + b*x + 1)/b

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (16) = 32\).
time = 0.42, size = 38, normalized size = 2.38 \begin {gather*} -\frac {b x + a + 1}{b} - \frac {2 \, \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2} {\left | b \right |}}\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 16, normalized size = 1.00 \begin {gather*} \frac {2\,\ln \left (a+b\,x+1\right )}{b}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*log(a + b*x + 1))/b - x

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