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

Optimal. Leaf size=182 \[ -\frac {2^{\frac {n+3}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^2 \left (1-n^2\right ) x \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{a^2 (n+1) x \sqrt {c-\frac {c}{a^2 x^2}}} \]

[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)/x/(c-c/a^2/x^2)^(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
)/x/(c-c/a^2/x^2)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {6160, 6150, 79, 69} \[ -\frac {2^{\frac {n+3}{2}} n \sqrt {1-a^2 x^2} (1-a x)^{\frac {1-n}{2}} \, _2F_1\left (\frac {1}{2} (-n-1),\frac {1-n}{2};\frac {3-n}{2};\frac {1}{2} (1-a x)\right )}{a^2 \left (1-n^2\right ) x \sqrt {c-\frac {c}{a^2 x^2}}}-\frac {\sqrt {1-a^2 x^2} (a x+1)^{\frac {n+1}{2}} (1-a x)^{\frac {1-n}{2}}}{a^2 (n+1) x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

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 6160

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

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a x)}}{\sqrt {c-\frac {c}{a^2 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-\frac {c}{a^2 x^2}} x}\\ &=\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-\frac {c}{a^2 x^2}} x}\\ &=-\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-\frac {c}{a^2 x^2}} x}+\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-\frac {c}{a^2 x^2}} x}\\ &=-\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-\frac {c}{a^2 x^2}} x}-\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-\frac {c}{a^2 x^2}} x}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 130, normalized size = 0.71 \[ \frac {\sqrt {1-a^2 x^2} (1-a x)^{\frac {1}{2}-\frac {n}{2}} \left (2^{\frac {n+3}{2}} n \, _2F_1\left (-\frac {n}{2}-\frac {1}{2},\frac {1}{2}-\frac {n}{2};\frac {3}{2}-\frac {n}{2};\frac {1}{2}-\frac {a x}{2}\right )-(n-1) (a x+1)^{\frac {n+1}{2}}\right )}{a^2 (n-1) (n+1) x \sqrt {c-\frac {c}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])/Sqrt[c - c/(a^2*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)*(1 + n)*Sqrt[c - c/(a^2*x^2)]*x)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} x^{2} \left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n} \sqrt {\frac {a^{2} c x^{2} - c}{a^{2} x^{2}}}}{a^{2} c x^{2} - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{\sqrt {c -\frac {c}{a^{2} x^{2}}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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