3.827 \(\int e^{2 \tanh ^{-1}(a+b x)} x^3 \, dx\)

Optimal. Leaf size=66 \[ -\frac {2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac {2 (1-a)^2 x}{b^3}-\frac {(1-a) x^2}{b^2}-\frac {2 x^3}{3 b}-\frac {x^4}{4} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 66, 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} \[ -\frac {(1-a) x^2}{b^2}-\frac {2 (1-a)^2 x}{b^3}-\frac {2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac {2 x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*ArcTanh[a + b*x])*x^3,x]

[Out]

(-2*(1 - a)^2*x)/b^3 - ((1 - a)*x^2)/b^2 - (2*x^3)/(3*b) - x^4/4 - (2*(1 - a)^3*Log[1 - a - b*x])/b^4

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^3 \, dx &=\int \frac {x^3 (1+a+b x)}{1-a-b x} \, dx\\ &=\int \left (-\frac {2 (-1+a)^2}{b^3}+\frac {2 (-1+a) x}{b^2}-\frac {2 x^2}{b}-x^3+\frac {2 (-1+a)^3}{b^3 (-1+a+b x)}\right ) \, dx\\ &=-\frac {2 (1-a)^2 x}{b^3}-\frac {(1-a) x^2}{b^2}-\frac {2 x^3}{3 b}-\frac {x^4}{4}-\frac {2 (1-a)^3 \log (1-a-b x)}{b^4}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 66, normalized size = 1.00 \[ -\frac {2 (1-a)^3 \log (-a-b x+1)}{b^4}-\frac {2 (1-a)^2 x}{b^3}-\frac {(1-a) x^2}{b^2}-\frac {2 x^3}{3 b}-\frac {x^4}{4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*ArcTanh[a + b*x])*x^3,x]

[Out]

(-2*(1 - a)^2*x)/b^3 - ((1 - a)*x^2)/b^2 - (2*x^3)/(3*b) - x^4/4 - (2*(1 - a)^3*Log[1 - a - b*x])/b^4

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fricas [A]  time = 0.70, size = 67, normalized size = 1.02 \[ -\frac {3 \, b^{4} x^{4} + 8 \, b^{3} x^{3} - 12 \, {\left (a - 1\right )} b^{2} x^{2} + 24 \, {\left (a^{2} - 2 \, a + 1\right )} b x - 24 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,x, algorithm="fricas")

[Out]

-1/12*(3*b^4*x^4 + 8*b^3*x^3 - 12*(a - 1)*b^2*x^2 + 24*(a^2 - 2*a + 1)*b*x - 24*(a^3 - 3*a^2 + 3*a - 1)*log(b*
x + a - 1))/b^4

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giac [A]  time = 0.15, size = 82, normalized size = 1.24 \[ \frac {2 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left ({\left | b x + a - 1 \right |}\right )}{b^{4}} - \frac {3 \, b^{4} x^{4} + 8 \, b^{3} x^{3} - 12 \, a b^{2} x^{2} + 24 \, a^{2} b x + 12 \, b^{2} x^{2} - 48 \, a b x + 24 \, b x}{12 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,x, algorithm="giac")

[Out]

2*(a^3 - 3*a^2 + 3*a - 1)*log(abs(b*x + a - 1))/b^4 - 1/12*(3*b^4*x^4 + 8*b^3*x^3 - 12*a*b^2*x^2 + 24*a^2*b*x
+ 12*b^2*x^2 - 48*a*b*x + 24*b*x)/b^4

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maple [A]  time = 0.03, size = 108, normalized size = 1.64 \[ -\frac {x^{4}}{4}-\frac {2 x^{3}}{3 b}+\frac {x^{2} a}{b^{2}}-\frac {x^{2}}{b^{2}}-\frac {2 a^{2} x}{b^{3}}+\frac {4 a x}{b^{3}}-\frac {2 x}{b^{3}}+\frac {2 \ln \left (b x +a -1\right ) a^{3}}{b^{4}}-\frac {6 \ln \left (b x +a -1\right ) a^{2}}{b^{4}}+\frac {6 \ln \left (b x +a -1\right ) a}{b^{4}}-\frac {2 \ln \left (b x +a -1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^4-2/3*x^3/b+1/b^2*x^2*a-1/b^2*x^2-2/b^3*a^2*x+4*a*x/b^3-2/b^3*x+2/b^4*ln(b*x+a-1)*a^3-6/b^4*ln(b*x+a-1)
*a^2+6/b^4*ln(b*x+a-1)*a-2/b^4*ln(b*x+a-1)

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maxima [A]  time = 0.30, size = 68, normalized size = 1.03 \[ -\frac {3 \, b^{3} x^{4} + 8 \, b^{2} x^{3} - 12 \, {\left (a - 1\right )} b x^{2} + 24 \, {\left (a^{2} - 2 \, a + 1\right )} x}{12 \, b^{3}} + \frac {2 \, {\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (b x + a - 1\right )}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)^2/(1-(b*x+a)^2)*x^3,x, algorithm="maxima")

[Out]

-1/12*(3*b^3*x^4 + 8*b^2*x^3 - 12*(a - 1)*b*x^2 + 24*(a^2 - 2*a + 1)*x)/b^3 + 2*(a^3 - 3*a^2 + 3*a - 1)*log(b*
x + a - 1)/b^4

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mupad [B]  time = 0.89, size = 106, normalized size = 1.61 \[ x^3\,\left (\frac {a-1}{3\,b}-\frac {a+1}{3\,b}\right )-\frac {x^4}{4}+\frac {\ln \left (a+b\,x-1\right )\,\left (2\,a^3-6\,a^2+6\,a-2\right )}{b^4}-\frac {x^2\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,\left (a-1\right )}{2\,b}+\frac {x\,\left (\frac {a-1}{b}-\frac {a+1}{b}\right )\,{\left (a-1\right )}^2}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*((a - 1)/(3*b) - (a + 1)/(3*b)) - x^4/4 + (log(a + b*x - 1)*(6*a - 6*a^2 + 2*a^3 - 2))/b^4 - (x^2*((a - 1)
/b - (a + 1)/b)*(a - 1))/(2*b) + (x*((a - 1)/b - (a + 1)/b)*(a - 1)^2)/b^2

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sympy [A]  time = 0.26, size = 66, normalized size = 1.00 \[ - \frac {x^{4}}{4} - x^{2} \left (- \frac {a}{b^{2}} + \frac {1}{b^{2}}\right ) - x \left (\frac {2 a^{2}}{b^{3}} - \frac {4 a}{b^{3}} + \frac {2}{b^{3}}\right ) - \frac {2 x^{3}}{3 b} + \frac {2 \left (a - 1\right )^{3} \log {\left (a + b x - 1 \right )}}{b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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