\(\int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx\) [186]

Optimal result

Integrand size = 18, antiderivative size = 125 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {152 \left (a+\frac {1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {79 \left (a+\frac {1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}} \]

[Out]

-152/1155*(a+1/x)^5/a^6/c^5/(1-1/a^2/x^2)^(5/2)+79/231*(a+1/x)^6/a^7/c^5/(1-1/a^2/x^2)^(7/2)-10/33*(a+1/x)^7/a
^8/c^5/(1-1/a^2/x^2)^(9/2)+1/11*(a+1/x)^8/a^9/c^5/(1-1/a^2/x^2)^(11/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 125, 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, 866, 1649, 803, 665} \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {79 \left (a+\frac {1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {152 \left (a+\frac {1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}} \]

[In]

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

[Out]

(-152*(a + x^(-1))^5)/(1155*a^6*c^5*(1 - 1/(a^2*x^2))^(5/2)) + (79*(a + x^(-1))^6)/(231*a^7*c^5*(1 - 1/(a^2*x^
2))^(7/2)) - (10*(a + x^(-1))^7)/(33*a^8*c^5*(1 - 1/(a^2*x^2))^(9/2)) + (a + x^(-1))^8/(11*a^9*c^5*(1 - 1/(a^2
*x^2))^(11/2))

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 803

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*(p + 1))), x] - Dist[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))
), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0]
&& LtQ[p, -1] && GtQ[m, 0]

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^2}{a^2}\right )^{3/2}}{\left (1-\frac {x}{a}\right )^8} \, dx,x,\frac {1}{x}\right )}{a^5 c^5} \\ & = \frac {\text {Subst}\left (\int \frac {x^3 \left (1+\frac {x}{a}\right )^8}{\left (1-\frac {x^2}{a^2}\right )^{13/2}} \, dx,x,\frac {1}{x}\right )}{a^5 c^5} \\ & = \frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^7 \left (8 a^3+11 a^2 x+11 a x^2\right )}{\left (1-\frac {x^2}{a^2}\right )^{11/2}} \, dx,x,\frac {1}{x}\right )}{11 a^5 c^5} \\ & = -\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}+\frac {\text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^6 \left (138 a^3+99 a^2 x\right )}{\left (1-\frac {x^2}{a^2}\right )^{9/2}} \, dx,x,\frac {1}{x}\right )}{99 a^5 c^5} \\ & = \frac {79 \left (a+\frac {1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}}-\frac {152 \text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^5}{\left (1-\frac {x^2}{a^2}\right )^{7/2}} \, dx,x,\frac {1}{x}\right )}{231 a^2 c^5} \\ & = -\frac {152 \left (a+\frac {1}{x}\right )^5}{1155 a^6 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}+\frac {79 \left (a+\frac {1}{x}\right )^6}{231 a^7 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{7/2}}-\frac {10 \left (a+\frac {1}{x}\right )^7}{33 a^8 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{9/2}}+\frac {\left (a+\frac {1}{x}\right )^8}{11 a^9 c^5 \left (1-\frac {1}{a^2 x^2}\right )^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.46 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\sqrt {1-\frac {1}{a^2 x^2}} x (1+a x)^2 \left (-152+61 a x-16 a^2 x^2+2 a^3 x^3\right )}{1155 c^5 (-1+a x)^6} \]

[In]

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

[Out]

-1/1155*(Sqrt[1 - 1/(a^2*x^2)]*x*(1 + a*x)^2*(-152 + 61*a*x - 16*a^2*x^2 + 2*a^3*x^3))/(c^5*(-1 + a*x)^6)

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.46

method	result	size
gosper	\(-\frac {\left (2 a^{3} x^{3}-16 a^{2} x^{2}+61 a x -152\right ) \left (a x +1\right )}{1155 \left (a x -1\right )^{4} c^{5} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(58\)
default	\(-\frac {\left (2 a^{3} x^{3}-16 a^{2} x^{2}+61 a x -152\right ) \left (a x +1\right )}{1155 \left (a x -1\right )^{4} c^{5} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(58\)
trager	\(-\frac {\left (a x +1\right ) \left (2 a^{5} x^{5}-12 a^{4} x^{4}+31 a^{3} x^{3}-46 a^{2} x^{2}-243 a x -152\right ) \sqrt {-\frac {-a x +1}{a x +1}}}{1155 a \,c^{5} \left (a x -1\right )^{6}}\)	\(76\)

[In]

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

[Out]

