3.9.81 \(\int \frac {e^{n \tanh ^{-1}(a+b x)}}{x^2} \, dx\) [881]

Optimal. Leaf size=92 \[ -\frac {4 b (1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {1}{2} (-2+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(1+a) (1-a-b x)}{(1-a) (1+a+b x)}\right )}{(1-a)^2 (2-n)} \]

[Out]

-4*b*(-b*x-a+1)^(1-1/2*n)*(b*x+a+1)^(-1+1/2*n)*hypergeom([2, 1-1/2*n],[2-1/2*n],(1+a)*(-b*x-a+1)/(1-a)/(b*x+a+
1))/(1-a)^2/(2-n)

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Rubi [A]
time = 0.03, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6298, 133} \begin {gather*} -\frac {4 b (-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n-2}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(a+1) (-a-b x+1)}{(1-a) (a+b x+1)}\right )}{(1-a)^2 (2-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])/x^2,x]

[Out]

(-4*b*(1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^((-2 + n)/2)*Hypergeometric2F1[2, 1 - n/2, 2 - n/2, ((1 + a)*(1 -
a - b*x))/((1 - a)*(1 + a + b*x))])/((1 - a)^2*(2 - n))

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 6298

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[(d + e*x)^m*((1
+ a*c + b*c*x)^(n/2)/(1 - a*c - b*c*x)^(n/2)), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rubi steps

\begin {align*} \int \frac {e^{n \tanh ^{-1}(a+b x)}}{x^2} \, dx &=\int \frac {(1-a-b x)^{-n/2} (1+a+b x)^{n/2}}{x^2} \, dx\\ &=-\frac {4 b (1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {1}{2} (-2+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(1+a) (1-a-b x)}{(1-a) (1+a+b x)}\right )}{(1-a)^2 (2-n)}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 83, normalized size = 0.90 \begin {gather*} \frac {4 b (1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{-1+\frac {n}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(1+a) (-1+a+b x)}{(-1+a) (1+a+b x)}\right )}{(-1+a)^2 (-2+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])/x^2,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(b*x+a))/x^2,x)

[Out]

int(exp(n*arctanh(b*x+a))/x^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(b*x+a))/x^2,x, algorithm="maxima")

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(b*x+a))/x**2,x)

[Out]

Integral(exp(n*atanh(a + b*x))/x**2, 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(b*x+a))/x^2,x, algorithm="giac")

[Out]

integrate((-(b*x + a + 1)/(b*x + a - 1))^(1/2*n)/x^2, 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+b\,x\right )}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a + b*x))/x^2,x)

[Out]

int(exp(n*atanh(a + b*x))/x^2, x)

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