\(\int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\) [228]

Optimal result

Integrand size = 18, antiderivative size = 115 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {64 a^2 c^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{105 (c-a c x)^{3/2}}+\frac {16 a^2 c^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}{35 \sqrt {c-a c x}}+\frac {2}{7} a^2 c^2 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3 \sqrt {c-a c x} \]

[Out]

64/105*a^2*c^4*(1-1/a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(3/2)+16/35*a^2*c^3*(1-1/a^2/x^2)^(3/2)*x^3/(-a*c*x+c)^(1/2)
+2/7*a^2*c^2*(1-1/a^2/x^2)^(3/2)*x^3*(-a*c*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {6311, 6316, 91, 79, 37} \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {142 \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{105 a^2 x \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {36 \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {2 x \left (\frac {1}{a x}+1\right )^{3/2} (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \]

[In]

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

[Out]

(-36*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^(5/2))/(35*a*(1 - 1/(a*x))^(5/2)) + (142*(1 + 1/(a*x))^(3/2)*(c - a*c*x)^
(5/2))/(105*a^2*(1 - 1/(a*x))^(5/2)*x) + (2*(1 + 1/(a*x))^(3/2)*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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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^{\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 )^2 \sqrt {1+\frac {x}{a}}}{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 )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {\left (2 \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \text {Subst}\left (\int \frac {\left (-\frac {9}{a}+\frac {7 x}{2 a^2}\right ) \sqrt {1+\frac {x}{a}}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \\ & = -\frac {36 \left (1+\frac {1}{a x}\right )^{3/2} (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 )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}}-\frac {\left (71 \left (\frac {1}{x}\right )^{5/2} (c-a c x)^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {x}{a}}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{35 a^2 \left (1-\frac {1}{a x}\right )^{5/2}} \\ & = -\frac {36 \left (1+\frac {1}{a x}\right )^{3/2} (c-a c x)^{5/2}}{35 a \left (1-\frac {1}{a x}\right )^{5/2}}+\frac {142 \left (1+\frac {1}{a x}\right )^{3/2} (c-a c x)^{5/2}}{105 a^2 \left (1-\frac {1}{a x}\right )^{5/2} x}+\frac {2 \left (1+\frac {1}{a x}\right )^{3/2} x (c-a c x)^{5/2}}{7 \left (1-\frac {1}{a x}\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.61 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 c^2 \sqrt {1+\frac {1}{a x}} \sqrt {c-a c x} \left (71+17 a x-39 a^2 x^2+15 a^3 x^3\right )}{105 a \sqrt {1-\frac {1}{a x}}} \]

[In]

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

[Out]

(2*c^2*Sqrt[1 + 1/(a*x)]*Sqrt[c - a*c*x]*(71 + 17*a*x - 39*a^2*x^2 + 15*a^3*x^3))/(105*a*Sqrt[1 - 1/(a*x)])

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.46

method	result	size
default	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, c^{2} \left (a x +1\right ) \left (15 a^{2} x^{2}-54 a x +71\right )}{105 \sqrt {\frac {a x -1}{a x +1}}\, a}\)	\(53\)
gosper	\(\frac {2 \left (a x +1\right ) \left (15 a^{2} x^{2}-54 a x +71\right ) \left (-a c x +c \right )^{\frac {5}{2}}}{105 a \left (a x -1\right )^{2} \sqrt {\frac {a x -1}{a x +1}}}\)	\(56\)
risch	\(-\frac {2 c^{3} \left (a x -1\right ) \left (15 a^{3} x^{3}-39 a^{2} x^{2}+17 a x +71\right )}{105 \sqrt {\frac {a x -1}{a x +1}}\, \sqrt {-c \left (a x -1\right )}\, a}\)	\(61\)

[In]

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

[Out]

2/105/((a*x-1)/(a*x+1))^(1/2)*(-c*(a*x-1))^(1/2)*c^2*(a*x+1)*(15*a^2*x^2-54*a*x+71)/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (15 \, a^{4} c^{2} x^{4} - 24 \, a^{3} c^{2} x^{3} - 22 \, a^{2} c^{2} x^{2} + 88 \, a c^{2} x + 71 \, c^{2}\right )} \sqrt {-a c x + c} \sqrt {\frac {a x - 1}{a x + 1}}}{105 \, {\left (a^{2} x - a\right )}} \]

[In]

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

[Out]

2/105*(15*a^4*c^2*x^4 - 24*a^3*c^2*x^3 - 22*a^2*c^2*x^2 + 88*a*c^2*x + 71*c^2)*sqrt(-a*c*x + c)*sqrt((a*x - 1)
/(a*x + 1))/(a^2*x - a)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (15 \, a^{3} \sqrt {-c} c^{2} x^{3} - 39 \, a^{2} \sqrt {-c} c^{2} x^{2} + 17 \, a \sqrt {-c} c^{2} x + 71 \, \sqrt {-c} c^{2}\right )} \sqrt {a x + 1}}{105 \, a} \]

[In]

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

[Out]

2/105*(15*a^3*sqrt(-c)*c^2*x^3 - 39*a^2*sqrt(-c)*c^2*x^2 + 17*a*sqrt(-c)*c^2*x + 71*sqrt(-c)*c^2)*sqrt(a*x + 1
)/a

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.84 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (64 \, \sqrt {2} \sqrt {-c} c - \frac {15 \, {\left (a c x + c\right )}^{3} \sqrt {-a c x - c} - 84 \, {\left (a c x + c\right )}^{2} \sqrt {-a c x - c} c - 140 \, {\left (-a c x - c\right )}^{\frac {3}{2}} c^{2}}{c^{2}}\right )} c^{2}}{105 \, a {\left | c \right |} \mathrm {sgn}\left (a x + 1\right )} \]

[In]

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

[Out]

2/105*(64*sqrt(2)*sqrt(-c)*c - (15*(a*c*x + c)^3*sqrt(-a*c*x - c) - 84*(a*c*x + c)^2*sqrt(-a*c*x - c)*c - 140*
(-a*c*x - c)^(3/2)*c^2)/c^2)*c^2/(a*abs(c)*sgn(a*x + 1))

Mupad [B] (verification not implemented)

Time = 4.40 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.52 \[ \int e^{\coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2\,c^2\,\sqrt {c-a\,c\,x}\,{\left (a\,x+1\right )}^2\,\sqrt {\frac {a\,x-1}{a\,x+1}}\,\left (15\,a^2\,x^2-54\,a\,x+71\right )}{105\,a\,\left (a\,x-1\right )} \]

[In]

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

[Out]

(2*c^2*(c - a*c*x)^(1/2)*(a*x + 1)^2*((a*x - 1)/(a*x + 1))^(1/2)*(15*a^2*x^2 - 54*a*x + 71))/(105*a*(a*x - 1))