\(\int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx\) [241]

Optimal result

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

[Out]

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

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

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {2 (1+5 a x)}{15 a c^2 (-1+a x)^2 \sqrt {c-a c x}} \]

[In]

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

[Out]

(-2*(1 + 5*a*x))/(15*a*c^2*(-1 + a*x)^2*Sqrt[c - a*c*x])

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.52

method	result	size
gosper	\(-\frac {2 \left (5 a x +1\right )}{15 \left (-a c x +c \right )^{\frac {5}{2}} a}\)	\(21\)
trager	\(\frac {2 \left (5 a x +1\right ) \sqrt {-a c x +c}}{15 c^{3} \left (a x -1\right )^{3} a}\)	\(31\)
pseudoelliptic	\(\frac {-\frac {2 a x}{3}-\frac {2}{15}}{c^{2} \left (a x -1\right )^{2} \sqrt {-c \left (a x -1\right )}\, a}\)	\(32\)
derivativedivides	\(-\frac {2 \left (\frac {2 c}{5 \left (-a c x +c \right )^{\frac {5}{2}}}-\frac {1}{3 \left (-a c x +c \right )^{\frac {3}{2}}}\right )}{c a}\)	\(33\)
default	\(\frac {\frac {2}{3 \left (-a c x +c \right )^{\frac {3}{2}}}-\frac {4 c}{5 \left (-a c x +c \right )^{\frac {5}{2}}}}{a c}\)	\(33\)

[In]

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

[Out]

-2/15*(5*a*x+1)/(-a*c*x+c)^(5/2)/a

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.40 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\frac {2 \, \sqrt {-a c x + c} {\left (5 \, a x + 1\right )}}{15 \, {\left (a^{4} c^{3} x^{3} - 3 \, a^{3} c^{3} x^{2} + 3 \, a^{2} c^{3} x - a c^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.74 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.45 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=\begin {cases} - \frac {2 \cdot \left (\frac {2 c}{5 \left (- a c x + c\right )^{\frac {5}{2}}} - \frac {1}{3 \left (- a c x + c\right )^{\frac {3}{2}}}\right )}{a c} & \text {for}\: a c \neq 0 \\\frac {\begin {cases} - x & \text {for}\: a = 0 \\\frac {a x + 2 \log {\left (a x - 1 \right )} - 1}{a} & \text {otherwise} \end {cases}}{c^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-2*(2*c/(5*(-a*c*x + c)**(5/2)) - 1/(3*(-a*c*x + c)**(3/2)))/(a*c), Ne(a*c, 0)), (Piecewise((-x, Eq
(a, 0)), ((a*x + 2*log(a*x - 1) - 1)/a, True))/c**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.60 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {2 \, {\left (5 \, a c x + c\right )}}{15 \, {\left (-a c x + c\right )}^{\frac {5}{2}} a c} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {2 \, {\left (5 \, a c x + c\right )}}{15 \, {\left (a c x - c\right )}^{2} \sqrt {-a c x + c} a c} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.50 \[ \int \frac {e^{2 \coth ^{-1}(a x)}}{(c-a c x)^{5/2}} \, dx=-\frac {10\,a\,x+2}{15\,a\,{\left (c-a\,c\,x\right )}^{5/2}} \]

[In]

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

[Out]

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