3.9.53 \(\int e^{-2 \tanh ^{-1}(a+b x)} x^2 \, dx\) [853]

Optimal. Leaf size=41 \[ -\frac {2 (1+a) x}{b^2}+\frac {x^2}{b}-\frac {x^3}{3}+\frac {2 (1+a)^2 \log (1+a+b x)}{b^3} \]

[Out]

-2*(1+a)*x/b^2+x^2/b-1/3*x^3+2*(1+a)^2*ln(b*x+a+1)/b^3

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Rubi [A]
time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6298, 78} \begin {gather*} \frac {2 (a+1)^2 \log (a+b x+1)}{b^3}-\frac {2 (a+1) x}{b^2}+\frac {x^2}{b}-\frac {x^3}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/E^(2*ArcTanh[a + b*x]),x]

[Out]

(-2*(1 + a)*x)/b^2 + x^2/b - x^3/3 + (2*(1 + a)^2*Log[1 + a + b*x])/b^3

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^2 \, dx &=\int \frac {x^2 (1-a-b x)}{1+a+b x} \, dx\\ &=\int \left (-\frac {2 (1+a)}{b^2}+\frac {2 x}{b}-x^2+\frac {2 (1+a)^2}{b^2 (1+a+b x)}\right ) \, dx\\ &=-\frac {2 (1+a) x}{b^2}+\frac {x^2}{b}-\frac {x^3}{3}+\frac {2 (1+a)^2 \log (1+a+b x)}{b^3}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 41, normalized size = 1.00 \begin {gather*} -\frac {2 (1+a) x}{b^2}+\frac {x^2}{b}-\frac {x^3}{3}+\frac {2 (1+a)^2 \log (1+a+b x)}{b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/E^(2*ArcTanh[a + b*x]),x]

[Out]

(-2*(1 + a)*x)/b^2 + x^2/b - x^3/3 + (2*(1 + a)^2*Log[1 + a + b*x])/b^3

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Maple [A]
time = 0.05, size = 50, normalized size = 1.22

method	result	size
default	\(-\frac {\frac {1}{3} b^{2} x^{3}-b \,x^{2}+2 a x +2 x}{b^{2}}+\frac {\left (2 a^{2}+4 a +2\right ) \ln \left (b x +a +1\right )}{b^{3}}\)	\(50\)
risch	\(-\frac {x^{3}}{3}+\frac {x^{2}}{b}-\frac {2 a x}{b^{2}}-\frac {2 x}{b^{2}}+\frac {2 \ln \left (b x +a +1\right ) a^{2}}{b^{3}}+\frac {4 \ln \left (b x +a +1\right ) a}{b^{3}}+\frac {2 \ln \left (b x +a +1\right )}{b^{3}}\)	\(67\)
norman	\(\frac {\left (-\frac {a}{3}+\frac {2}{3}\right ) x^{3}+\frac {2 a^{3}+6 a^{2}+6 a +2}{b^{3}}-\frac {b \,x^{4}}{3}-\frac {\left (1+a \right ) x^{2}}{b}}{b x +a +1}+\frac {2 \left (a^{2}+2 a +1\right ) \ln \left (b x +a +1\right )}{b^{3}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 46, normalized size = 1.12 \begin {gather*} -\frac {b^{2} x^{3} - 3 \, b x^{2} + 6 \, {\left (a + 1\right )} x}{3 \, b^{2}} + \frac {2 \, {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^2*x^3 - 3*b*x^2 + 6*(a + 1)*x)/b^2 + 2*(a^2 + 2*a + 1)*log(b*x + a + 1)/b^3

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Fricas [A]
time = 0.35, size = 45, normalized size = 1.10 \begin {gather*} -\frac {b^{3} x^{3} - 3 \, b^{2} x^{2} + 6 \, {\left (a + 1\right )} b x - 6 \, {\left (a^{2} + 2 \, a + 1\right )} \log \left (b x + a + 1\right )}{3 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^3*x^3 - 3*b^2*x^2 + 6*(a + 1)*b*x - 6*(a^2 + 2*a + 1)*log(b*x + a + 1))/b^3

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Sympy [A]
time = 0.10, size = 41, normalized size = 1.00 \begin {gather*} - \frac {x^{3}}{3} - x \left (\frac {2 a}{b^{2}} + \frac {2}{b^{2}}\right ) + \frac {x^{2}}{b} + \frac {2 \left (a + 1\right )^{2} \log {\left (a + b x + 1 \right )}}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (39) = 78\).
time = 0.41, size = 102, normalized size = 2.49 \begin {gather*} \frac {{\left (b x + a + 1\right )}^{3} {\left (\frac {3 \, {\left (a b + 2 \, b\right )}}{{\left (b x + a + 1\right )} b} - \frac {3 \, {\left (a^{2} b^{2} + 6 \, a b^{2} + 5 \, b^{2}\right )}}{{\left (b x + a + 1\right )}^{2} b^{2}} - 1\right )}}{3 \, b^{3}} - \frac {2 \, {\left (a^{2} + 2 \, a + 1\right )} \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left (b x + a + 1\right )}^{2} {\left | b \right |}}\right )}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x + a + 1)^3*(3*(a*b + 2*b)/((b*x + a + 1)*b) - 3*(a^2*b^2 + 6*a*b^2 + 5*b^2)/((b*x + a + 1)^2*b^2) - 1
)/b^3 - 2*(a^2 + 2*a + 1)*log(abs(b*x + a + 1)/((b*x + a + 1)^2*abs(b)))/b^3

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Mupad [B]
time = 0.06, size = 73, normalized size = 1.78 \begin {gather*} \frac {\ln \left (a+b\,x+1\right )\,\left (2\,a^2+4\,a+2\right )}{b^3}-\frac {x^3}{3}-x^2\,\left (\frac {a-1}{2\,b}-\frac {a+1}{2\,b}\right )+\frac {x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,\left (a+1\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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