3.928 \(\int e^{n \coth ^{-1}(a x)} (c-\frac {c}{a^2 x^2}) \, dx\)

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

[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 [C]  time = 0.07, antiderivative size = 81, normalized size of antiderivative = 0.53, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6194, 136} \[ -\frac {c 2^{2-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+4}{2}} F_1\left (\frac {n+4}{2};\frac {n-2}{2},2;\frac {n+6}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (n+4)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(2 - n/2)*c*(1 + 1/(a*x))^((4 + n)/2)*AppellF1[(4 + n)/2, (-2 + n)/2, 2, (6 + n)/2, (a + x^(-1))/(2*a), 1
+ 1/(a*x)])/(a*(4 + n)))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 6194

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 \operatorname {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 = 0.26, size = 123, normalized size = 0.80 \[ \frac {c e^{n \coth ^{-1}(a x)} \left (n e^{2 \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;e^{2 \coth ^{-1}(a x)}\right )+(n+2) \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;e^{2 \coth ^{-1}(a x)}\right )+4 e^{2 \coth ^{-1}(a x)} \, _2F_1\left (2,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \coth ^{-1}(a x)}\right )+a n x+2 a x\right )}{a (n+2)} \]

Warning: Unable to verify antiderivative.

[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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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} c x^{2} - c\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a^{2} x^{2}}, x\right ) \]

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

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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maple [F]  time = 0.06, 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 \[ \int {\left (c - \frac {c}{a^{2} x^{2}}\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}\,{d x} \]

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

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

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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