3.1280 \(\int e^{-3 \tanh ^{-1}(a x)} (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=86 \[ -\frac {2^{p-\frac {1}{2}} (1-a x)^{p+\frac {5}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2}-p,p+\frac {5}{2};p+\frac {7}{2};\frac {1}{2} (1-a x)\right )}{a (2 p+5)} \]

[Out]

-2^(-1/2+p)*(-a*x+1)^(5/2+p)*(-a^2*c*x^2+c)^p*hypergeom([5/2+p, 3/2-p],[7/2+p],-1/2*a*x+1/2)/a/(5+2*p)/((-a^2*
x^2+1)^p)

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Rubi [A]  time = 0.09, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6143, 6140, 69} \[ -\frac {2^{p-\frac {1}{2}} (1-a x)^{p+\frac {5}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2}-p,p+\frac {5}{2};p+\frac {7}{2};\frac {1}{2} (1-a x)\right )}{a (2 p+5)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(-1/2 + p)*(1 - a*x)^(5/2 + p)*(c - a^2*c*x^2)^p*Hypergeometric2F1[3/2 - p, 5/2 + p, 7/2 + p, (1 - a*x)/2
])/(a*(5 + 2*p)*(1 - a^2*x^2)^p))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), 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 6140

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])

Rule 6143

Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(c^IntPart[p]*(c + d*x^2)^Frac
Part[p])/(1 - a^2*x^2)^FracPart[p], Int[(1 - a^2*x^2)^p*E^(n*ArcTanh[a*x]), 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^{-3 \tanh ^{-1}(a x)} \left (c-a^2 c x^2\right )^p \, dx &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int e^{-3 \tanh ^{-1}(a x)} \left (1-a^2 x^2\right )^p \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int (1-a x)^{\frac {3}{2}+p} (1+a x)^{-\frac {3}{2}+p} \, dx\\ &=-\frac {2^{-\frac {1}{2}+p} (1-a x)^{\frac {5}{2}+p} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2}-p,\frac {5}{2}+p;\frac {7}{2}+p;\frac {1}{2} (1-a x)\right )}{a (5+2 p)}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 86, normalized size = 1.00 \[ -\frac {2^{p-\frac {3}{2}} (1-a x)^{p+\frac {5}{2}} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {3}{2}-p,p+\frac {5}{2};p+\frac {7}{2};\frac {1}{2} (1-a x)\right )}{a \left (p+\frac {5}{2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2^(-3/2 + p)*(1 - a*x)^(5/2 + p)*(c - a^2*c*x^2)^p*Hypergeometric2F1[3/2 - p, 5/2 + p, 7/2 + p, (1 - a*x)/2
])/(a*(5/2 + p)*(1 - a^2*x^2)^p))

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} {\left (a x - 1\right )} {\left (-a^{2} c x^{2} + c\right )}^{p}}{a^{2} x^{2} + 2 \, a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (-a^{2} c x^{2} + c\right )}^{p}}{{\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*(-a^2*c*x^2 + c)^p/(a*x + 1)^3, x)

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maple [F]  time = 0.45, size = 0, normalized size = 0.00 \[ \int \frac {\left (-a^{2} c \,x^{2}+c \right )^{p} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}{\left (a x +1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-a^{2} x^{2} + 1\right )}^{\frac {3}{2}} {\left (-a^{2} c x^{2} + c\right )}^{p}}{{\left (a x + 1\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((-a^2*x^2 + 1)^(3/2)*(-a^2*c*x^2 + c)^p/(a*x + 1)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-a^2\,c\,x^2\right )}^p\,{\left (1-a^2\,x^2\right )}^{3/2}}{{\left (a\,x+1\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}} \left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{p}}{\left (a x + 1\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-(a*x - 1)*(a*x + 1))**(3/2)*(-c*(a*x - 1)*(a*x + 1))**p/(a*x + 1)**3, x)

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