3.994 \(\int e^{\tanh ^{-1}(a x)} x^m \sqrt{1-a^2 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac{a x^{m+2}}{m+2}+\frac{x^{m+1}}{m+1} \]

[Out]

x^(1 + m)/(1 + m) + (a*x^(2 + m))/(2 + m)

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Rubi [A]  time = 0.0799803, antiderivative size = 24, 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, 43} \[ \frac{a x^{m+2}}{m+2}+\frac{x^{m+1}}{m+1} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a*x]*x^m*Sqrt[1 - a^2*x^2],x]

[Out]

x^(1 + m)/(1 + m) + (a*x^(2 + m))/(2 + 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0132547, size = 20, normalized size = 0.83 \[ x^{m+1} \left (\frac{a x}{m+2}+\frac{1}{m+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcTanh[a*x]*x^m*Sqrt[1 - a^2*x^2],x]

[Out]

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

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Maple [A]  time = 0.027, size = 27, normalized size = 1.1 \begin{align*}{\frac{{x}^{1+m} \left ( amx+ax+m+2 \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)*x^m,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.13452, size = 66, normalized size = 2.75 \begin{align*} \frac{{\left ({\left (a m + a\right )} x^{2} +{\left (m + 2\right )} x\right )} x^{m}}{m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*m + a)*x^2 + (m + 2)*x)*x^m/(m^2 + 3*m + 2)

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Sympy [A]  time = 0.277716, size = 82, normalized size = 3.42 \begin{align*} \begin{cases} a \log{\left (x \right )} - \frac{1}{x} & \text{for}\: m = -2 \\a x + \log{\left (x \right )} & \text{for}\: m = -1 \\\frac{a m x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac{a x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac{m x x^{m}}{m^{2} + 3 m + 2} + \frac{2 x x^{m}}{m^{2} + 3 m + 2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*log(x) - 1/x, Eq(m, -2)), (a*x + log(x), Eq(m, -1)), (a*m*x**2*x**m/(m**2 + 3*m + 2) + a*x**2*x**
m/(m**2 + 3*m + 2) + m*x*x**m/(m**2 + 3*m + 2) + 2*x*x**m/(m**2 + 3*m + 2), True))

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Giac [A]  time = 1.13644, size = 55, normalized size = 2.29 \begin{align*} \frac{a m x^{2} x^{m} + a x^{2} x^{m} + m x x^{m} + 2 \, x x^{m}}{m^{2} + 3 \, m + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*m*x^2*x^m + a*x^2*x^m + m*x*x^m + 2*x*x^m)/(m^2 + 3*m + 2)