\(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx\) [178]

Optimal result

Integrand size = 18, antiderivative size = 105 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=\frac {3}{8} a c^4 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5-\frac {3 c^4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6310, 6313, 821, 272, 43, 65, 214} \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=-\frac {3 c^4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a}+\frac {3}{8} a c^4 x^2 \sqrt {1-\frac {1}{a^2 x^2}}+\frac {1}{5} a^4 c^4 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}-\frac {1}{4} a^3 c^4 x^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} \]

[In]

Int[E^(3*ArcCoth[a*x])*(c - a*c*x)^4,x]

[Out]

(3*a*c^4*Sqrt[1 - 1/(a^2*x^2)]*x^2)/8 - (a^3*c^4*(1 - 1/(a^2*x^2))^(3/2)*x^4)/4 + (a^4*c^4*(1 - 1/(a^2*x^2))^(
5/2)*x^5)/5 - (3*c^4*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 821

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 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 (a^4 c^4\right ) \int e^{3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^4 x^4 \, dx \\ & = -\left (\left (a^4 c^4\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right ) \left (1-\frac {x^2}{a^2}\right )^{3/2}}{x^6} \, dx,x,\frac {1}{x}\right )\right ) \\ & = \frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5+\left (a^3 c^4\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}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5+\frac {1}{2} \left (a^3 c^4\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^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5-\frac {1}{8} \left (3 a c^4\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^4 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5+\frac {\left (3 c^4\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^4 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5-\frac {1}{8} \left (3 a c^4\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^4 \sqrt {1-\frac {1}{a^2 x^2}} x^2-\frac {1}{4} a^3 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^4+\frac {1}{5} a^4 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5-\frac {3 c^4 \text {arctanh}\left (\sqrt {1-\frac {1}{a^2 x^2}}\right )}{8 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.76 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=\frac {c^4 \left (a \sqrt {1-\frac {1}{a^2 x^2}} x \left (8+25 a x-16 a^2 x^2-10 a^3 x^3+8 a^4 x^4\right )-15 \log \left (a \left (1+\sqrt {1-\frac {1}{a^2 x^2}}\right ) x\right )\right )}{40 a} \]

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a*c*x)^4,x]

[Out]

(c^4*(a*Sqrt[1 - 1/(a^2*x^2)]*x*(8 + 25*a*x - 16*a^2*x^2 - 10*a^3*x^3 + 8*a^4*x^4) - 15*Log[a*(1 + Sqrt[1 - 1/
(a^2*x^2)])*x]))/(40*a)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.22

method	result	size
risch	\(\frac {\left (8 a^{4} x^{4}-10 a^{3} x^{3}-16 a^{2} x^{2}+25 a x +8\right ) \left (a x -1\right ) c^{4}}{40 a \sqrt {\frac {a x -1}{a x +1}}}-\frac {3 \ln \left (\frac {a^{2} x}{\sqrt {a^{2}}}+\sqrt {a^{2} x^{2}-1}\right ) c^{4} \sqrt {\left (a x -1\right ) \left (a x +1\right )}}{8 \sqrt {a^{2}}\, \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(128\)
default	\(\frac {\left (a x -1\right )^{2} c^{4} \left (24 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a^{2} x^{2}-30 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}\, a x +16 \left (a^{2} x^{2}-1\right )^{\frac {3}{2}} \sqrt {a^{2}}+45 \sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}\, a x -40 \left (\left (a x -1\right ) \left (a x +1\right )\right )^{\frac {3}{2}} \sqrt {a^{2}}-45 \ln \left (\frac {a^{2} x +\sqrt {a^{2} x^{2}-1}\, \sqrt {a^{2}}}{\sqrt {a^{2}}}\right ) a \right )}{120 a \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (a x +1\right ) \sqrt {\left (a x -1\right ) \left (a x +1\right )}\, \sqrt {a^{2}}}\)	\(192\)

[In]

int(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

1/40*(8*a^4*x^4-10*a^3*x^3-16*a^2*x^2+25*a*x+8)*(a*x-1)/a*c^4/((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^4/(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 = 126, normalized size of antiderivative = 1.20 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=-\frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right ) - 15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right ) - {\left (8 \, a^{5} c^{4} x^{5} - 2 \, a^{4} c^{4} x^{4} - 26 \, a^{3} c^{4} x^{3} + 9 \, a^{2} c^{4} x^{2} + 33 \, a c^{4} x + 8 \, c^{4}\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{40 \, a} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^4,x, algorithm="fricas")

[Out]

-1/40*(15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) + 1) - 15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) - 1) - (8*a^5*c^4*x^5
- 2*a^4*c^4*x^4 - 26*a^3*c^4*x^3 + 9*a^2*c^4*x^2 + 33*a*c^4*x + 8*c^4)*sqrt((a*x - 1)/(a*x + 1)))/a

Sympy [F]

\[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=c^{4} \left (\int \left (- \frac {4 a x}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\right )\, dx + \int \frac {6 a^{2} x^{2}}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx + \int \left (- \frac {4 a^{3} x^{3}}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\right )\, dx + \int \frac {a^{4} x^{4}}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx + \int \frac {1}{\frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx\right ) \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(3/2)*(-a*c*x+c)**4,x)

[Out]

c**4*(Integral(-4*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(6*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(-4*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(a**4*x**4/(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. 259 vs. \(2 (89) = 178\).

