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

Optimal. Leaf size=49 \[ c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {2 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}-\frac {c \csc ^{-1}(a x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 49, 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, 1807, 844, 216, 266, 63, 208} \[ c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {2 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{a}-\frac {c \csc ^{-1}(a x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c/(a*x))/E^ArcCoth[a*x],x]

[Out]

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

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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rule 1807

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

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

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Mathematica [A]  time = 0.11, size = 73, normalized size = 1.49 \[ \frac {c \left (a x \sqrt {1-\frac {1}{a^2 x^2}}-2 \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )-2 \sin ^{-1}\left (\frac {\sqrt {1-\frac {1}{a x}}}{\sqrt {2}}\right )-2 \sin ^{-1}\left (\frac {1}{a x}\right )\right )}{a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.52, size = 88, normalized size = 1.80 \[ \frac {2 \, c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right ) - 2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) + 2 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a c x + c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 85, normalized size = 1.73 \[ \frac {2 \, c \arctan \left (-x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1}\right ) \mathrm {sgn}\left (a x + 1\right )}{a} + \frac {2 \, c \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{{\left | a \right |}} + \frac {\sqrt {a^{2} x^{2} - 1} c \mathrm {sgn}\left (a x + 1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.05, size = 137, normalized size = 2.80 \[ \frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c \left (2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}-\arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right ) \sqrt {a^{2}}-2 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.41, size = 114, normalized size = 2.33 \[ -2 \, a {\left (\frac {c \sqrt {\frac {a x - 1}{a x + 1}}}{\frac {{\left (a x - 1\right )} a^{2}}{a x + 1} - a^{2}} - \frac {c \arctan \left (\sqrt {\frac {a x - 1}{a x + 1}}\right )}{a^{2}} + \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.07, size = 82, normalized size = 1.67 \[ \frac {2\,c\,\mathrm {atan}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}-\frac {4\,c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {2\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a-\frac {a\,\left (a\,x-1\right )}{a\,x+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c*atan(((a*x - 1)/(a*x + 1))^(1/2)))/a - (4*c*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a + (2*c*((a*x - 1)/(a*x
+ 1))^(1/2))/(a - (a*(a*x - 1))/(a*x + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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