\(\int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx\) [237]

Optimal result

Integrand size = 20, antiderivative size = 40 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\frac {4 (c-a c x)^{3/2}}{3 a}-\frac {2 (c-a c x)^{5/2}}{5 a c} \]

[Out]

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

[In]

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

[Out]

(4*(c - a*c*x)^(3/2))/(3*a) - (2*(c - a*c*x)^(5/2))/(5*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)^{3/2} \, dx \\ & = -\int \frac {(1+a x) (c-a c x)^{3/2}}{1-a x} \, dx \\ & = -\left (c \int (1+a x) \sqrt {c-a c x} \, dx\right ) \\ & = -\left (c \int \left (2 \sqrt {c-a c x}-\frac {(c-a c x)^{3/2}}{c}\right ) \, dx\right ) \\ & = \frac {4 (c-a c x)^{3/2}}{3 a}-\frac {2 (c-a c x)^{5/2}}{5 a c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.75 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {2 c (-1+a x) (7+3 a x) \sqrt {c-a c x}}{15 a} \]

[In]

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

[Out]

(-2*c*(-1 + a*x)*(7 + 3*a*x)*Sqrt[c - a*c*x])/(15*a)

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {2 \left (-a c x +c \right )^{\frac {3}{2}} \left (3 a x +7\right )}{15 a}\)	\(21\)
pseudoelliptic	\(-\frac {2 \sqrt {-c \left (a x -1\right )}\, \left (a x +\frac {7}{3}\right ) \left (a x -1\right ) c}{5 a}\)	\(27\)
trager	\(-\frac {2 c \left (3 a^{2} x^{2}+4 a x -7\right ) \sqrt {-a c x +c}}{15 a}\)	\(30\)
derivativedivides	\(-\frac {2 \left (\frac {\left (-a c x +c \right )^{\frac {5}{2}}}{5}-\frac {2 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}\right )}{c a}\)	\(33\)
default	\(\frac {-\frac {2 \left (-a c x +c \right )^{\frac {5}{2}}}{5}+\frac {4 c \left (-a c x +c \right )^{\frac {3}{2}}}{3}}{a c}\)	\(33\)
risch	\(\frac {2 c^{2} \left (3 a^{2} x^{2}+4 a x -7\right ) \left (a x -1\right )}{15 a \sqrt {-c \left (a x -1\right )}}\)	\(38\)

[In]

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

[Out]

2/15*(-a*c*x+c)^(3/2)*(3*a*x+7)/a

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {2 \, {\left (3 \, a^{2} c x^{2} + 4 \, a c x - 7 \, c\right )} \sqrt {-a c x + c}}{15 \, a} \]

[In]

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

[Out]

-2/15*(3*a^2*c*x^2 + 4*a*c*x - 7*c)*sqrt(-a*c*x + c)/a

Sympy [A] (verification not implemented)

Time = 2.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\begin {cases} - \frac {2 \left (- \frac {2 c \left (- a c x + c\right )^{\frac {3}{2}}}{3} + \frac {\left (- a c x + c\right )^{\frac {5}{2}}}{5}\right )}{a c} & \text {for}\: a c \neq 0 \\c^{\frac {3}{2}} \left (\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x + 2 \log {\left (a x - 1 \right )} - 1}{a} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*(-2*c*(-a*c*x + c)**(3/2)/3 + (-a*c*x + c)**(5/2)/5)/(a*c), Ne(a*c, 0)), (c**(3/2)*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 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=-\frac {2 \, {\left (3 \, {\left (-a c x + c\right )}^{\frac {5}{2}} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c\right )}}{15 \, a c} \]

[In]

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

[Out]

-2/15*(3*(-a*c*x + c)^(5/2) - 10*(-a*c*x + c)^(3/2)*c)/(a*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (32) = 64\).

Time = 0.26 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.78 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\frac {2 \, {\left (15 \, \sqrt {-a c x + c} c - \frac {3 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} - 10 \, {\left (-a c x + c\right )}^{\frac {3}{2}} c + 15 \, \sqrt {-a c x + c} c^{2}}{c}\right )}}{15 \, a} \]

[In]

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

[Out]

2/15*(15*sqrt(-a*c*x + c)*c - (3*(a*c*x - c)^2*sqrt(-a*c*x + c) - 10*(-a*c*x + c)^(3/2)*c + 15*sqrt(-a*c*x + c
)*c^2)/c)/a

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80 \[ \int e^{2 \coth ^{-1}(a x)} (c-a c x)^{3/2} \, dx=\frac {4\,{\left (c-a\,c\,x\right )}^{3/2}}{3\,a}-\frac {2\,{\left (c-a\,c\,x\right )}^{5/2}}{5\,a\,c} \]

[In]

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

[Out]

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