3.993 \(\int e^{\tanh ^{-1}(a x)} x^m (1-a^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=54 \[ -\frac{a^2 x^{m+3}}{m+3}-\frac{a^3 x^{m+4}}{m+4}+\frac{a x^{m+2}}{m+2}+\frac{x^{m+1}}{m+1} \]

[Out]

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

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Rubi [A]  time = 0.0987892, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {6150, 75} \[ -\frac{a^2 x^{m+3}}{m+3}-\frac{a^3 x^{m+4}}{m+4}+\frac{a x^{m+2}}{m+2}+\frac{x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^m*(1 - a^2*x^2)^(3/2),x]

[Out]

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

Rule 6150

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*(x_)^(m_.)*((c_) + (d_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[c^p, Int[x^m*(1 -
a*x)^(p - n/2)*(1 + a*x)^(p + n/2), x], x] /; FreeQ[{a, c, d, m, n, p}, x] && EqQ[a^2*c + d, 0] && (IntegerQ[p
] || GtQ[c, 0])

Rule 75

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps

\begin{align*} \int e^{\tanh ^{-1}(a x)} x^m \left (1-a^2 x^2\right )^{3/2} \, dx &=\int x^m (1-a x) (1+a x)^2 \, dx\\ &=\int \left (x^m+a x^{1+m}-a^2 x^{2+m}-a^3 x^{3+m}\right ) \, dx\\ &=\frac{x^{1+m}}{1+m}+\frac{a x^{2+m}}{2+m}-\frac{a^2 x^{3+m}}{3+m}-\frac{a^3 x^{4+m}}{4+m}\\ \end{align*}

Mathematica [A]  time = 0.0937093, size = 54, normalized size = 1. \[ \frac{x^{m+1} \left ((2 m+5) \left (\frac{a^2 x^2}{m+3}+\frac{2 a x}{m+2}+\frac{1}{m+1}\right )-(a x+1)^3\right )}{m+4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^m*(1 - a^2*x^2)^(3/2),x]

[Out]

(x^(1 + m)*(-(1 + a*x)^3 + (5 + 2*m)*((1 + m)^(-1) + (2*a*x)/(2 + m) + (a^2*x^2)/(3 + m))))/(4 + m)

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Maple [B]  time = 0.031, size = 142, normalized size = 2.6 \begin{align*} -{\frac{{x}^{1+m} \left ({a}^{3}{m}^{3}{x}^{3}+6\,{a}^{3}{m}^{2}{x}^{3}+11\,{a}^{3}m{x}^{3}+{a}^{2}{m}^{3}{x}^{2}+6\,{x}^{3}{a}^{3}+7\,{a}^{2}{m}^{2}{x}^{2}+14\,{a}^{2}m{x}^{2}-a{m}^{3}x+8\,{a}^{2}{x}^{2}-8\,a{m}^{2}x-19\,amx-{m}^{3}-12\,ax-9\,{m}^{2}-26\,m-24 \right ) }{ \left ( 4+m \right ) \left ( 3+m \right ) \left ( 2+m \right ) \left ( 1+m \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^(1+m)*(a^3*m^3*x^3+6*a^3*m^2*x^3+11*a^3*m*x^3+a^2*m^3*x^2+6*a^3*x^3+7*a^2*m^2*x^2+14*a^2*m*x^2-a*m^3*x+8*a^
2*x^2-8*a*m^2*x-19*a*m*x-m^3-12*a*x-9*m^2-26*m-24)/(4+m)/(3+m)/(2+m)/(1+m)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.98979, size = 278, normalized size = 5.15 \begin{align*} -\frac{{\left ({\left (a^{3} m^{3} + 6 \, a^{3} m^{2} + 11 \, a^{3} m + 6 \, a^{3}\right )} x^{4} +{\left (a^{2} m^{3} + 7 \, a^{2} m^{2} + 14 \, a^{2} m + 8 \, a^{2}\right )} x^{3} -{\left (a m^{3} + 8 \, a m^{2} + 19 \, a m + 12 \, a\right )} x^{2} -{\left (m^{3} + 9 \, m^{2} + 26 \, m + 24\right )} x\right )} x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^3*m^3 + 6*a^3*m^2 + 11*a^3*m + 6*a^3)*x^4 + (a^2*m^3 + 7*a^2*m^2 + 14*a^2*m + 8*a^2)*x^3 - (a*m^3 + 8*a*m
^2 + 19*a*m + 12*a)*x^2 - (m^3 + 9*m^2 + 26*m + 24)*x)*x^m/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)

