\(\int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\) [245]

Optimal result

Integrand size = 20, antiderivative size = 89 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {18 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \]

[Out]

-18/35*(1+1/a/x)^(5/2)*(-a*c*x+c)^(5/2)/a/(1-1/a/x)^(5/2)+2/7*(1+1/a/x)^(5/2)*x*(-a*c*x+c)^(5/2)/(1-1/a/x)^(5/
2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6311, 6316, 79, 37} \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 x \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {18 \left (\frac {1}{a x}+1\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}} \]

[In]

Int[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(-18*(1 + 1/(a*x))^(5/2)*(c - a*c*x)^(5/2))/(35*a*(1 - 1/(a*x))^(5/2)) + (2*(1 + 1/(a*x))^(5/2)*x*(c - a*c*x)^
(5/2))/(7*(1 - 1/(a*x))^(5/2))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 6311

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6316

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> Dist[(-c^p)*x^m*(1/x)^m, S
ubst[Int[(1 + d*(x/c))^p*((1 + x/a)^(n/2)/(x^(m + 2)*(1 - x/a)^(n/2))), x], x, 1/x], x] /; FreeQ[{a, c, d, m,
n, p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {(c-a c x)^{5/2} \int e^{3 \coth ^{-1}(a x)} \left (1-\frac {1}{a x}\right )^{5/2} x^{5/2} \, dx}{\left (1-\frac {1}{a x}\right )^{5/2} x^{5/2}} \\ & = -\frac {\left (\left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x}{a}\right ) \left (1+\frac {x}{a}\right )^{3/2}}{x^{9/2}} \, dx,x,\frac {1}{x}\right )}{\left (1-\frac {1}{a x}\right )^{5/2}} \\ & = \frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {\left (9 \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {x}{a}\right )^{3/2}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{7 a \left (1-\frac {1}{a x}\right )^{5/2}} \\ & = -\frac {18 \left (1+\frac {1}{a x}\right )^{5/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {2 \left (1+\frac {1}{a x}\right )^{5/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.66 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \sqrt {1+\frac {1}{a x}} (-9+5 a x) \sqrt {c-a c x} (c+a c x)^2}{35 a \sqrt {1-\frac {1}{a x}}} \]

[In]

Integrate[E^(3*ArcCoth[a*x])*(c - a*c*x)^(5/2),x]

[Out]

(2*Sqrt[1 + 1/(a*x)]*(-9 + 5*a*x)*Sqrt[c - a*c*x]*(c + a*c*x)^2)/(35*a*Sqrt[1 - 1/(a*x)])

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.54

method	result	size
gosper	\(\frac {2 \left (a x +1\right ) \left (5 a x -9\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{35 a \left (a x -1\right ) \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(48\)
default	\(\frac {2 \left (a x -1\right ) \left (a x +1\right ) \sqrt {-c \left (a x -1\right )}\, c^{2} \left (5 a x -9\right )}{35 \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}} a}\)	\(50\)
risch	\(-\frac {2 c^{3} \left (a x -1\right ) \left (5 a^{3} x^{3}+a^{2} x^{2}-13 a x -9\right )}{35 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a}\)	\(60\)

[In]

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

[Out]

2/35*(a*x+1)*(5*a*x-9)*(-a*c*x+c)^(5/2)/a/(a*x-1)/((a*x-1)/(a*x+1))^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{4} c^{2} x^{4} + 6 \, a^{3} c^{2} x^{3} - 12 \, a^{2} c^{2} x^{2} - 22 \, a c^{2} x - 9 \, c^{2}\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{35 \, {\left (a^{2} x - a\right )}} \]

[In]

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

[Out]

2/35*(5*a^4*c^2*x^4 + 6*a^3*c^2*x^3 - 12*a^2*c^2*x^2 - 22*a*c^2*x - 9*c^2)*sqrt(-a*c*x + c)*sqrt((a*x - 1)/(a*
x + 1))/(a^2*x - a)

Sympy [F(-1)]

Timed out. \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.83 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{3} \sqrt {-c} c^{2} x^{3} - 9 \, a^{2} \sqrt {-c} c^{2} x^{2} - 5 \, a \sqrt {-c} c^{2} x + 9 \, \sqrt {-c} c^{2}\right )} {\left (a x + 1\right )}^{\frac {3}{2}}}{35 \, {\left (a x - 1\right )} a} \]

[In]

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

[Out]

2/35*(5*a^3*sqrt(-c)*c^2*x^3 - 9*a^2*sqrt(-c)*c^2*x^2 - 5*a*sqrt(-c)*c^2*x + 9*sqrt(-c)*c^2)*(a*x + 1)^(3/2)/(
(a*x - 1)*a)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.90 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (16 \, \sqrt {2} \sqrt {-c} c + \frac {5 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} - 14 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c}{c^{2}}\right )} c^{2}}{35 \, a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

-2/35*(16*sqrt(2)*sqrt(-c)*c + (5*(a*c*x + c)^3*sqrt(-a*c*x - c) - 14*(a*c*x + c)^2*sqrt(-a*c*x - c)*c)/c^2)*c
^2/(a*abs(c)*sgn(a*x + 1))

Mupad [B] (verification not implemented)

Time = 4.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int e^{3 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2\,c^2\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (-5\,a^3\,x^3-11\,a^2\,x^2+a\,x+23\right )}{35\,a}-\frac {64\,c^2\,\sqrt {c-a\,c\,x}\,\sqrt {\frac {a\,x-1}{a\,x+1}}}{35\,a\,\left (a\,x-1\right )} \]

[In]

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

[Out]

- (2*c^2*(c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2)*(a*x - 11*a^2*x^2 - 5*a^3*x^3 + 23))/(35*a) - (64*c^2*(
c - a*c*x)^(1/2)*((a*x - 1)/(a*x + 1))^(1/2))/(35*a*(a*x - 1))