3.1215 \(\int e^{-\tanh ^{-1}(a x)} x^m (c-a^2 c x^2)^p \, dx\)

Optimal. Leaf size=137 \[ \frac {x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6153, 6149, 808, 364} \[ \frac {x^{m+1} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+3}{2};a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+4}{2};a^2 x^2\right )}{m+2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(c - a^2*c*x^2)^p*Hypergeometric2F1[(1 + m)/2, 1/2 - p, (3 + m)/2, a^2*x^2])/((1 + m)*(1 - a^2*x^2)
^p) - (a*x^(2 + m)*(c - a^2*c*x^2)^p*Hypergeometric2F1[(2 + m)/2, 1/2 - p, (4 + m)/2, a^2*x^2])/((2 + m)*(1 -
a^2*x^2)^p)

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 808

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 6149

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

Rule 6153

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

Rubi steps

\begin {align*} \int e^{-\tanh ^{-1}(a x)} x^m \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^{-\tanh ^{-1}(a x)} x^m \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 x^m (1-a x) \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\left (\left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^m \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx-\left (a \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p\right ) \int x^{1+m} \left (1-a^2 x^2\right )^{-\frac {1}{2}+p} \, dx\\ &=\frac {x^{1+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {1+m}{2},\frac {1}{2}-p;\frac {3+m}{2};a^2 x^2\right )}{1+m}-\frac {a x^{2+m} \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \, _2F_1\left (\frac {2+m}{2},\frac {1}{2}-p;\frac {4+m}{2};a^2 x^2\right )}{2+m}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 115, normalized size = 0.84 \[ \left (1-a^2 x^2\right )^{-p} \left (c-a^2 c x^2\right )^p \left (\frac {x^{m+1} \, _2F_1\left (\frac {m+1}{2},\frac {1}{2}-p;\frac {m+1}{2}+1;a^2 x^2\right )}{m+1}-\frac {a x^{m+2} \, _2F_1\left (\frac {m+2}{2},\frac {1}{2}-p;\frac {m+2}{2}+1;a^2 x^2\right )}{m+2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{a x + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.41, size = 0, normalized size = 0.00 \[ \int \frac {x^{m} \left (-a^{2} c \,x^{2}+c \right )^{p} \sqrt {-a^{2} x^{2}+1}}{a x +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m*(-a^2*c*x^2+c)^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 \[ \int \frac {\sqrt {-a^{2} x^{2} + 1} {\left (-a^{2} c x^{2} + c\right )}^{p} x^{m}}{a x + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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