3.10.28 \(\int e^{n \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\) [928]

Optimal. Leaf size=154 \[ \frac {4 c \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (1+\frac {1}{a x}\right )^{\frac {1}{2} (-2+n)} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (2-n)}-\frac {2^{1+\frac {n}{2}} c \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{a (2-n)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6329, 130, 71, 133} \begin {gather*} \frac {4 c \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n-2}{2}} \, _2F_1\left (2,1-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a (2-n)}-\frac {c 2^{\frac {n}{2}+1} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{a (2-n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(n*ArcCoth[a*x])*(c - c/(a^2*x^2)),x]

[Out]

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

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 130

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_))^2, x_Symbol] :> Dist[b*(d/f^2),
Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x] + Dist[(b*e - a*f)*((d*e - c*f)/f^2), Int[(a + b*x)^(m - 1)*(
(c + d*x)^(n - 1)/(e + f*x)^2), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[m + n, 0] && EqQ[2*b*d*e
- f*(b*c + a*d), 0]

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 6329

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[-c^p, Subst[Int[(1 - x/a)^(p
- n/2)*((1 + x/a)^(p + n/2)/x^2), x], x, 1/x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[c + a^2*d, 0] &&  !Integ
erQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegersQ[2*p, p + n/2]

Rubi steps

\begin {align*} \int e^{n \coth ^{-1}(a x)} \left (c-\frac {c}{a^2 x^2}\right ) \, dx &=-\left (c \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right )^{1-\frac {n}{2}} \left (1+\frac {x}{a}\right )^{1+\frac {n}{2}}}{x^2} \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {2^{2-\frac {n}{2}} c \left (1+\frac {1}{a x}\right )^{\frac {4+n}{2}} F_1\left (\frac {4+n}{2};\frac {1}{2} (-2+n),2;\frac {6+n}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (4+n)}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 123, normalized size = 0.80 \begin {gather*} \frac {c e^{n \coth ^{-1}(a x)} \left (2 a x+a n x+e^{2 \coth ^{-1}(a x)} n \, _2F_1\left (1,1+\frac {n}{2};2+\frac {n}{2};e^{2 \coth ^{-1}(a x)}\right )+(2+n) \, _2F_1\left (1,\frac {n}{2};1+\frac {n}{2};e^{2 \coth ^{-1}(a x)}\right )+4 e^{2 \coth ^{-1}(a x)} \, _2F_1\left (2,1+\frac {n}{2};2+\frac {n}{2};-e^{2 \coth ^{-1}(a x)}\right )\right )}{a (2+n)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(n*ArcCoth[a*x])*(c - c/(a^2*x^2)),x]

[Out]

(c*E^(n*ArcCoth[a*x])*(2*a*x + a*n*x + E^(2*ArcCoth[a*x])*n*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcCot
h[a*x])] + (2 + n)*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcCoth[a*x])] + 4*E^(2*ArcCoth[a*x])*Hypergeometri
c2F1[2, 1 + n/2, 2 + n/2, -E^(2*ArcCoth[a*x])]))/(a*(2 + n))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (c -\frac {c}{a^{2} x^{2}}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(exp(n*arccoth(a*x))*(c-c/a^2/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*arccoth(a*x))*(c-c/a^2/x^2),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {c \left (\int a^{2} e^{n \operatorname {acoth}{\left (a x \right )}}\, dx + \int \left (- \frac {e^{n \operatorname {acoth}{\left (a x \right )}}}{x^{2}}\right )\, dx\right )}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(Integral(a**2*exp(n*acoth(a*x)), x) + Integral(-exp(n*acoth(a*x))/x**2, x))/a**2

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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*arccoth(a*x))*(c-c/a^2/x^2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,\left (c-\frac {c}{a^2\,x^2}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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