3.219 \(\int e^{-3 \coth ^{-1}(a x)} (c-a c x) \, dx\)

Optimal. Leaf size=92 \[ -\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}}+4 c x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {15 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

[Out]

-15/2*c*arctanh((1-1/a^2/x^2)^(1/2))/a+8*c*(a-1/x)/a^2/(1-1/a^2/x^2)^(1/2)+4*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 = 0.24, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6175, 6178, 1805, 1807, 807, 266, 63, 208} \[ -\frac {1}{2} a c x^2 \sqrt {1-\frac {1}{a^2 x^2}}+4 c x \sqrt {1-\frac {1}{a^2 x^2}}+\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\frac {15 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[(c - a*c*x)/E^(3*ArcCoth[a*x]),x]

[Out]

(8*c*(a - x^(-1)))/(a^2*Sqrt[1 - 1/(a^2*x^2)]) + 4*c*Sqrt[1 - 1/(a^2*x^2)]*x - (a*c*Sqrt[1 - 1/(a^2*x^2)]*x^2)
/2 - (15*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 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 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^{-3 \coth ^{-1}(a x)} (c-a c x) \, dx &=-\left ((a c) \int e^{-3 \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 )^4}{x^3 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-(a c) \operatorname {Subst}\left (\int \frac {-1+\frac {4 x}{a}-\frac {7 x^2}{a^2}}{x^3 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}-\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 {8}{a}+\frac {15 x}{a^2}}{x^2 \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+4 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 {(15 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x^2}{a^2}}} \, dx,x,\frac {1}{x}\right )}{2 a}\\ &=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+4 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 {(15 c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{4 a}\\ &=\frac {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+4 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} (15 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 {8 c \left (a-\frac {1}{x}\right )}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+4 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 {15 c \tanh ^{-1}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{2 a}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 68, normalized size = 0.74 \[ \frac {1}{2} c \left (\frac {x \sqrt {1-\frac {1}{a^2 x^2}} \left (-a^2 x^2+7 a x+24\right )}{a x+1}-\frac {15 \log \left (a x \left (\sqrt {1-\frac {1}{a^2 x^2}}+1\right )\right )}{a}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c - a*c*x)/E^(3*ArcCoth[a*x]),x]

[Out]

(c*((Sqrt[1 - 1/(a^2*x^2)]*x*(24 + 7*a*x - a^2*x^2))/(1 + a*x) - (15*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x])/a))
/2

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fricas [A]  time = 0.53, size = 81, normalized size = 0.88 \[ -\frac {15 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 15 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (a^{2} c x^{2} - 7 \, a c x - 24 \, c\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="fricas")

[Out]

-1/2*(15*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (a^2*c*x^2 - 7*a*c*x
- 24*c)*sqrt((a*x - 1)/(a*x + 1)))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {undef} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="giac")

[Out]

undef

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maple [B]  time = 0.05, size = 422, normalized size = 4.59 \[ -\frac {\left (\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{3} a^{3}-16 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x^{2} a^{2}+2 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x^{2} a^{2}-\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+16 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x^{2} a^{3}+8 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-32 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}\, x a +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, x a -2 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}+32 \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) x \,a^{2}-16 \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}-\ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a +16 a \ln \left (\frac {a^{2} x +\sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right ) c \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}{2 a \sqrt {a^{2}}\, \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \left (a x -1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x^3*a^3-16*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)*x^2*a^2+2*(a^2*x^2-1)^(1/2)
*(a^2)^(1/2)*x^2*a^2-ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+16*ln((a^2*x+((a*x-1)*(a*x+
1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x^2*a^3+8*((a*x-1)*(a*x+1))^(3/2)*(a^2)^(1/2)-32*((a*x-1)*(a*x+1))^(1/2)*(
a^2)^(1/2)*x*a+(a^2*x^2-1)^(1/2)*(a^2)^(1/2)*x*a-2*ln((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2
+32*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*x*a^2-16*((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2)-l
n((a^2*x+(a^2*x^2-1)^(1/2)*(a^2)^(1/2))/(a^2)^(1/2))*a+16*a*ln((a^2*x+((a*x-1)*(a*x+1))^(1/2)*(a^2)^(1/2))/(a^
2)^(1/2)))/a*c*((a*x-1)/(a*x+1))^(3/2)/(a^2)^(1/2)/((a*x-1)*(a*x+1))^(1/2)/(a*x-1)

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maxima [A]  time = 0.31, size = 156, normalized size = 1.70 \[ \frac {1}{2} \, a {\left (\frac {2 \, {\left (9 \, c \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 7 \, c \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {2 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {{\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} - a^{2}} - \frac {15 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} + \frac {15 \, c \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {16 \, c \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)*((a*x-1)/(a*x+1))^(3/2),x, algorithm="maxima")

[Out]

1/2*a*(2*(9*c*((a*x - 1)/(a*x + 1))^(3/2) - 7*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) - 15*c*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 + 15*c*log(sqrt((a*x - 1)/(a*x + 1)
) - 1)/a^2 + 16*c*sqrt((a*x - 1)/(a*x + 1))/a^2)

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mupad [B]  time = 0.07, size = 117, normalized size = 1.27 \[ \frac {7\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}-9\,c\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{a-\frac {2\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}}-\frac {15\,c\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{a}+\frac {8\,c\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(7*c*((a*x - 1)/(a*x + 1))^(1/2) - 9*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) - (15*c*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/a + (8*c*((a*x - 1)/(a*x + 1))^(1/2))/a

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - c \left (\int \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx + \int \left (- \frac {2 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\right )\, dx + \int \frac {a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-a*c*x+c)*((a*x-1)/(a*x+1))**(3/2),x)

[Out]

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

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