3.882 \(\int \frac {e^{n \tanh ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac {(a+b x+1)^{\frac {n+2}{2}} (-a-b x+1)^{1-\frac {n}{2}}}{2 \left (1-a^2\right ) x^2}-\frac {2 b^2 (2 a+n) (a+b x+1)^{\frac {n-2}{2}} (-a-b x+1)^{1-\frac {n}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(a+1) (-a-b x+1)}{(1-a) (a+b x+1)}\right )}{(1-a)^3 (a+1) (2-n)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6163, 96, 131} \[ -\frac {(a+b x+1)^{\frac {n+2}{2}} (-a-b x+1)^{1-\frac {n}{2}}}{2 \left (1-a^2\right ) x^2}-\frac {2 b^2 (2 a+n) (a+b x+1)^{\frac {n-2}{2}} (-a-b x+1)^{1-\frac {n}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(a+1) (-a-b x+1)}{(1-a) (a+b x+1)}\right )}{(1-a)^3 (a+1) (2-n)} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcTanh[a + b*x])/x^3,x]

[Out]

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

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 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 6163

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^3} \, dx &=\int \frac {(1-a-b x)^{-n/2} (1+a+b x)^{n/2}}{x^3} \, dx\\ &=-\frac {(1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {2+n}{2}}}{2 \left (1-a^2\right ) x^2}+\frac {(b (2 a+n)) \int \frac {(1-a-b x)^{-n/2} (1+a+b x)^{n/2}}{x^2} \, dx}{2 \left (1-a^2\right )}\\ &=-\frac {(1-a-b x)^{1-\frac {n}{2}} (1+a+b x)^{\frac {2+n}{2}}}{2 \left (1-a^2\right ) x^2}-\frac {2 b^2 (2 a+n) (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)^3 (1+a) (2-n)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 123, normalized size = 0.81 \[ \frac {(-a-b x+1)^{1-\frac {n}{2}} (a+b x+1)^{\frac {n}{2}-1} \left ((a-1)^2 (n-2) (a+b x+1)^2-4 b^2 x^2 (2 a+n) \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {(a+1) (a+b x-1)}{(a-1) (a+b x+1)}\right )\right )}{2 (a-1)^3 (a+1) (n-2) x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcTanh[a + b*x])/x^3,x]

[Out]

((1 - a - b*x)^(1 - n/2)*(1 + a + b*x)^(-1 + n/2)*((-1 + a)^2*(-2 + n)*(1 + a + b*x)^2 - 4*b^2*(2*a + n)*x^2*H
ypergeometric2F1[2, 1 - n/2, 2 - n/2, ((1 + a)*(-1 + a + b*x))/((-1 + a)*(1 + a + b*x))]))/(2*(-1 + a)^3*(1 +
a)*(-2 + n)*x^2)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))/x^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))/x^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \arctanh \left (b x +a \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*arctanh(b*x+a))/x^3,x)

[Out]

int(exp(n*arctanh(b*x+a))/x^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\frac {b x + a + 1}{b x + a - 1}\right )^{\frac {1}{2} \, n}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arctanh(b*x+a))/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {e}}^{n\,\mathrm {atanh}\left (a+b\,x\right )}}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*atanh(a + b*x))/x^3,x)

[Out]

int(exp(n*atanh(a + b*x))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {e^{n \operatorname {atanh}{\left (a + b x \right )}}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*atanh(b*x+a))/x**3,x)

[Out]

Integral(exp(n*atanh(a + b*x))/x**3, x)

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