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Sympy [A]  time = 0.808258, size = 585, normalized size = 10.83 \begin{align*} \begin{cases} - a^{3} \log{\left (x \right )} + \frac{a^{2}}{x} - \frac{a}{2 x^{2}} - \frac{1}{3 x^{3}} & \text{for}\: m = -4 \\- a^{3} x - a^{2} \log{\left (x \right )} - \frac{a}{x} - \frac{1}{2 x^{2}} & \text{for}\: m = -3 \\- \frac{a^{3} x^{2}}{2} - a^{2} x + a \log{\left (x \right )} - \frac{1}{x} & \text{for}\: m = -2 \\- \frac{a^{3} x^{3}}{3} - \frac{a^{2} x^{2}}{2} + a x + \log{\left (x \right )} & \text{for}\: m = -1 \\- \frac{a^{3} m^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{6 a^{3} m^{2} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{11 a^{3} m x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{6 a^{3} x^{4} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{a^{2} m^{3} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{7 a^{2} m^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{14 a^{2} m x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} - \frac{8 a^{2} x^{3} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{a m^{3} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{8 a m^{2} x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{19 a m x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{12 a x^{2} x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{m^{3} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{9 m^{2} x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{26 m x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} + \frac{24 x x^{m}}{m^{4} + 10 m^{3} + 35 m^{2} + 50 m + 24} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x+1)*(-a**2*x**2+1)*x**m,x)

[Out]

Piecewise((-a**3*log(x) + a**2/x - a/(2*x**2) - 1/(3*x**3), Eq(m, -4)), (-a**3*x - a**2*log(x) - a/x - 1/(2*x*
*2), Eq(m, -3)), (-a**3*x**2/2 - a**2*x + a*log(x) - 1/x, Eq(m, -2)), (-a**3*x**3/3 - a**2*x**2/2 + a*x + log(
x), Eq(m, -1)), (-a**3*m**3*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 6*a**3*m**2*x**4*x**m/(m**4 + 1
0*m**3 + 35*m**2 + 50*m + 24) - 11*a**3*m*x**4*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 6*a**3*x**4*x**m/
(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - a**2*m**3*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 7*a**2*m
**2*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) - 14*a**2*m*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m +
24) - 8*a**2*x**3*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + a*m**3*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 5
0*m + 24) + 8*a*m**2*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 19*a*m*x**2*x**m/(m**4 + 10*m**3 + 35*
m**2 + 50*m + 24) + 12*a*x**2*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + m**3*x*x**m/(m**4 + 10*m**3 + 35*m
**2 + 50*m + 24) + 9*m**2*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24) + 26*m*x*x**m/(m**4 + 10*m**3 + 35*m**
2 + 50*m + 24) + 24*x*x**m/(m**4 + 10*m**3 + 35*m**2 + 50*m + 24), True))

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Giac [B]  time = 1.17111, size = 266, normalized size = 4.93 \begin{align*} -\frac{a^{3} m^{3} x^{4} x^{m} + 6 \, a^{3} m^{2} x^{4} x^{m} + a^{2} m^{3} x^{3} x^{m} + 11 \, a^{3} m x^{4} x^{m} + 7 \, a^{2} m^{2} x^{3} x^{m} + 6 \, a^{3} x^{4} x^{m} - a m^{3} x^{2} x^{m} + 14 \, a^{2} m x^{3} x^{m} - 8 \, a m^{2} x^{2} x^{m} + 8 \, a^{2} x^{3} x^{m} - m^{3} x x^{m} - 19 \, a m x^{2} x^{m} - 9 \, m^{2} x x^{m} - 12 \, a x^{2} x^{m} - 26 \, m x x^{m} - 24 \, x x^{m}}{m^{4} + 10 \, m^{3} + 35 \, m^{2} + 50 \, m + 24} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^3*m^3*x^4*x^m + 6*a^3*m^2*x^4*x^m + a^2*m^3*x^3*x^m + 11*a^3*m*x^4*x^m + 7*a^2*m^2*x^3*x^m + 6*a^3*x^4*x^m
- a*m^3*x^2*x^m + 14*a^2*m*x^3*x^m - 8*a*m^2*x^2*x^m + 8*a^2*x^3*x^m - m^3*x*x^m - 19*a*m*x^2*x^m - 9*m^2*x*x
^m - 12*a*x^2*x^m - 26*m*x*x^m - 24*x*x^m)/(m^4 + 10*m^3 + 35*m^2 + 50*m + 24)