-1/1155*(2*a^3*x^3-16*a^2*x^2+61*a*x-152)*(a*x+1)/(a*x-1)^4/c^5/((a*x-1)/(a*x+1))^(3/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.07 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {{\left (2 \, a^{6} x^{6} - 10 \, a^{5} x^{5} + 19 \, a^{4} x^{4} - 15 \, a^{3} x^{3} - 289 \, a^{2} x^{2} - 395 \, a x - 152\right )} \sqrt {\frac {a x - 1}{a x + 1}}}{1155 \, {\left (a^{7} c^{5} x^{6} - 6 \, a^{6} c^{5} x^{5} + 15 \, a^{5} c^{5} x^{4} - 20 \, a^{4} c^{5} x^{3} + 15 \, a^{3} c^{5} x^{2} - 6 \, a^{2} c^{5} x + a c^{5}\right )}} \]

[In]

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

[Out]

-1/1155*(2*a^6*x^6 - 10*a^5*x^5 + 19*a^4*x^4 - 15*a^3*x^3 - 289*a^2*x^2 - 395*a*x - 152)*sqrt((a*x - 1)/(a*x +
1))/(a^7*c^5*x^6 - 6*a^6*c^5*x^5 + 15*a^5*c^5*x^4 - 20*a^4*c^5*x^3 + 15*a^3*c^5*x^2 - 6*a^2*c^5*x + a*c^5)

Sympy [F]

\[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=- \frac {\int \frac {1}{\frac {a^{6} x^{6} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {6 a^{5} x^{5} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {15 a^{4} x^{4} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {20 a^{3} x^{3} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {15 a^{2} x^{2} \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} - \frac {6 a x \sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1} + \frac {\sqrt {\frac {a x}{a x + 1} - \frac {1}{a x + 1}}}{a x + 1}}\, dx}{c^{5}} \]

[In]

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

[Out]

-Integral(1/(a**6*x**6*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 6*a**5*x**5*sqrt(a*x/(a*x + 1) - 1/(a*x +
1))/(a*x + 1) + 15*a**4*x**4*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 20*a**3*x**3*sqrt(a*x/(a*x + 1) -
1/(a*x + 1))/(a*x + 1) + 15*a**2*x**2*sqrt(a*x/(a*x + 1) - 1/(a*x + 1))/(a*x + 1) - 6*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)/c**5

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.57 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=-\frac {\frac {385 \, {\left (a x - 1\right )}}{a x + 1} - \frac {495 \, {\left (a x - 1\right )}^{2}}{{\left (a x + 1\right )}^{2}} + \frac {231 \, {\left (a x - 1\right )}^{3}}{{\left (a x + 1\right )}^{3}} - 105}{9240 \, a c^{5} \left (\frac {a x - 1}{a x + 1}\right )^{\frac {11}{2}}} \]

[In]

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

[Out]

-1/9240*(385*(a*x - 1)/(a*x + 1) - 495*(a*x - 1)^2/(a*x + 1)^2 + 231*(a*x - 1)^3/(a*x + 1)^3 - 105)/(a*c^5*((a
*x - 1)/(a*x + 1))^(11/2))

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.32 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {4 \, {\left (1155 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{7} x^{7} + 2079 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{6} x^{6} + 2541 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{5} x^{5} + 825 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{4} x^{4} + 165 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{3} x^{3} - 55 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )}^{2} x^{2} + 11 \, {\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}}{1155 \, {\left ({\left (a + \sqrt {a^{2} - \frac {1}{x^{2}}}\right )} x - 1\right )}^{11} a c^{5}} \]

[In]

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

[Out]

4/1155*(1155*(a + sqrt(a^2 - 1/x^2))^7*x^7 + 2079*(a + sqrt(a^2 - 1/x^2))^6*x^6 + 2541*(a + sqrt(a^2 - 1/x^2))
^5*x^5 + 825*(a + sqrt(a^2 - 1/x^2))^4*x^4 + 165*(a + sqrt(a^2 - 1/x^2))^3*x^3 - 55*(a + sqrt(a^2 - 1/x^2))^2*
x^2 + 11*(a + sqrt(a^2 - 1/x^2))*x - 1)/(((a + sqrt(a^2 - 1/x^2))*x - 1)^11*a*c^5)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.58 \[ \int \frac {e^{3 \coth ^{-1}(a x)}}{(c-a c x)^5} \, dx=\frac {\frac {3\,{\left (a\,x-1\right )}^2}{7\,{\left (a\,x+1\right )}^2}-\frac {{\left (a\,x-1\right )}^3}{5\,{\left (a\,x+1\right )}^3}-\frac {a\,x-1}{3\,\left (a\,x+1\right )}+\frac {1}{11}}{8\,a\,c^5\,{\left (\frac {a\,x-1}{a\,x+1}\right )}^{11/2}} \]

[In]

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

[Out]

((3*(a*x - 1)^2)/(7*(a*x + 1)^2) - (a*x - 1)^3/(5*(a*x + 1)^3) - (a*x - 1)/(3*(a*x + 1)) + 1/11)/(8*a*c^5*((a*
x - 1)/(a*x + 1))^(11/2))