3.14.38 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x \sqrt {c-a^2 c x^2}} \, dx\) [1338]

Optimal. Leaf size=101 \[ -\frac {2 (1-a x)^{\frac {1-n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \, _2F_1\left (1,\frac {1-n}{2};\frac {3-n}{2};\frac {1-a x}{1+a x}\right )}{(1-n) \sqrt {c-a^2 c x^2}} \]

[Out]

-2*(-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 = 0.15, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6288, 6285, 133} \begin {gather*} -\frac {2 \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} (a x+1)^{\frac {n-1}{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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(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 133

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)/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)))*Hypergeometric2F1[m + 1, -n, m + 2,
(-(d*e - c*f))*((a + b*x)/((b*c - a*d)*(e + f*x)))], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0] && (SumSimplerQ[m, 1] ||  !SumSimplerQ[p, 1]) &&  !ILtQ[m, 0]

Rule 6285

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 6288

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 \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 {2 (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 = 0.03, size = 99, normalized size = 0.98 \begin {gather*} \frac {2 (1-a x)^{\frac {1}{2}-\frac {n}{2}} (1+a x)^{\frac {1}{2} (-1+n)} \sqrt {1-a^2 x^2} \, _2F_1\left (1,\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1-a x}{1+a x}\right )}{(-1+n) \sqrt {c-a^2 c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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