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

Optimal. Leaf size=150 \[ \frac {2 \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2} \, _2F_1\left (1,\frac {n}{2};\frac {n+2}{2};\frac {a+\frac {1}{x}}{a-\frac {1}{x}}\right )}{a c}-\frac {(n+1) \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2}}{a c n}+\frac {x \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2}}{c} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 164, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {6194, 129, 155, 12, 131} \[ \frac {2 n \left (\frac {1}{a x}+1\right )^{\frac {n-2}{2}} \left (1-\frac {1}{a x}\right )^{1-\frac {n}{2}} \, _2F_1\left (1,1-\frac {n}{2};2-\frac {n}{2};\frac {a-\frac {1}{x}}{a+\frac {1}{x}}\right )}{a c (2-n)}-\frac {(n+1) \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2}}{a c n}+\frac {x \left (\frac {1}{a x}+1\right )^{n/2} \left (1-\frac {1}{a x}\right )^{-n/2}}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 129

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[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*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 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 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

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

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Mathematica [A]  time = 0.27, size = 94, normalized size = 0.63 \[ \frac {e^{n \coth ^{-1}(a x)} \left (n^2 e^{2 \coth ^{-1}(a x)} \, _2F_1\left (1,\frac {n}{2}+1;\frac {n}{2}+2;e^{2 \coth ^{-1}(a x)}\right )+(n+2) \left (n \, _2F_1\left (1,\frac {n}{2};\frac {n}{2}+1;e^{2 \coth ^{-1}(a x)}\right )+a n x-1\right )\right )}{a c n (n+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} x^{2} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {1}{2} \, n}}{a^{2} c x^{2} - c}, x\right ) \]

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

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

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {{\mathrm e}^{n \,\mathrm {arccoth}\left (a x \right )}}{c -\frac {c}{a^{2} x^{2}}}\, dx \]

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

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

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

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 \[ \frac {a^{2} \int \frac {x^{2} e^{n \operatorname {acoth}{\left (a x \right )}}}{a^{2} x^{2} - 1}\, dx}{c} \]

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]

a**2*Integral(x**2*exp(n*acoth(a*x))/(a**2*x**2 - 1), x)/c

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