\(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx\) [179]

Optimal result

Integrand size = 18, antiderivative size = 78 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {3}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4-\frac {3 c^3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \]

[Out]

-1/4*a^3*c^3*(1-1/a^2/x^2)^(3/2)*x^4-3/8*c^3*arctanh((1-1/a^2/x^2)^(1/2))/a+3/8*a*c^3*x^2*(1-1/a^2/x^2)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6310, 6313, 272, 43, 65, 214} \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {3 c^3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a}+\frac {3}{8} a c^3 x^2 \sqrt {1-\frac {1}{a^2 x^2}}-\frac {1}{4} a^3 c^3 x^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \]

[In]

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

[Out]

(3*a*c^3*Sqrt[1 - 1/(a^2*x^2)]*x^2)/8 - (a^3*c^3*(1 - 1/(a^2*x^2))^(3/2)*x^4)/4 - (3*c^3*ArcTanh[Sqrt[1 - 1/(a
^2*x^2)]])/(8*a)

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

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 214

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

Rule 272

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 6310

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 6313

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*} \text {integral}& = -\left (\left (a^3 c^3\right ) \int e^{3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^3 x^3 \, dx\right ) \\ & = \left (a^3 c^3\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x^2}{a^2}\right )^{3/2}}{x^5} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {1}{2} \left (a^3 c^3\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a^2}\right )^{3/2}}{x^3} \, dx,x,\frac {1}{x^2}\right ) \\ & = -\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4-\frac {1}{8} \left (3 a c^3\right ) \text {Subst}\left (\int \frac {\sqrt {1-\frac {x}{a^2}}}{x^2} \, dx,x,\frac {1}{x^2}\right ) \\ & = \frac {3}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {\left (3 c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {1-\frac {x}{a^2}}} \, dx,x,\frac {1}{x^2}\right )}{16 a} \\ & = \frac {3}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4-\frac {1}{8} \left (3 a c^3\right ) \text {Subst}\left (\int \frac {1}{a^2-a^2 x^2} \, dx,x,\sqrt {1-\frac {1}{a^2 x^2}}\right ) \\ & = \frac {3}{8} a c^3 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4-\frac {3 c^3 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.82 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {c^3 \left (a^2 \sqrt {1-\frac {1}{a^2 x^2}} x^2 \left (5-2 a^2 x^2\right )-3 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{8 a} \]

[In]

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

[Out]

(c^3*(a^2*Sqrt[1 - 1/(a^2*x^2)]*x^2*(5 - 2*a^2*x^2) - 3*Log[a*(1 + Sqrt[1 - 1/(a^2*x^2)])*x]))/(8*a)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.36

method	result	size
risch	\(-\frac {x \left (2 a^{2} x^{2}-5\right ) \left (a x -1\right ) c^{3}}{8 \sqrt {\frac {a x -1}{a x +1}}}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c^{3} \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 \sqrt {a^{2}}\, \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(106\)
default	\(-\frac {\left (a x -1\right )^{2} c^{3} \left (2 x \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}-3 x \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}+3 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right )\right )}{8 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}\)	\(124\)

[In]

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

[Out]

-1/8*x*(2*a^2*x^2-5)*(a*x-1)*c^3/((a*x-1)/(a*x+1))^(1/2)-3/8*ln(a^2*x/(a^2)^(1/2)+(a^2*x^2-1)^(1/2))/(a^2)^(1/
2)*c^3/(a*x+1)/((a*x-1)/(a*x+1))^(1/2)*((a*x-1)*(a*x+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.40 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {3 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 3 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) + {\left (2 \, a^{4} c^{3} x^{4} + 2 \, a^{3} c^{3} x^{3} - 5 \, a^{2} c^{3} x^{2} - 5 \, a c^{3} x\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{8 \, a} \]

[In]

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

[Out]

-1/8*(3*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 3*c^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1) + (2*a^4*c^3*x^4 + 2
*a^3*c^3*x^3 - 5*a^2*c^3*x^2 - 5*a*c^3*x)*sqrt((a*x - 1)/(a*x + 1)))/a

Sympy [F]

\[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=- c^{3} \left (\int \frac {3 a x}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx + \int \left (- \frac {3 a^{2} x^{2}}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\right )\, dx + \int \frac {a^{3} x^{3}}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx + \int \left (- \frac {1}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\right )\, dx\right ) \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 221 vs. \(2 (66) = 132\).

Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.83 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{8} \, {\left (\frac {3 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {3 \, c^{3} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} + \frac {2 \, {\left (3 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 11 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} - 11 \, c^{3} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} + 3 \, c^{3} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {4 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {6 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {4 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {{\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} - a^{2}}\right )} a \]

[In]

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

[Out]

-1/8*(3*c^3*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 3*c^3*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 + 2*(3*c^3*(
(a*x - 1)/(a*x + 1))^(7/2) - 11*c^3*((a*x - 1)/(a*x + 1))^(5/2) - 11*c^3*((a*x - 1)/(a*x + 1))^(3/2) + 3*c^3*s
qrt((a*x - 1)/(a*x + 1)))/(4*(a*x - 1)*a^2/(a*x + 1) - 6*(a*x - 1)^2*a^2/(a*x + 1)^2 + 4*(a*x - 1)^3*a^2/(a*x
+ 1)^3 - (a*x - 1)^4*a^2/(a*x + 1)^4 - a^2))*a

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=-\frac {1}{8} \, {\left (\frac {2 \, a^{2} c^{3} x^{2}}{\mathrm {sgn}\left (a x + 1\right )} - \frac {5 \, c^{3}}{\mathrm {sgn}\left (a x + 1\right )}\right )} \sqrt {a^{2} x^{2} - 1} x + \frac {3 \, c^{3} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{8 \, {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

-1/8*(2*a^2*c^3*x^2/sgn(a*x + 1) - 5*c^3/sgn(a*x + 1))*sqrt(a^2*x^2 - 1)*x + 3/8*c^3*log(abs(-x*abs(a) + sqrt(
a^2*x^2 - 1)))/(abs(a)*sgn(a*x + 1))

Mupad [B] (verification not implemented)

Time = 4.33 (sec) , antiderivative size = 176, normalized size of antiderivative = 2.26 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^3 \, dx=\frac {\frac {3\,c^3\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {11\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{4}-\frac {11\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{4}+\frac {3\,c^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{4}}{a-\frac {4\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {6\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {4\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}}-\frac {3\,c^3\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a} \]

[In]

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

[Out]

((3*c^3*((a*x - 1)/(a*x + 1))^(1/2))/4 - (11*c^3*((a*x - 1)/(a*x + 1))^(3/2))/4 - (11*c^3*((a*x - 1)/(a*x + 1)
)^(5/2))/4 + (3*c^3*((a*x - 1)/(a*x + 1))^(7/2))/4)/(a - (4*a*(a*x - 1))/(a*x + 1) + (6*a*(a*x - 1)^2)/(a*x +
1)^2 - (4*a*(a*x - 1)^3)/(a*x + 1)^3 + (a*(a*x - 1)^4)/(a*x + 1)^4) - (3*c^3*atanh(((a*x - 1)/(a*x + 1))^(1/2)
))/(4*a)