3.1336 \(\int \frac {e^{n \tanh ^{-1}(a x)} x}{\sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=176 \[ -\frac {2^{\frac {n+3}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^2 \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{a^2 (n+1) \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6153, 6150, 79, 69} \[ -\frac {2^{\frac {n+3}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^2 \left (1-n^2\right ) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{a^2 (n+1) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^Simplify[p + 1], x], x] /; FreeQ[{a, b, c,
d, e, f, n, p}, x] &&  !RationalQ[p] && SumSimplerQ[p, 1]

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

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Mathematica [A]  time = 0.10, size = 124, normalized size = 0.70 \[ \frac {\sqrt {1-a^2 x^2} (1-a x)^{\frac {1}{2}-\frac {n}{2}} \left (2^{\frac {n+3}{2}} n \, _2F_1\left (-\frac {n}{2}-\frac {1}{2},\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )-(n-1) (a x+1)^{\frac {n+1}{2}}\right )}{a^2 \left (n^2-1\right ) \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(1/2 - n/2)*Sqrt[1 - a^2*x^2]*(-((-1 + n)*(1 + a*x)^((1 + n)/2)) + 2^((3 + n)/2)*n*Hypergeometric2F
1[-1/2 - n/2, 1/2 - n/2, 3/2 - n/2, 1/2 - (a*x)/2]))/(a^2*(-1 + n^2)*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} x \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.28, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )} x}{\sqrt {-a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(1/2),x)

[Out]

int((x*exp(n*atanh(a*x)))/(c - a^2*c*x^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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