3.9.54 \(\int e^{-2 \tanh ^{-1}(a+b x)} x \, dx\) [854]

Optimal. Leaf size=29 \[ \frac {2 x}{b}-\frac {x^2}{2}-\frac {2 (1+a) \log (1+a+b x)}{b^2} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6298, 78} \begin {gather*} -\frac {2 (a+1) \log (a+b x+1)}{b^2}+\frac {2 x}{b}-\frac {x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/E^(2*ArcTanh[a + b*x]),x]

[Out]

(2*x)/b - x^2/2 - (2*(1 + a)*Log[1 + a + b*x])/b^2

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 6298

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

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 x}{b}-\frac {x^2}{2}-\frac {2 (1+a) \log (1+a+b x)}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/E^(2*ArcTanh[a + b*x]),x]

[Out]

(2*x)/b - x^2/2 - (2*(1 + a)*Log[1 + a + b*x])/b^2

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Maple [A]
time = 0.06, size = 33, normalized size = 1.14

method	result	size
default	\(-\frac {\frac {1}{2} b \,x^{2}-2 x}{b}+\frac {\left (-2-2 a \right ) \ln \left (b x +a +1\right )}{b^{2}}\)	\(33\)
risch	\(-\frac {x^{2}}{2}+\frac {2 x}{b}-\frac {2 \ln \left (b x +a +1\right )}{b^{2}}-\frac {2 \ln \left (b x +a +1\right ) a}{b^{2}}\)	\(38\)
norman	\(\frac {\left (-\frac {a}{2}+\frac {3}{2}\right ) x^{2}-\frac {b \,x^{3}}{2}-\frac {2 a^{2}+4 a +2}{b^{2}}}{b x +a +1}-\frac {2 \left (1+a \right ) \ln \left (b x +a +1\right )}{b^{2}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 30, normalized size = 1.03 \begin {gather*} -\frac {b x^{2} - 4 \, x}{2 \, b} - \frac {2 \, {\left (a + 1\right )} \log \left (b x + a + 1\right )}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x^2 - 4*x)/b - 2*(a + 1)*log(b*x + a + 1)/b^2

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Fricas [A]
time = 0.34, size = 29, normalized size = 1.00 \begin {gather*} -\frac {b^{2} x^{2} - 4 \, b x + 4 \, {\left (a + 1\right )} \log \left (b x + a + 1\right )}{2 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^2*x^2 - 4*b*x + 4*(a + 1)*log(b*x + a + 1))/b^2

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Sympy [A]
time = 0.07, size = 26, normalized size = 0.90 \begin {gather*} - \frac {x^{2}}{2} + \frac {2 x}{b} - \frac {2 \left (a + 1\right ) \log {\left (a + b x + 1 \right )}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (27) = 54\).
time = 0.41, size = 69, normalized size = 2.38 \begin {gather*} \frac {\frac {{\left (b x + a + 1\right )}^{2} {\left (\frac {2 \, {\left (a b + 3 \, b\right )}}{{\left (b x + a + 1\right )} b} - 1\right )}}{b} + \frac {4 \, {\left (a + 1\right )} \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2} {\left | b \right |}}\right )}{b}}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((b*x + a + 1)^2*(2*(a*b + 3*b)/((b*x + a + 1)*b) - 1)/b + 4*(a + 1)*log(abs(b*x + a + 1)/((b*x + a + 1)^2
*abs(b)))/b)/b

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Mupad [B]
time = 0.89, size = 42, normalized size = 1.45 \begin {gather*} -\frac {x^2}{2}-x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )-\frac {\ln \left (a+b\,x+1\right )\,\left (2\,a+2\right )}{b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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