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

Optimal. Leaf size=185 \[ -\frac {c 2^{n/2} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1-\frac {n}{2},1-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{2 a}\right )}{a (2-n)}-\frac {2 c (1-n) \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2} \, _2F_1\left (1,\frac {n}{2};\frac {n+2}{2};\frac {a+\frac {1}{x}}{a-\frac {1}{x}}\right )}{a n}+c x \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \]

[Out]

c*(1-1/a/x)^(1-1/2*n)*(1+1/a/x)^(1/2*n)*x-2*c*(1-n)*(1+1/a/x)^(1/2*n)*hypergeom([1, 1/2*n],[1+1/2*n],(a+1/x)/(
a-1/x))/a/n/((1-1/a/x)^(1/2*n))-2^(1/2*n)*c*(1-1/a/x)^(1-1/2*n)*hypergeom([1-1/2*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 = 0.06, antiderivative size = 81, normalized size of antiderivative = 0.44, 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 = {6179, 136} \[ -\frac {c 2^{2-\frac {n}{2}} \left (\frac {1}{a x}+1\right )^{\frac {n+2}{2}} F_1\left (\frac {n+2}{2};\frac {n-2}{2},2;\frac {n+4}{2};\frac {a+\frac {1}{x}}{2 a},1+\frac {1}{a x}\right )}{a (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((2^(2 - n/2)*c*(1 + 1/(a*x))^((2 + n)/2)*AppellF1[(2 + n)/2, (-2 + n)/2, 2, (4 + n)/2, (a + x^(-1))/(2*a), 1
+ 1/(a*x)])/(a*(2 + 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 6179

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

Rubi steps

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

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Mathematica [A]  time = 0.38, size = 155, normalized size = 0.84 \[ \frac {c e^{n \coth ^{-1}(a x)} \left (n \left (-e^{2 \coth ^{-1}(a x)}\right ) \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;-e^{2 \coth ^{-1}(a x)}\right )+(n-1) n e^{2 \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;e^{2 \coth ^{-1}(a x)}\right )+(n+2) \left (\, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;-e^{2 \coth ^{-1}(a x)}\right )+(n-1) \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;e^{2 \coth ^{-1}(a x)}\right )+a n x\right )\right )}{a n (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*E^(n*ArcCoth[a*x])*(-(E^(2*ArcCoth[a*x])*n*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, -E^(2*ArcCoth[a*x])]) + E
^(2*ArcCoth[a*x])*(-1 + n)*n*Hypergeometric2F1[1, 1 + n/2, 2 + n/2, E^(2*ArcCoth[a*x])] + (2 + n)*(a*n*x + Hyp
ergeometric2F1[1, n/2, 1 + n/2, -E^(2*ArcCoth[a*x])] + (-1 + n)*Hypergeometric2F1[1, n/2, 1 + n/2, E^(2*ArcCot
h[a*x])])))/(a*n*(2 + n))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a c x - c\right )} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))*(c-c/a/x),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*arccoth(a*x))*(c-c/a/x),x, algorithm="giac")

[Out]

integrate((c - c/(a*x))*((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 \[ \int {\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )} \left (c -\frac {c}{a x}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c - c/(a*x))*((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 \[ \int {\mathrm {e}}^{n\,\mathrm {acoth}\left (a\,x\right )}\,\left (c-\frac {c}{a\,x}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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