3.828 \(\int e^{2 \tanh ^{-1}(a+b x)} x^2 \, dx\)

Optimal. Leaf size=49 \[ -\frac {2 (1-a)^2 \log (-a-b x+1)}{b^3}-\frac {2 (1-a) x}{b^2}-\frac {x^2}{b}-\frac {x^3}{3} \]

[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 = 0.05, antiderivative size = 49, 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 = {6163, 77} \[ -\frac {2 (1-a) x}{b^2}-\frac {2 (1-a)^2 \log (-a-b x+1)}{b^3}-\frac {x^2}{b}-\frac {x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])*x^2,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 77

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 6163

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 = 0.03, size = 44, normalized size = 0.90 \[ -\frac {b x \left (-6 a+b^2 x^2+3 b x+6\right )+6 (a-1)^2 \log (-a-b x+1)}{3 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b*x*(6 - 6*a + 3*b*x + b^2*x^2) + 6*(-1 + a)^2*Log[1 - a - b*x])/b^3

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fricas [A]  time = 0.97, size = 45, normalized size = 0.92 \[ -\frac {b^{3} x^{3} + 3 \, b^{2} x^{2} - 6 \, {\left (a - 1\right )} b x + 6 \, {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^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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giac [A]  time = 0.15, size = 52, normalized size = 1.06 \[ -\frac {2 \, {\left (a^{2} - 2 \, a + 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{3}} - \frac {b^{3} x^{3} + 3 \, b^{2} x^{2} - 6 \, a b x + 6 \, b x}{3 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 68, normalized size = 1.39 \[ -\frac {x^{3}}{3}-\frac {x^{2}}{b}+\frac {2 a x}{b^{2}}-\frac {2 x}{b^{2}}-\frac {2 \ln \left (b x +a -1\right ) a^{2}}{b^{3}}+\frac {4 \ln \left (b x +a -1\right ) a}{b^{3}}-\frac {2 \ln \left (b x +a -1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 46, normalized size = 0.94 \[ -\frac {b^{2} x^{3} + 3 \, b x^{2} - 6 \, {\left (a - 1\right )} x}{3 \, b^{2}} - \frac {2 \, {\left (a^{2} - 2 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^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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mupad [B]  time = 0.06, size = 74, normalized size = 1.51 \[ x^2\,\left (\frac {a-1}{2\,b}-\frac {a+1}{2\,b}\right )-\frac {x^3}{3}-\frac {\ln \left (a+b\,x-1\right )\,\left (2\,a^2-4\,a+2\right )}{b^3}-\frac {x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,\left (a-1\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.21, size = 42, normalized size = 0.86 \[ - \frac {x^{3}}{3} - x \left (- \frac {2 a}{b^{2}} + \frac {2}{b^{2}}\right ) - \frac {x^{2}}{b} - \frac {2 \left (a - 1\right )^{2} \log {\left (a + b x - 1 \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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