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

Optimal. Leaf size=104 \[ \frac {x \sqrt {\frac {1}{a x}+1}}{c \sqrt {1-\frac {1}{a x}}}-\frac {2 \sqrt {\frac {1}{a x}+1}}{a c \sqrt {1-\frac {1}{a x}}}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}\right )}{a c} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {6194, 103, 21, 94, 92, 208} \[ \frac {x \sqrt {\frac {1}{a x}+1}}{c \sqrt {1-\frac {1}{a x}}}-\frac {2 \sqrt {\frac {1}{a x}+1}}{a c \sqrt {1-\frac {1}{a x}}}+\frac {\tanh ^{-1}\left (\sqrt {1-\frac {1}{a x}} \sqrt {\frac {1}{a x}+1}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 92

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

Rule 94

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

Rule 103

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, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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

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Mathematica [A]  time = 0.14, size = 56, normalized size = 0.54 \[ \frac {\frac {x \sqrt {1-\frac {1}{a^2 x^2}} (a x-2)}{a x-1}+\frac {\log \left (x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )}{a}}{c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 93, normalized size = 0.89 \[ \frac {{\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - {\left (a x - 1\right )} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} x^{2} - a x - 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c x - a c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 127, normalized size = 1.22 \[ a {\left (\frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} - \frac {\log \left ({\left | \sqrt {\frac {a x - 1}{a x + 1}} - 1 \right |}\right )}{a^{2} c} - \frac {\frac {3 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c {\left (\frac {{\left (a x - 1\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a x + 1} - \sqrt {\frac {a x - 1}{a x + 1}}\right )}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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) - (3*(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 = 251, normalized size = 2.41 \[ \frac {2 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+3 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}-4 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}-\left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-6 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x a +2 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )+3 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{2 a \sqrt {a^{2}}\, \left (a x -1\right ) c \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {\frac {a x -1}{a x +1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(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+3*((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)-6*((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))
+3*((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 = 0.32, size = 116, normalized size = 1.12 \[ -a {\left (\frac {\frac {3 \, {\left (a x - 1\right )}}{a x + 1} - 1}{a^{2} c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - a^{2} c \sqrt {\frac {a x - 1}{a x + 1}}} - \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2} c} + \frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2} c}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*((3*(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(sq
rt((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 = 0.08, size = 62, normalized size = 0.60 \[ \frac {2\,a\,x+4\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )\,\sqrt {\frac {a\,x-1}{a\,x+1}}-4}{2\,a\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*x + 4*atanh(((a*x - 1)/(a*x + 1))^(1/2))*((a*x - 1)/(a*x + 1))^(1/2) - 4)/(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 \[ \frac {a^{2} \int \frac {x^{2}}{a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}} - \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}\, dx}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*Integral(x**2/(a**2*x**2*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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