\(\int \frac {e^{\coth ^{-1}(a x)} x^3}{(c-a^2 c x^2)^{5/2}} \, dx\) [701]

Optimal result

Integrand size = 25, antiderivative size = 176 \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5}{8 (1+a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {3 a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[Out]

-1/8*a*(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*(1-1/a^2/x^2)^(5/2)*x^5/(-a*x+1)/(-a^2*c*
x^2+c)^(5/2)-1/8*a*(1-1/a^2/x^2)^(5/2)*x^5/(a*x+1)/(-a^2*c*x^2+c)^(5/2)-3/8*a*(1-1/a^2/x^2)^(5/2)*x^5*arctanh(
a*x)/(-a^2*c*x^2+c)^(5/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6327, 6328, 90, 213} \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}}+\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{2 (1-a x) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (a x+1) \left (c-a^2 c x^2\right )^{5/2}}-\frac {a x^5 \left (1-\frac {1}{a^2 x^2}\right )^{5/2}}{8 (1-a x)^2 \left (c-a^2 c x^2\right )^{5/2}} \]

[In]

Int[(E^ArcCoth[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]

[Out]

-1/8*(a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/((1 - a*x)^2*(c - a^2*c*x^2)^(5/2)) + (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(2*
(1 - a*x)*(c - a^2*c*x^2)^(5/2)) - (a*(1 - 1/(a^2*x^2))^(5/2)*x^5)/(8*(1 + a*x)*(c - a^2*c*x^2)^(5/2)) - (3*a*
(1 - 1/(a^2*x^2))^(5/2)*x^5*ArcTanh[a*x])/(8*(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 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

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^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 \frac {x^3}{(-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}{4 a^3 (-1+a x)^3}+\frac {1}{2 a^3 (-1+a x)^2}+\frac {1}{8 a^3 (1+a x)^2}+\frac {3}{8 a^3 \left (-1+a^2 x^2\right )}\right ) \, dx}{\left (c-a^2 c x^2\right )^{5/2}} \\ & = -\frac {a \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 \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 \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 {\left (3 a^2 \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5\right ) \int \frac {1}{-1+a^2 x^2} \, dx}{8 \left (c-a^2 c x^2\right )^{5/2}} \\ & = -\frac {a \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 \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 \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 {3 a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \text {arctanh}(a x)}{8 \left (c-a^2 c x^2\right )^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.41 \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {a \left (1-\frac {1}{a^2 x^2}\right )^{5/2} x^5 \left (\frac {2+a x-5 a^2 x^2}{(-1+a x)^2 (1+a x)}-3 \text {arctanh}(a x)\right )}{8 \left (c-a^2 c x^2\right )^{5/2}} \]

[In]

Integrate[(E^ArcCoth[a*x]*x^3)/(c - a^2*c*x^2)^(5/2),x]

[Out]

(a*(1 - 1/(a^2*x^2))^(5/2)*x^5*((2 + a*x - 5*a^2*x^2)/((-1 + a*x)^2*(1 + a*x)) - 3*ArcTanh[a*x]))/(8*(c - a^2*
c*x^2)^(5/2))

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.96

method	result	size
default	\(\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (3 a^{3} \ln \left (a x +1\right ) x^{3}-3 a^{3} \ln \left (a x -1\right ) x^{3}-3 a^{2} \ln \left (a x +1\right ) x^{2}+3 a^{2} \ln \left (a x -1\right ) x^{2}+10 a^{2} x^{2}-3 a \ln \left (a x +1\right ) x +3 a \ln \left (a x -1\right ) x -2 a x +3 \ln \left (a x +1\right )-3 \ln \left (a x -1\right )-4\right )}{16 \sqrt {\frac {a x -1}{a x +1}}\, \left (a x -1\right ) \left (a^{2} x^{2}-1\right ) c^{3} a^{4} \left (a x +1\right )}\)	\(169\)

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-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)*(3*a^3*ln(a*x+1)*x^3-3*a^3*ln(a*x-1)*x^3-3*a^2*ln(
a*x+1)*x^2+3*a^2*ln(a*x-1)*x^2+10*a^2*x^2-3*a*ln(a*x+1)*x+3*a*ln(a*x-1)*x-2*a*x+3*ln(a*x+1)-3*ln(a*x-1)-4)/(a^
2*x^2-1)/c^3/a^4/(a*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.77 \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {3 \, {\left (a^{4} x^{3} - a^{3} x^{2} - a^{2} x + a\right )} \sqrt {-c} \log \left (\frac {a^{2} c x^{2} + 2 \, \sqrt {-a^{2} c} \sqrt {-c} x + c}{a^{2} x^{2} - 1}\right ) - 2 \, {\left (5 \, a^{2} x^{2} - a x - 2\right )} \sqrt {-a^{2} c}}{16 \, {\left (a^{8} c^{3} x^{3} - a^{7} c^{3} x^{2} - a^{6} c^{3} x + a^{5} c^{3}\right )}} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

-1/16*(3*(a^4*x^3 - a^3*x^2 - a^2*x + a)*sqrt(-c)*log((a^2*c*x^2 + 2*sqrt(-a^2*c)*sqrt(-c)*x + c)/(a^2*x^2 - 1
)) - 2*(5*a^2*x^2 - a*x - 2)*sqrt(-a^2*c))/(a^8*c^3*x^3 - a^7*c^3*x^2 - a^6*c^3*x + a^5*c^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Timed out} \]

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)*x**3/(-a**2*c*x**2+c)**(5/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

integrate(x^3/((-a^2*c*x^2 + c)^(5/2)*sqrt((a*x - 1)/(a*x + 1))), x)

Giac [F(-2)]

Exception generated. \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x^3/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{\coth ^{-1}(a x)} x^3}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {x^3}{{\left (c-a^2\,c\,x^2\right )}^{5/2}\,\sqrt {\frac {a\,x-1}{a\,x+1}}} \,d x \]

[In]

int(x^3/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)),x)

[Out]

int(x^3/((c - a^2*c*x^2)^(5/2)*((a*x - 1)/(a*x + 1))^(1/2)), x)