3.153 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x} \, dx\)

Optimal. Leaf size=111 \[ \frac {2 (1-a x)^{-n/2} (a x+1)^{n/2} \, _2F_1\left (1,-\frac {n}{2};1-\frac {n}{2};\frac {1-a x}{a x+1}\right )}{n}-\frac {2^{\frac {n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{n} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6126, 105, 69, 131} \[ \frac {2 (1-a x)^{-n/2} (a x+1)^{n/2} \, _2F_1\left (1,-\frac {n}{2};1-\frac {n}{2};\frac {1-a x}{a x+1}\right )}{n}-\frac {2^{\frac {n}{2}+1} (1-a x)^{-n/2} \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a*x])/x,x]

[Out]

(2*(1 + a*x)^(n/2)*Hypergeometric2F1[1, -n/2, 1 - n/2, (1 - a*x)/(1 + a*x)])/(n*(1 - a*x)^(n/2)) - (2^(1 + n/2
)*Hypergeometric2F1[-n/2, -n/2, 1 - n/2, (1 - a*x)/2])/(n*(1 - a*x)^(n/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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -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 6126

Int[E^(ArcTanh[(a_.)*(x_)]*(n_))*(x_)^(m_.), x_Symbol] :> Int[(x^m*(1 + a*x)^(n/2))/(1 - a*x)^(n/2), x] /; Fre
eQ[{a, m, n}, x] &&  !IntegerQ[(n - 1)/2]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 95, normalized size = 0.86 \[ \frac {2 (1-a x)^{-n/2} \left ((a x+1)^{n/2} \, _2F_1\left (1,-\frac {n}{2};1-\frac {n}{2};\frac {1-a x}{a x+1}\right )-2^{n/2} \, _2F_1\left (-\frac {n}{2},-\frac {n}{2};1-\frac {n}{2};\frac {1}{2} (1-a x)\right )\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a*x])/x,x]

[Out]

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

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fricas [F]  time = 0.92, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {a x + 1}{a x - 1}\right )^{\frac {1}{2} \, n}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/x,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/x,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (a x \right )}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(a*x))/x,x)

[Out]

int(exp(n*arctanh(a*x))/x,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(a*x))/x,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a*x))/x,x)

[Out]

int(exp(n*atanh(a*x))/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(a*x))/x,x)

[Out]

Integral(exp(n*atanh(a*x))/x, x)

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