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

Optimal. Leaf size=70 \[ -\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c}+\frac {2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {6177, 852, 1805, 807, 266, 63, 208} \[ -\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x \sqrt {1-\frac {1}{a^2 x^2}}}{c}+\frac {2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]/(c - c/(a*x)),x]

[Out]

(-2*(a + x^(-1)))/(a^2*c*Sqrt[1 - 1/(a^2*x^2)]) + (Sqrt[1 - 1/(a^2*x^2)]*x)/c + (2*ArcTanh[Sqrt[1 - 1/(a^2*x^2
)]])/(a*c)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 852

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[((f + g*x)^n*(a + c*x^2)^(m + p))/(d - e*x)^m, x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 6177

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

Rubi steps

\begin {align*} \int \frac {e^{\coth ^{-1}(a x)}}{c-\frac {c}{a x}} \, dx &=-\left (c \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{x^2 \left (c-\frac {c x}{a}\right )^2} \, dx,x,\frac {1}{x}\right )\right )\\ &=-\frac {\operatorname {Subst}\left (\int \frac {\left (c+\frac {c x}{a}\right )^2}{x^2 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=-\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\operatorname {Subst}\left (\int \frac {-c^2-\frac {2 c^2 x}{a}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{c^3}\\ &=-\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{a c}\\ &=-\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{a c}\\ &=-\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right )}{c}\\ &=-\frac {2 \left (a+\frac {1}{x}\right )}{a^2 c \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\sqrt {1-\frac {1}{a^2 x^2}} x}{c}+\frac {2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a c}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 63, normalized size = 0.90 \[ \frac {a x \sqrt {1-\frac {1}{a^2 x^2}} (a x-3)+2 (a x-1) \log \left (x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )}{a c (a x-1)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*Sqrt[1 - 1/(a^2*x^2)]*x*(-3 + a*x) + 2*(-1 + a*x)*Log[(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/(a*c*(-1 + a*x))

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fricas [A]  time = 0.54, size = 94, normalized size = 1.34 \[ \frac {2 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 2 \, {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} x^{2} - 2 \, a x - 3\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a*x - 1)*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 2*(a*x - 1)*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (a^2*x^2 -
2*a*x - 3)*sqrt((a*x - 1)/(a*x + 1)))/(a^2*c*x - a*c)

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giac [A]  time = 0.15, size = 128, normalized size = 1.83 \[ 2 \, a {\left (\frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac {\log \left ({\left | \sqrt {\frac {a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2} c} - \frac {\frac {2 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c {\left (\frac {{\left (a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1} - \sqrt {\frac {a x - 1}{a x + 1}}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*(log(sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c) - log(abs(sqrt((a*x - 1)/(a*x + 1)) - 1))/(a^2*c) - (2*(a*x -
1)/(a*x + 1) - 1)/(a^2*c*((a*x - 1)*sqrt((a*x - 1)/(a*x + 1))/(a*x + 1) - sqrt((a*x - 1)/(a*x + 1)))))

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maple [B]  time = 0.06, size = 250, normalized size = 3.57 \[ \frac {2 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}-4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}-\left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x a +2 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )+2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{a \sqrt {a^{2}}\, \left (a x -1\right ) c \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {\frac {a x -1}{a x +1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+2*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x
^2*a^2-4*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)
-4*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x*a+2*a*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))+2*(
(a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/a/(a^2)^(1/2)/(a*x-1)/c/((a*x-1)*(a*x+1))^(1/2)/((a*x-1)/(a*x+1))^(1/2)

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maxima [A]  time = 0.30, size = 116, normalized size = 1.66 \[ -2 \, a {\left (\frac {\frac {2 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - a^{2} c \sqrt {\frac {a x - 1}{a x + 1}}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*((2*(a*x - 1)/(a*x + 1) - 1)/(a^2*c*((a*x - 1)/(a*x + 1))^(3/2) - a^2*c*sqrt((a*x - 1)/(a*x + 1))) - log(
sqrt((a*x - 1)/(a*x + 1)) + 1)/(a^2*c) + log(sqrt((a*x - 1)/(a*x + 1)) - 1)/(a^2*c))

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mupad [B]  time = 1.22, size = 62, normalized size = 0.89 \[ \frac {2\,a\,x+8\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}-6}{2\,a\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x + 8*atanh(((a*x - 1)/(a*x + 1))^(1/2))*((a*x - 1)/(a*x + 1))^(1/2) - 6)/(2*a*c*((a*x - 1)/(a*x + 1))^(1
/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a \int \frac {x}{a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Integral(x/(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - sqrt(a*x/(a*x + 1) - 1/(a*x + 1))), x)/c

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