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 = 0.08, antiderivative size = 24, normalized size of antiderivative = 1.00, 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 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])

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])

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.61, size = 29, normalized size = 1.21 \[ \frac {{\left ({\left (a m + a\right )} x^{2} + {\left (m + 2\right )} x\right )} x^{m}}{m^{2} + 3 \, m + 2} \]

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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giac [A]  time = 0.24, size = 41, normalized size = 1.71 \[ \frac {a m x^{2} x^{m} + a x^{2} x^{m} + m x x^{m} + 2 \, x x^{m}}{m^{2} + 3 \, m + 2} \]

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)

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maple [A]  time = 0.02, size = 27, normalized size = 1.12 \[ \frac {x^{1+m} \left (a m x +a x +m +2\right )}{\left (2+m \right ) \left (1+m \right )} \]

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 [A]  time = 0.30, size = 24, normalized size = 1.00 \[ \frac {a x^{m + 2}}{m + 2} + \frac {x^{m + 1}}{m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.10, size = 26, normalized size = 1.08 \[ \frac {x^{m+1}\,\left (m+a\,x+a\,m\,x+2\right )}{m^2+3\,m+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(a*x + 1),x)

[Out]

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

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sympy [A]  time = 0.25, size = 82, normalized size = 3.42 \[ \begin {cases} a \log {\relax (x )} - \frac {1}{x} & \text {for}\: m = -2 \\a x + \log {\relax (x )} & \text {for}\: m = -1 \\\frac {a m x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac {a x^{2} x^{m}}{m^{2} + 3 m + 2} + \frac {m x x^{m}}{m^{2} + 3 m + 2} + \frac {2 x x^{m}}{m^{2} + 3 m + 2} & \text {otherwise} \end {cases} \]

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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