Time = 0.21 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.47 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=-\frac {1}{40} \, {\left (\frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} + 1\right )}{a^{2}} - \frac {15 \, c^{4} \log \left (\sqrt {\frac {a x - 1}{a x + 1}} - 1\right )}{a^{2}} - \frac {2 \, {\left (15 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {9}{2}} - 70 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}} - 128 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}} + 70 \, c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}} - 15 \, c^{4} \sqrt {\frac {a x - 1}{a x + 1}}\right )}}{\frac {5 \, {\left (a x - 1\right )} a^{2}}{a x + 1} - \frac {10 \, {\left (a x - 1\right )}^{2} a^{2}}{{\left (a x + 1\right )}^{2}} + \frac {10 \, {\left (a x - 1\right )}^{3} a^{2}}{{\left (a x + 1\right )}^{3}} - \frac {5 \, {\left (a x - 1\right )}^{4} a^{2}}{{\left (a x + 1\right )}^{4}} + \frac {{\left (a x - 1\right )}^{5} a^{2}}{{\left (a x + 1\right )}^{5}} - a^{2}}\right )} a \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^4,x, algorithm="maxima")

[Out]

-1/40*(15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) + 1)/a^2 - 15*c^4*log(sqrt((a*x - 1)/(a*x + 1)) - 1)/a^2 - 2*(15*c
^4*((a*x - 1)/(a*x + 1))^(9/2) - 70*c^4*((a*x - 1)/(a*x + 1))^(7/2) - 128*c^4*((a*x - 1)/(a*x + 1))^(5/2) + 70
*c^4*((a*x - 1)/(a*x + 1))^(3/2) - 15*c^4*sqrt((a*x - 1)/(a*x + 1)))/(5*(a*x - 1)*a^2/(a*x + 1) - 10*(a*x - 1)
^2*a^2/(a*x + 1)^2 + 10*(a*x - 1)^3*a^2/(a*x + 1)^3 - 5*(a*x - 1)^4*a^2/(a*x + 1)^4 + (a*x - 1)^5*a^2/(a*x + 1
)^5 - a^2))*a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.31 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=\frac {3 \, c^{4} \log \left ({\left | -x {\left | a \right |} + \sqrt {a^{2} x^{2} - 1} \right |}\right )}{8 \, {\left | a \right |} \mathrm {sgn}\left (a x + 1\right )} + \frac {1}{40} \, \sqrt {a^{2} x^{2} - 1} {\left ({\left (\frac {25 \, c^{4}}{\mathrm {sgn}\left (a x + 1\right )} - 2 \, {\left (\frac {8 \, a c^{4}}{\mathrm {sgn}\left (a x + 1\right )} - {\left (\frac {4 \, a^{3} c^{4} x}{\mathrm {sgn}\left (a x + 1\right )} - \frac {5 \, a^{2} c^{4}}{\mathrm {sgn}\left (a x + 1\right )}\right )} x\right )} x\right )} x + \frac {8 \, c^{4}}{a \mathrm {sgn}\left (a x + 1\right )}\right )} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*(-a*c*x+c)^4,x, algorithm="giac")

[Out]

3/8*c^4*log(abs(-x*abs(a) + sqrt(a^2*x^2 - 1)))/(abs(a)*sgn(a*x + 1)) + 1/40*sqrt(a^2*x^2 - 1)*((25*c^4/sgn(a*
x + 1) - 2*(8*a*c^4/sgn(a*x + 1) - (4*a^3*c^4*x/sgn(a*x + 1) - 5*a^2*c^4/sgn(a*x + 1))*x)*x)*x + 8*c^4/(a*sgn(
a*x + 1)))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 214, normalized size of antiderivative = 2.04 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^4 \, dx=\frac {\frac {3\,c^4\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{4}-\frac {7\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}}{2}+\frac {32\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}}{5}+\frac {7\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}}{2}-\frac {3\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{9/2}}{4}}{a-\frac {5\,a\,\left (a\,x-1\right )}{a\,x+1}+\frac {10\,a\,{\left (a\,x-1\right )}^2}{{\left (a\,x+1\right )}^2}-\frac {10\,a\,{\left (a\,x-1\right )}^3}{{\left (a\,x+1\right )}^3}+\frac {5\,a\,{\left (a\,x-1\right )}^4}{{\left (a\,x+1\right )}^4}-\frac {a\,{\left (a\,x-1\right )}^5}{{\left (a\,x+1\right )}^5}}-\frac {3\,c^4\,\mathrm {atanh}\left (\sqrt {\frac {a\,x-1}{a\,x+1}}\right )}{4\,a} \]

[In]

int((c - a*c*x)^4/((a*x - 1)/(a*x + 1))^(3/2),x)

[Out]

((3*c^4*((a*x - 1)/(a*x + 1))^(1/2))/4 - (7*c^4*((a*x - 1)/(a*x + 1))^(3/2))/2 + (32*c^4*((a*x - 1)/(a*x + 1))
^(5/2))/5 + (7*c^4*((a*x - 1)/(a*x + 1))^(7/2))/2 - (3*c^4*((a*x - 1)/(a*x + 1))^(9/2))/4)/(a - (5*a*(a*x - 1)
)/(a*x + 1) + (10*a*(a*x - 1)^2)/(a*x + 1)^2 - (10*a*(a*x - 1)^3)/(a*x + 1)^3 + (5*a*(a*x - 1)^4)/(a*x + 1)^4
- (a*(a*x - 1)^5)/(a*x + 1)^5) - (3*c^4*atanh(((a*x - 1)/(a*x + 1))^(1/2)))/(4*a)