\(\int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx\) [236]

Optimal result

Integrand size = 20, antiderivative size = 40 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6302, 6265, 21, 45} \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c} \]

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6265

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

Rule 6302

Int[E^(ArcCoth[(a_.)*(x_)]*(n_))*(u_.), x_Symbol] :> Dist[(-1)^(n/2), Int[u*E^(n*ArcTanh[a*x]), x], x] /; Free
Q[a, x] && IntegerQ[n/2]

Rubi steps \begin{align*} \text {integral}& = -\int e^{2 \text {arctanh}(a x)} (c-a c x)^{5/2} \, dx \\ & = -\int \frac {(1+a x) (c-a c x)^{5/2}}{1-a x} \, dx \\ & = -\left (c \int (1+a x) (c-a c x)^{3/2} \, dx\right ) \\ & = -\left (c \int \left (2 (c-a c x)^{3/2}-\frac {(c-a c x)^{5/2}}{c}\right ) \, dx\right ) \\ & = \frac {4 (c-a c x)^{5/2}}{5 a}-\frac {2 (c-a c x)^{7/2}}{7 a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 c^2 (-1+a x)^2 (9+5 a x) \sqrt {c-a c x}}{35 a} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.52

method	result	size
gosper	\(\frac {2 \left (-a c x +c \right )^{\frac {5}{2}} \left (5 a x +9\right )}{35 a}\)	\(21\)
pseudoelliptic	\(\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (a x +\frac {9}{5}\right ) \left (a x -1\right )^{2} c^{2}}{7 a}\)	\(31\)
derivativedivides	\(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {7}{2}}}{7}-\frac {2 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}\right )}{c a}\)	\(33\)
default	\(\frac {-\frac {2 \left (-a c x +c \right )^{\frac {7}{2}}}{7}+\frac {4 c \left (-a c x +c \right )^{\frac {5}{2}}}{5}}{a c}\)	\(33\)
trager	\(\frac {2 c^{2} \left (5 a^{3} x^{3}-a^{2} x^{2}-13 a x +9\right ) \sqrt {-a c x +c}}{35 a}\)	\(40\)
risch	\(-\frac {2 c^{3} \left (5 a^{3} x^{3}-a^{2} x^{2}-13 a x +9\right ) \left (a x -1\right )}{35 a \sqrt {-c \left (a x -1\right )}}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.22 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{3} c^{2} x^{3} - a^{2} c^{2} x^{2} - 13 \, a c^{2} x + 9 \, c^{2}\right )} \sqrt {-a c x + c}}{35 \, a} \]

[In]

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

[Out]

2/35*(5*a^3*c^2*x^3 - a^2*c^2*x^2 - 13*a*c^2*x + 9*c^2)*sqrt(-a*c*x + c)/a

Sympy [A] (verification not implemented)

Time = 2.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\begin {cases} - \frac {2 \left (- \frac {2 c \left (- a c x + c\right )^{\frac {5}{2}}}{5} + \frac {\left (- a c x + c\right )^{\frac {7}{2}}}{7}\right )}{a c} & \text {for}\: a c \neq 0 \\c^{\frac {5}{2}} \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x + 2 \log {\left (a x - 1 \right )} - 1}{a} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*(-2*c*(-a*c*x + c)**(5/2)/5 + (-a*c*x + c)**(7/2)/7)/(a*c), Ne(a*c, 0)), (c**(5/2)*Piecewise((-x
, Eq(a, 0)), ((a*x + 2*log(a*x - 1) - 1)/a, True)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (5 \, {\left (-a c x + c\right )}^{\frac {7}{2}} - 14 \, {\left (-a c x + c\right )}^{\frac {5}{2}} c\right )}}{35 \, a c} \]

[In]

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

[Out]

-2/35*(5*(-a*c*x + c)^(7/2) - 14*(-a*c*x + c)^(5/2)*c)/(a*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (32) = 64\).

Time = 0.26 (sec) , antiderivative size = 141, normalized size of antiderivative = 3.52 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=-\frac {2 \, {\left (21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 70 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c - 35 \, {\left ({\left (-a c x + c\right )}^{\frac {3}{2}} - 3 \, \sqrt {-a c x + c} c\right )} c - \frac {3 \, {\left (5 \, {\left (a c x - c\right )}^{3} \sqrt {-a c x + c} + 21 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} c - 35 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c^{2} + 35 \, \sqrt {-a c x + c} c^{3}\right )}}{c}\right )}}{105 \, a} \]

[In]

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

[Out]

-2/105*(21*(a*c*x - c)^2*sqrt(-a*c*x + c) - 70*(-a*c*x + c)^(3/2)*c - 35*((-a*c*x + c)^(3/2) - 3*sqrt(-a*c*x +
c)*c)*c - 3*(5*(a*c*x - c)^3*sqrt(-a*c*x + c) + 21*(a*c*x - c)^2*sqrt(-a*c*x + c)*c - 35*(-a*c*x + c)^(3/2)*c
^2 + 35*sqrt(-a*c*x + c)*c^3)/c)/a

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{5/2} \, dx=\frac {4\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a}-\frac {2\,{\left (c-a\,c\,x\right )}^{7/2}}{7\,a\,c} \]

[In]

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

[Out]

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