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

Optimal. Leaf size=65 \[ -\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}}+2 c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {3 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6175, 6178, 1807, 807, 266, 63, 208} \[ -\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}}+2 c x \sqrt {1-\frac {1}{a^2 x^2}}-\frac {3 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)/E^ArcCoth[a*x],x]

[Out]

2*c*Sqrt[1 - 1/(a^2*x^2)]*x - (a*c*Sqrt[1 - 1/(a^2*x^2)]*x^2)/2 - (3*c*ArcTanh[Sqrt[1 - 1/(a^2*x^2)]])/(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 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 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 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 {\left (1-\frac {x}{a}\right )^2}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=-\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 {\frac {4}{a}-\frac {3 x}{a^2}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=2 c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {(3 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=2 c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2+\frac {(3 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a}\\ &=2 c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{2} (3 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 )\\ &=2 c \sqrt {1-\frac {1}{a^2 x^2}} x-\frac {1}{2} a c \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {3 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 53, normalized size = 0.82 \[ -\frac {c \left (a x \sqrt {1-\frac {1}{a^2 x^2}} (a x-4)+3 \log \left (a x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )\right )}{2 a} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/2*(c*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(-4 + a*x) + 3*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/a

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fricas [A]  time = 0.60, size = 81, normalized size = 1.25 \[ -\frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} c x^{2} - 3 \, a c x - 4 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 68, normalized size = 1.05 \[ \frac {3 \, c \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right ) \mathrm {sgn}\left (a x + 1\right )}{2 \, {\left | a \right |}} - \frac {1}{2} \, \sqrt {a^{2} x^{2} - 1} {\left (c x \mathrm {sgn}\left (a x + 1\right ) - \frac {4 \, c \mathrm {sgn}\left (a x + 1\right )}{a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 153, normalized size = 2.35 \[ \frac {\sqrt {\frac {a x -1}{a x +1}}\, \left (a x +1\right ) c \left (-\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a +4 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}+\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a -4 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{2 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, a \sqrt {a^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x-1)/(a*x+1))^(1/2)*(a*x+1)*c*(-(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x*a+4*((a*x-1)*(a*x+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-4*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 = 0.31, size = 135, normalized size = 2.08 \[ \frac {1}{2} \, a {\left (\frac {2 \, {\left (5 \, c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 3 \, c \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} - \frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {3 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(2*(5*c*((a*x - 1)/(a*x + 1))^(3/2) - 3*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) - 3*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 3*c*log(sqrt((a*x - 1)/(a*x + 1))
- 1)/a^2)

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mupad [B]  time = 0.06, size = 96, normalized size = 1.48 \[ \frac {3\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}-5\,c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a-\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}}-\frac {3\,c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a} \]

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]

(3*c*((a*x - 1)/(a*x + 1))^(1/2) - 5*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) - (3*c*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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