\(\int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx\) [165]

Optimal result

Integrand size = 16, antiderivative size = 100 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {12 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac {1}{x}\right )^4}-\frac {23 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac {1}{x}\right )^3} \]

[Out]

-1/7*a^4*(1-1/a^2/x^2)^(3/2)/c^4/(a-1/x)^5+12/35*a^3*(1-1/a^2/x^2)^(3/2)/c^4/(a-1/x)^4-23/105*a^2*(1-1/a^2/x^2
)^(3/2)/c^4/(a-1/x)^3

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6310, 6313, 1653, 807, 673, 665} \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {23 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac {1}{x}\right )^3}-\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {12 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac {1}{x}\right )^4} \]

[In]

Int[E^ArcCoth[a*x]/(c - a*c*x)^4,x]

[Out]

-1/7*(a^4*(1 - 1/(a^2*x^2))^(3/2))/(c^4*(a - x^(-1))^5) + (12*a^3*(1 - 1/(a^2*x^2))^(3/2))/(35*c^4*(a - x^(-1)
)^4) - (23*a^2*(1 - 1/(a^2*x^2))^(3/2))/(105*c^4*(a - x^(-1))^3)

Rule 665

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a + c*x^2)^(p + 1)/
(2*c*d*(p + 1))), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

Rule 673

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*c*d*(m + p + 1))), x] + Dist[Simplify[m + 2*p + 2]/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^
p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p +
2], 0]

Rule 807

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

Rule 1653

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff
[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x)^(m + q - 1)*((a + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q + 2*p + 1))), x]
+ Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*
f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - c*d*x), x], x], x] /; NeQ[m + q +
2*p + 1, 0]] /; FreeQ[{a, c, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IGtQ[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^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a x}\right )^4 x^4} \, dx}{a^4 c^4} \\ & = -\frac {\text {Subst}\left (\int \frac {x^2 \sqrt {1-\frac {x^2}{a^2}}}{\left (1-\frac {x}{a}\right )^5} \, dx,x,\frac {1}{x}\right )}{a^4 c^4} \\ & = \frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{c^4 \left (a-\frac {1}{x}\right )^4}-\frac {\text {Subst}\left (\int \frac {\left (\frac {4}{a^2}-\frac {3 x}{a^3}\right ) \sqrt {1-\frac {x^2}{a^2}}}{\left (1-\frac {x}{a}\right )^5} \, dx,x,\frac {1}{x}\right )}{c^4} \\ & = -\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{c^4 \left (a-\frac {1}{x}\right )^4}-\frac {23 \text {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{\left (1-\frac {x}{a}\right )^4} \, dx,x,\frac {1}{x}\right )}{7 a^2 c^4} \\ & = -\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {12 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac {1}{x}\right )^4}-\frac {23 \text {Subst}\left (\int \frac {\sqrt {1-\frac {x^2}{a^2}}}{\left (1-\frac {x}{a}\right )^3} \, dx,x,\frac {1}{x}\right )}{35 a^2 c^4} \\ & = -\frac {a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{7 c^4 \left (a-\frac {1}{x}\right )^5}+\frac {12 a^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{35 c^4 \left (a-\frac {1}{x}\right )^4}-\frac {23 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}{105 c^4 \left (a-\frac {1}{x}\right )^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.51 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x \left (23+13 a x-8 a^2 x^2+2 a^3 x^3\right )}{105 c^4 (-1+a x)^4} \]

[In]

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

[Out]

-1/105*(Sqrt[1 - 1/(a^2*x^2)]*x*(23 + 13*a*x - 8*a^2*x^2 + 2*a^3*x^3))/(c^4*(-1 + a*x)^4)

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.50

method	result	size
gosper	\(-\frac {\left (2 a^{2} x^{2}-10 a x +23\right ) \left (a x +1\right )}{105 \left (a x -1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(50\)
default	\(-\frac {\left (2 a^{2} x^{2}-10 a x +23\right ) \left (a x +1\right )}{105 \left (a x -1\right )^{3} c^{4} \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(50\)
trager	\(-\frac {\left (a x +1\right ) \left (2 a^{3} x^{3}-8 a^{2} x^{2}+13 a x +23\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{105 a \,c^{4} \left (a x -1\right )^{4}}\)	\(60\)

[In]

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

[Out]

-1/105*(2*a^2*x^2-10*a*x+23)*(a*x+1)/(a*x-1)^3/c^4/((a*x-1)/(a*x+1))^(1/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {{\left (2 \, a^{4} x^{4} - 6 \, a^{3} x^{3} + 5 \, a^{2} x^{2} + 36 \, a x + 23\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{105 \, {\left (a^{5} c^{4} x^{4} - 4 \, a^{4} c^{4} x^{3} + 6 \, a^{3} c^{4} x^{2} - 4 \, a^{2} c^{4} x + a c^{4}\right )}} \]

[In]

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

[Out]

-1/105*(2*a^4*x^4 - 6*a^3*x^3 + 5*a^2*x^2 + 36*a*x + 23)*sqrt((a*x - 1)/(a*x + 1))/(a^5*c^4*x^4 - 4*a^4*c^4*x^
3 + 6*a^3*c^4*x^2 - 4*a^2*c^4*x + a*c^4)

Sympy [F]

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

[In]

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

[Out]

Integral(1/(a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a**3*x**3*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + 6*a*
*2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) - 4*a*x*sqrt(a*x/(a*x + 1) - 1/(a*x + 1)) + sqrt(a*x/(a*x + 1) - 1/(
a*x + 1))), x)/c**4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.55 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=\frac {\frac {42 \, {\left (a x - 1\right )}}{a x + 1} - \frac {35 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} - 15}{420 \, a c^{4} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {7}{2}}} \]

[In]

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

[Out]

1/420*(42*(a*x - 1)/(a*x + 1) - 35*(a*x - 1)^2/(a*x + 1)^2 - 15)/(a*c^4*((a*x - 1)/(a*x + 1))^(7/2))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {4 \, {\left (70 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 35 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} + 21 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} - 7 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x + 1\right )}}{105 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{7} a c^{4}} \]

[In]

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

[Out]

-4/105*(70*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 35*(a + sqrt(a^2 - 1/x^2))^3*x^3 + 21*(a + sqrt(a^2 - 1/x^2))^2*x^2
- 7*(a + sqrt(a^2 - 1/x^2))*x + 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^7*a*c^4)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.56 \[ \int \frac {e^{\coth ^{-1}(a x)}}{(c-a c x)^4} \, dx=-\frac {\frac {{\left (a\,x-1\right )}^2}{3\,{\left (a\,x+1\right )}^2}-\frac {2\,\left (a\,x-1\right )}{5\,\left (a\,x+1\right )}+\frac {1}{7}}{4\,a\,c^4\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{7/2}} \]

[In]

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

[Out]

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