3.13.19 \(\int e^{-\tanh ^{-1}(a x)} (1-a^2 x^2)^p \, dx\) [1219]

Optimal. Leaf size=59 \[ -\frac {2^{\frac {1}{2}+p} (1-a x)^{\frac {3}{2}+p} \, _2F_1\left (\frac {1}{2}-p,\frac {3}{2}+p;\frac {5}{2}+p;\frac {1}{2} (1-a x)\right )}{a (3+2 p)} \]

[Out]

-2^(1/2+p)*(-a*x+1)^(3/2+p)*hypergeom([3/2+p, 1/2-p],[5/2+p],-1/2*a*x+1/2)/a/(3+2*p)

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6275, 71} \begin {gather*} -\frac {2^{p+\frac {1}{2}} (1-a x)^{p+\frac {3}{2}} \, _2F_1\left (\frac {1}{2}-p,p+\frac {3}{2};p+\frac {5}{2};\frac {1}{2} (1-a x)\right )}{a (2 p+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - a^2*x^2)^p/E^ArcTanh[a*x],x]

[Out]

-((2^(1/2 + p)*(1 - a*x)^(3/2 + p)*Hypergeometric2F1[1/2 - p, 3/2 + p, 5/2 + p, (1 - a*x)/2])/(a*(3 + 2*p)))

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 6275

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

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^p \, dx &=\int (1-a x)^{\frac {1}{2}+p} (1+a x)^{-\frac {1}{2}+p} \, dx\\ &=-\frac {2^{\frac {1}{2}+p} (1-a x)^{\frac {3}{2}+p} \, _2F_1\left (\frac {1}{2}-p,\frac {3}{2}+p;\frac {5}{2}+p;\frac {1}{2} (1-a x)\right )}{a (3+2 p)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 56, normalized size = 0.95 \begin {gather*} \frac {(2-2 a x)^{\frac {1}{2}+p} (-1+a x) \, _2F_1\left (\frac {1}{2}-p,\frac {3}{2}+p;\frac {5}{2}+p;\frac {1}{2} (1-a x)\right )}{a (3+2 p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2 - 2*a*x)^(1/2 + p)*(-1 + a*x)*Hypergeometric2F1[1/2 - p, 3/2 + p, 5/2 + p, (1 - a*x)/2])/(a*(3 + 2*p))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (-a^{2} x^{2}+1\right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*x^2 + 1)^(p + 1/2)/(a*x + 1), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p/(a*x + 1), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )} \left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{a x + 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-(a*x - 1)*(a*x + 1))*(-(a*x - 1)*(a*x + 1))**p/(a*x + 1), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-a^2*x^2 + 1)*(-a^2*x^2 + 1)^p/(a*x + 1), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (1-a^2\,x^2\right )}^p\,\sqrt {1-a^2\,x^2}}{a\,x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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