3.1339 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x^2 \sqrt {c-a^2 c x^2}} \, dx\)

Optimal. Leaf size=167 \[ -\frac {2 a n \sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {1-a x}{a x+1}\right )}{(1-n) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{x \sqrt {c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6153, 6150, 96, 131} \[ -\frac {2 a n \sqrt {1-a^2 x^2} (a x+1)^{\frac {n-1}{2}} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {1-a x}{a x+1}\right )}{(1-n) \sqrt {c-a^2 c x^2}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{x \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

Rule 96

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))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

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

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Mathematica [A]  time = 0.08, size = 117, normalized size = 0.70 \[ \frac {\sqrt {1-a^2 x^2} (1-a x)^{\frac {1}{2}-\frac {n}{2}} (a x+1)^{\frac {n-1}{2}} \left (2 a n x \, _2F_1\left (1,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1-a x}{a x+1}\right )-(n-1) (a x+1)\right )}{(n-1) x \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - a*x)^(1/2 - n/2)*(1 + a*x)^((-1 + n)/2)*Sqrt[1 - a^2*x^2]*(-((-1 + n)*(1 + a*x)) + 2*a*n*x*Hypergeometri
c2F1[1, 1/2 - n/2, 3/2 - n/2, (1 - a*x)/(1 + a*x)]))/((-1 + n)*x*Sqrt[c - a^2*c*x^2])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} c x^{2} + c} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{4} - c x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*c*x^2 + c)*((a*x + 1)/(a*x - 1))^(1/2*n)/(a^2*c*x^4 - c*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.27, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{x^{2} \sqrt {-a^{2} c \,x^{2}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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