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

Optimal. Leaf size=47 \[ \frac {c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}-\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6175, 6178, 266, 47, 63, 208} \[ \frac {c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}-\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[a*x]*(c - a*c*x),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 6175

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

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(a x)} (c-a c x) \, dx &=-\left ((a c) \int e^{\coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right ) x \, dx\right )\\ &=(a c) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{x^3} \, dx,x,\frac {1}{x}\right )\\ &=\frac {1}{2} (a c) \operatorname {Subst}\left (\int \frac {\sqrt {1-\frac {x}{a^2}}}{x^2} \, dx,x,\frac {1}{x^2}\right )\\ &=-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {c \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a}\\ &=-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {1}{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 )\\ &=-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 51, normalized size = 1.09 \[ \frac {c \left (\log \left (a x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )-a^2 x^2 \sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 77, normalized size = 1.64 \[ \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (a^{2} c x^{2} + a c x\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.20, size = 154, normalized size = 3.28 \[ \frac {1}{4} \, a c {\left (\frac {\log \left (\sqrt {\frac {a x - 1}{a x + 1}} + \frac {1}{\sqrt {\frac {a x - 1}{a x + 1}}} + 2\right )}{a^{2}} - \frac {\log \left ({\left | \sqrt {\frac {a x - 1}{a x + 1}} + \frac {1}{\sqrt {\frac {a x - 1}{a x + 1}}} - 2 \right |}\right )}{a^{2}} - \frac {4 \, {\left (\sqrt {\frac {a x - 1}{a x + 1}} + \frac {1}{\sqrt {\frac {a x - 1}{a x + 1}}}\right )}}{{\left ({\left (\sqrt {\frac {a x - 1}{a x + 1}} + \frac {1}{\sqrt {\frac {a x - 1}{a x + 1}}}\right )}^{2} - 4\right )} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*c*(log(sqrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)) + 2)/a^2 - log(abs(sqrt((a*x - 1)/(a*x +
1)) + 1/sqrt((a*x - 1)/(a*x + 1)) - 2))/a^2 - 4*(sqrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)))/(((s
qrt((a*x - 1)/(a*x + 1)) + 1/sqrt((a*x - 1)/(a*x + 1)))^2 - 4)*a^2))

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maple [B]  time = 0.04, size = 93, normalized size = 1.98 \[ -\frac {\left (a x -1\right ) c \left (x \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}-\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{2 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.31, size = 132, normalized size = 2.81 \[ \frac {1}{2} \, a {\left (\frac {2 \, {\left (c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + c \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} + \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(2*(c*((a*x - 1)/(a*x + 1))^(3/2) + c*sqrt((a*x - 1)/(a*x + 1)))/(2*(a*x - 1)*a^2/(a*x + 1) - (a*x - 1)^
2*a^2/(a*x + 1)^2 - 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 = 1.20, size = 94, normalized size = 2.00 \[ \frac {c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}-\frac {c\,\sqrt {\frac {a\,x-1}{a\,x+1}}+c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a-\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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