3.2.53 \(\int \frac {e^{n \tanh ^{-1}(a x)}}{x} \, dx\) [153]

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

[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 = 0.03, 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 = {6261, 132, 71, 133} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 71

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], 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 132

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

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 6261

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 = 0.01, size = 95, normalized size = 0.86 \begin {gather*} \frac {2 (1-a x)^{-n/2} \left ((1+a x)^{n/2} \, _2F_1\left (1,-\frac {n}{2};1-\frac {n}{2};\frac {1-a x}{1+a x}\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} \end {gather*}

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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Maple [F]
time = 0.01, 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 \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,x, algorithm="maxima")

[Out]

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

[Out]

integral((-(a*x + 1)/(a*x - 1))^(1/2*n)/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}\, dx \end {gather*}

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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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,x, algorithm="giac")

[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 \begin {gather*} \int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a\,x\right )}}{x} \,d x \end {gather*}

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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