3.1334 \(\int \frac{e^{n \tanh ^{-1}(a x)} x^3}{\sqrt{c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=275 \[ -\frac{2^{\frac{n-1}{2}} n \left (n^2+5\right ) \sqrt{1-a^2 x^2} (1-a x)^{\frac{3-n}{2}} \text{Hypergeometric2F1}\left (\frac{1-n}{2},\frac{3-n}{2},\frac{5-n}{2},\frac{1}{2} (1-a x)\right )}{3 a^4 (1-n) (3-n) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} (a x+1)^{\frac{n+1}{2}} \left (a (1-n) n x+n^2+n+4\right ) (1-a x)^{\frac{1-n}{2}}}{6 a^4 (1-n) \sqrt{c-a^2 c x^2}}-\frac{x^2 \sqrt{1-a^2 x^2} (a x+1)^{\frac{n+1}{2}} (1-a x)^{\frac{1-n}{2}}}{3 a^2 \sqrt{c-a^2 c x^2}} \]

[Out]

-(x^2*(1 - a*x)^((1 - n)/2)*(1 + a*x)^((1 + n)/2)*Sqrt[1 - a^2*x^2])/(3*a^2*Sqrt[c - a^2*c*x^2]) - ((1 - a*x)^
((1 - n)/2)*(1 + a*x)^((1 + n)/2)*(4 + n + n^2 + a*(1 - n)*n*x)*Sqrt[1 - a^2*x^2])/(6*a^4*(1 - n)*Sqrt[c - a^2
*c*x^2]) - (2^((-1 + n)/2)*n*(5 + n^2)*(1 - a*x)^((3 - n)/2)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[(1 - n)/2, (3
- n)/2, (5 - n)/2, (1 - a*x)/2])/(3*a^4*(1 - n)*(3 - n)*Sqrt[c - a^2*c*x^2])

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Rubi [A]  time = 0.380919, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {6153, 6150, 100, 146, 69} \[ -\frac{2^{\frac{n-1}{2}} n \left (n^2+5\right ) \sqrt{1-a^2 x^2} (1-a x)^{\frac{3-n}{2}} \, _2F_1\left (\frac{1-n}{2},\frac{3-n}{2};\frac{5-n}{2};\frac{1}{2} (1-a x)\right )}{3 a^4 (1-n) (3-n) \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} (a x+1)^{\frac{n+1}{2}} \left (a (1-n) n x+n^2+n+4\right ) (1-a x)^{\frac{1-n}{2}}}{6 a^4 (1-n) \sqrt{c-a^2 c x^2}}-\frac{x^2 \sqrt{1-a^2 x^2} (a x+1)^{\frac{n+1}{2}} (1-a x)^{\frac{1-n}{2}}}{3 a^2 \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^2*(1 - a*x)^((1 - n)/2)*(1 + a*x)^((1 + n)/2)*Sqrt[1 - a^2*x^2])/(3*a^2*Sqrt[c - a^2*c*x^2]) - ((1 - a*x)^
((1 - n)/2)*(1 + a*x)^((1 + n)/2)*(4 + n + n^2 + a*(1 - n)*n*x)*Sqrt[1 - a^2*x^2])/(6*a^4*(1 - n)*Sqrt[c - a^2
*c*x^2]) - (2^((-1 + n)/2)*n*(5 + n^2)*(1 - a*x)^((3 - n)/2)*Sqrt[1 - a^2*x^2]*Hypergeometric2F1[(1 - n)/2, (3
- n)/2, (5 - n)/2, (1 - a*x)/2])/(3*a^4*(1 - n)*(3 - n)*Sqrt[c - a^2*c*x^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]

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(m + n + 3)), x] - Dist[
(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m +
1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && NeQ[m + n + 3, 0]

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]))

