\(\int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) [224]

Optimal result

Integrand size = 18, antiderivative size = 94 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {4 \left (a+\frac {1}{x}\right )}{5 a^2 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {5 a+\frac {2}{x}}{5 a^2 c^5 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6310, 6313, 866, 1649, 651} \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {4 \left (a+\frac {1}{x}\right )}{5 a^2 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {5 a+\frac {2}{x}}{5 a^2 c^5 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \]

[In]

Int[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^5),x]

[Out]

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

Rule 651

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[((-a)*e + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 866

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[d^(2*m)/a
^m, Int[(f + g*x)^n*((a + c*x^2)^(m + p)/(d - e*x)^m), x], x] /; FreeQ[{a, c, d, e, f, g, n, p}, x] && NeQ[e*f
- d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[f, 0] && ILtQ[m, -1] &&  !(IGtQ[n, 0] && ILtQ[m +
n, 0] &&  !GtQ[p, 1])

Rule 1649

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a*e + c*d*x, x], f = PolynomialRemainder[Pq, a*e + c*d*x, x]}, Simp[(-d)*f*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2
*a*e*(p + 1))), x] + Dist[d/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*ExpandToSum[2*a*e*(p + 1)
*Q + f*(m + 2*p + 2), x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] && ILtQ[p
+ 1/2, 0] && GtQ[m, 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}& = -\frac {\int \frac {e^{-3 \coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^5 x^5} \, dx}{a^5 c^5} \\ & = \frac {\text {Subst}\left (\int \frac {x^3}{\left (1-\frac {x}{a}\right )^2 \left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{a^5 c^5} \\ & = \frac {\text {Subst}\left (\int \frac {x^3 \left (1+\frac {x}{a}\right )^2}{\left (1-\frac {x^2}{a^2}\right )^{7/2}} \, dx,x,\frac {1}{x}\right )}{a^5 c^5} \\ & = \frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}-\frac {\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right ) \left (2 a^3+5 a^2 x+5 a x^2\right )}{\left (1-\frac {x^2}{a^2}\right )^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 a^5 c^5} \\ & = -\frac {4 \left (a+\frac {1}{x}\right )}{5 a^2 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {\text {Subst}\left (\int \frac {6 a^3+15 a^2 x}{\left (1-\frac {x^2}{a^2}\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{15 a^5 c^5} \\ & = -\frac {4 \left (a+\frac {1}{x}\right )}{5 a^2 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}+\frac {\left (a+\frac {1}{x}\right )^2}{5 a^3 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {5 a+\frac {2}{x}}{5 a^2 c^5 \sqrt {1-\frac {1}{a^2 x^2}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (2+a x-4 a^2 x^2+2 a^3 x^3\right )}{5 c^5 (-1+a x)^3 (1+a x)} \]

[In]

Integrate[1/(E^(3*ArcCoth[a*x])*(c - a*c*x)^5),x]

[Out]

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

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.57

method	result	size
trager	\(\frac {\left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{5 a \,c^{5} \left (a x -1\right )^{3}}\)	\(54\)
gosper	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{3} x^{3}-4 a^{2} x^{2}+a x +2\right ) \left (a x +1\right )}{5 \left (a x -1\right )^{4} c^{5} a}\)	\(57\)
default	\(\frac {\left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} \left (2 a^{4} x^{4}-2 a^{3} x^{3}-3 a^{2} x^{2}+3 a x +2\right )}{5 \left (a x -1\right )^{4} c^{5} a}\)	\(61\)

[In]

int(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {{\left (2 \, a^{3} x^{3} - 4 \, a^{2} x^{2} + a x + 2\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{5 \, {\left (a^{4} c^{5} x^{3} - 3 \, a^{3} c^{5} x^{2} + 3 \, a^{2} c^{5} x - a c^{5}\right )}} \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^5,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \left (- \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{6} x^{6} - 4 a^{5} x^{5} + 5 a^{4} x^{4} - 5 a^{2} x^{2} + 4 a x - 1}\right )\, dx + \int \frac {a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a^{6} x^{6} - 4 a^{5} x^{5} + 5 a^{4} x^{4} - 5 a^{2} x^{2} + 4 a x - 1}\, dx}{c^{5}} \]

[In]

integrate(((a*x-1)/(a*x+1))**(3/2)/(-a*c*x+c)**5,x)

[Out]

-(Integral(-sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**6*x**6 - 4*a**5*x**5 + 5*a**4*x**4 - 5*a**2*x**2 + 4*a*x - 1
), x) + Integral(a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a**6*x**6 - 4*a**5*x**5 + 5*a**4*x**4 - 5*a**2*x**2 +
4*a*x - 1), x))/c**5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {1}{40} \, a {\left (\frac {5 \, \sqrt {\frac {a x - 1}{a x + 1}}}{a^{2} c^{5}} - \frac {\frac {5 \, {\left (a x - 1\right )}}{a x + 1} - \frac {15 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 1}{a^{2} c^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {5}{2}}}\right )} \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^5,x, algorithm="maxima")

[Out]

1/40*a*(5*sqrt((a*x - 1)/(a*x + 1))/(a^2*c^5) - (5*(a*x - 1)/(a*x + 1) - 15*(a*x - 1)^2/(a*x + 1)^2 - 1)/(a^2*
c^5*((a*x - 1)/(a*x + 1))^(5/2)))

Giac [F]

\[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\int { -\frac {\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}{{\left (a c x - c\right )}^{5}} \,d x } \]

[In]

integrate(((a*x-1)/(a*x+1))^(3/2)/(-a*c*x+c)^5,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.15 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.54 \[ \int \frac {e^{-3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {2\,a^3\,x^3-4\,a^2\,x^2+a\,x+2}{5\,a\,c^5\,{\left (a\,x+1\right )}^3\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{5/2}} \]

[In]

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

[Out]

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