\(\int \frac {e^{\coth ^{-1}(a x)}}{x (c-a^2 c x^2)^{5/2}} \, dx\) [705]

Optimal result

Integrand size = 25, antiderivative size = 271 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/8*a^5*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)^2/(-a^2*c*x^2+c)^(5/2)-1/2*a^5*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^
2*c*x^2+c)^(5/2)-1/8*a^5*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-a^5*(1-1/a^2/x^2)^(5/2)*x^5*ln(x
)/(-a^2*c*x^2+c)^(5/2)+11/16*a^5*(1-1/a^2/x^2)^(5/2)*x^5*ln(-a*x+1)/(-a^2*c*x^2+c)^(5/2)+5/16*a^5*(1-1/a^2/x^2
)^(5/2)*x^5*ln(a*x+1)/(-a^2*c*x^2+c)^(5/2)

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {6327, 6328, 90} \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \log (a x+1)}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

[In]

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

[Out]

-1/8*(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/((1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)
/(2*(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - (a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) -
(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[x])/(c - a^2*c*x^2)^(5/2) + (11*a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 -
a*x])/(16*(c - a^2*c*x^2)^(5/2)) + (5*a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*Log[1 + a*x])/(16*(c - a^2*c*x^2)^(5/2))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6327

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

Rule 6328

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

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {e^{\coth ^{-1}(a x)}}{\left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^6} \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = \frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{x (-1+a x)^3 (1+a x)^2} \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = \frac {\left (a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \left (-\frac {1}{x}+\frac {a}{4 (-1+a x)^3}-\frac {a}{2 (-1+a x)^2}+\frac {11 a}{16 (-1+a x)}+\frac {a}{8 (1+a x)^2}+\frac {5 a}{16 (1+a x)}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = -\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (x)}{\left (c-a^2 c x^2\right )^{5/2}}+\frac {11 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1-a x)}{16 \left (c-a^2 c x^2\right )^{5/2}}+\frac {5 a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \log (1+a x)}{16 \left (c-a^2 c x^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.32 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (-\frac {2}{(-1+a x)^2}+\frac {8}{-1+a x}-\frac {2}{1+a x}-16 \log (x)+11 \log (1-a x)+5 \log (1+a x)\right )}{16 \left (c-a^2 c x^2\right )^{5/2}} \]

[In]

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

[Out]

(a^5*(1 - 1/(a^2*x^2))^(5/2)*x^5*(-2/(-1 + a*x)^2 + 8/(-1 + a*x) - 2/(1 + a*x) - 16*Log[x] + 11*Log[1 - a*x] +
5*Log[1 + a*x]))/(16*(c - a^2*c*x^2)^(5/2))

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.72

method	result	size
default	\(-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (5 a^{3} \ln \left (a x +1\right ) x^{3}-16 a^{3} \ln \left (x \right ) x^{3}+11 a^{3} \ln \left (a x -1\right ) x^{3}-5 a^{2} \ln \left (a x +1\right ) x^{2}+16 a^{2} \ln \left (x \right ) x^{2}-11 a^{2} \ln \left (a x -1\right ) x^{2}+6 a^{2} x^{2}-5 a \ln \left (a x +1\right ) x +16 a \ln \left (x \right ) x -11 a \ln \left (a x -1\right ) x +2 a x +5 \ln \left (a x +1\right )-16 \ln \left (x \right )+11 \ln \left (a x -1\right )-12\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} \left (a x +1\right )}\)	\(196\)

[In]

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

[Out]

-1/16/((a*x-1)/(a*x+1))^(1/2)/(a*x-1)*(-c*(a^2*x^2-1))^(1/2)*(5*a^3*ln(a*x+1)*x^3-16*a^3*ln(x)*x^3+11*a^3*ln(a
*x-1)*x^3-5*a^2*ln(a*x+1)*x^2+16*a^2*ln(x)*x^2-11*a^2*ln(a*x-1)*x^2+6*a^2*x^2-5*a*ln(a*x+1)*x+16*a*ln(x)*x-11*
a*ln(a*x-1)*x+2*a*x+5*ln(a*x+1)-16*ln(x)+11*ln(a*x-1)-12)/(a^2*x^2-1)/c^3/(a*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.54 \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {{\left (6 \, a^{2} x^{2} + 2 \, a x + 5 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x + 1\right ) + 11 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (a x - 1\right ) - 16 \, {\left (a^{3} x^{3} - a^{2} x^{2} - a x + 1\right )} \log \left (x\right ) - 12\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{4} c^{3} x^{3} - a^{3} c^{3} x^{2} - a^{2} c^{3} x + a c^{3}\right )}} \]

[In]

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

[Out]

-1/16*(6*a^2*x^2 + 2*a*x + 5*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(a*x + 1) + 11*(a^3*x^3 - a^2*x^2 - a*x + 1)*log
(a*x - 1) - 16*(a^3*x^3 - a^2*x^2 - a*x + 1)*log(x) - 12)*sqrt(-a^2*c)/(a^4*c^3*x^3 - a^3*c^3*x^2 - a^2*c^3*x
+ a*c^3)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int { \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {5}{2}} x \sqrt {\frac {a x - 1}{a x + 1}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)}}{x \left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {1}{x\,{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

[In]

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

[Out]

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