Rubi steps

\begin{align*} \int \frac{e^{n \tanh ^{-1}(a x)} x^3}{\sqrt{c-a^2 c x^2}} \, dx &=\frac{\sqrt{1-a^2 x^2} \int \frac{e^{n \tanh ^{-1}(a x)} x^3}{\sqrt{1-a^2 x^2}} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=\frac{\sqrt{1-a^2 x^2} \int x^3 (1-a x)^{-\frac{1}{2}-\frac{n}{2}} (1+a x)^{-\frac{1}{2}+\frac{n}{2}} \, dx}{\sqrt{c-a^2 c x^2}}\\ &=-\frac{x^2 (1-a x)^{\frac{1-n}{2}} (1+a x)^{\frac{1+n}{2}} \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{\sqrt{1-a^2 x^2} \int x (1-a x)^{-\frac{1}{2}-\frac{n}{2}} (1+a x)^{-\frac{1}{2}+\frac{n}{2}} (-2-a n x) \, dx}{3 a^2 \sqrt{c-a^2 c x^2}}\\ &=-\frac{x^2 (1-a x)^{\frac{1-n}{2}} (1+a x)^{\frac{1+n}{2}} \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{(1-a x)^{\frac{1-n}{2}} (1+a x)^{\frac{1+n}{2}} \left (4+n+n^2+a (1-n) n x\right ) \sqrt{1-a^2 x^2}}{6 a^4 (1-n) \sqrt{c-a^2 c x^2}}+\frac{\left (n \left (5+n^2\right ) \sqrt{1-a^2 x^2}\right ) \int (1-a x)^{\frac{1}{2}-\frac{n}{2}} (1+a x)^{-\frac{1}{2}+\frac{n}{2}} \, dx}{6 a^3 (1-n) \sqrt{c-a^2 c x^2}}\\ &=-\frac{x^2 (1-a x)^{\frac{1-n}{2}} (1+a x)^{\frac{1+n}{2}} \sqrt{1-a^2 x^2}}{3 a^2 \sqrt{c-a^2 c x^2}}-\frac{(1-a x)^{\frac{1-n}{2}} (1+a x)^{\frac{1+n}{2}} \left (4+n+n^2+a (1-n) n x\right ) \sqrt{1-a^2 x^2}}{6 a^4 (1-n) \sqrt{c-a^2 c x^2}}-\frac{2^{\frac{1}{2} (-1+n)} n \left (5+n^2\right ) (1-a x)^{\frac{3-n}{2}} \sqrt{1-a^2 x^2} \, _2F_1\left (\frac{1-n}{2},\frac{3-n}{2};\frac{5-n}{2};\frac{1}{2} (1-a x)\right )}{3 a^4 (1-n) (3-n) \sqrt{c-a^2 c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.261604, size = 187, normalized size = 0.68 \[ \frac{\sqrt{1-a^2 x^2} (1-a x)^{\frac{1}{2}-\frac{n}{2}} \left (2^{\frac{n}{2}+1} n \left (n^2+5\right ) (a x-1) \text{Hypergeometric2F1}\left (\frac{1}{2}-\frac{n}{2},\frac{3}{2}-\frac{n}{2},\frac{5}{2}-\frac{n}{2},\frac{1}{2}-\frac{a x}{2}\right )-\sqrt{2} (n-3) (a x+1)^{\frac{n+1}{2}} \left (n \left (2 a^2 x^2-a x-1\right )-2 \left (a^2 x^2+2\right )+n^2 (a x-1)\right )\right )}{6 \sqrt{2} a^4 (n-3) (n-1) \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(1/2 - n/2)*Sqrt[1 - a^2*x^2]*(-(Sqrt[2]*(-3 + n)*(1 + a*x)^((1 + n)/2)*(n^2*(-1 + a*x) - 2*(2 + a^
2*x^2) + n*(-1 - a*x + 2*a^2*x^2))) + 2^(1 + n/2)*n*(5 + n^2)*(-1 + a*x)*Hypergeometric2F1[1/2 - n/2, 3/2 - n/
2, 5/2 - n/2, 1/2 - (a*x)/2]))/(6*Sqrt[2]*a^4*(-3 + n)*(-1 + n)*Sqrt[c - a^2*c*x^2])

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Maple [F]  time = 0.209, size = 0, normalized size = 0. \begin{align*} \int{{{\rm e}^{n{\it Artanh} \left ( ax \right ) }}{x}^{3}{\frac{1}{\sqrt{-{a}^{2}c{x}^{2}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(1/2),x)

[Out]

int(exp(n*arctanh(a*x))*x^3/(-a^2*c*x^2+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{\sqrt{-a^{2} c x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{-a^{2} c x^{2} + c} x^{3} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*x^3*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c*x^2 - c), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} e^{n \operatorname{atanh}{\left (a x \right )}}}{\sqrt{- c \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))*x**3/(-a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**3*exp(n*atanh(a*x))/sqrt(-c*(a*x - 1)*(a*x + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3} \left (\frac{a x + 1}{a x - 1}\right )^{\frac{1}{2} \, n}}{\sqrt{-a^{2} c x^{2} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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