\(\int e^{\coth ^{-1}(a x)} (c-a^2 c x^2)^{3/2} \, dx\) [617]

Optimal result

Integrand size = 22, antiderivative size = 93 \[ \int e^{\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {2 (1+a x)^3 \left (c-a^2 c x^2\right )^{3/2}}{3 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3}+\frac {(1+a x)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 \left (1-\frac {1}{a^2 x^2}\right )^{3/2} x^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6327, 6328, 45} \[ \int e^{\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=\frac {(a x+1)^4 \left (c-a^2 c x^2\right )^{3/2}}{4 a^4 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}}-\frac {2 (a x+1)^3 \left (c-a^2 c x^2\right )^{3/2}}{3 a^4 x^3 \left (1-\frac {1}{a^2 x^2}\right )^{3/2}} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.57 \[ \int e^{\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {c (1+a x)^3 (-5+3 a x) \sqrt {c-a^2 c x^2}}{12 a^2 \sqrt {1-\frac {1}{a^2 x^2}} x} \]

[In]

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

[Out]

-1/12*(c*(1 + a*x)^3*(-5 + 3*a*x)*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]*x)

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {\left (3 a^{3} x^{3}+4 a^{2} x^{2}-6 a x -12\right ) x c \sqrt {-c \left (a^{2} x^{2}-1\right )}}{12 \left (a x +1\right ) \sqrt {\frac {a x -1}{a x +1}}}\)	\(63\)
gosper	\(\frac {x \left (3 a^{3} x^{3}+4 a^{2} x^{2}-6 a x -12\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {3}{2}}}{12 \left (a x -1\right ) \left (a x +1\right )^{2} \sqrt {\frac {a x -1}{a x +1}}}\)	\(68\)

[In]

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

[Out]

-1/12*(3*a^3*x^3+4*a^2*x^2-6*a*x-12)*x*c*(-c*(a^2*x^2-1))^(1/2)/(a*x+1)/((a*x-1)/(a*x+1))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.46 \[ \int e^{\coth ^{-1}(a x)} \left (c-a^2 c x^2\right )^{3/2} \, dx=-\frac {{\left (3 \, a^{3} c x^{4} + 4 \, a^{2} c x^{3} - 6 \, a c x^{2} - 12 \, c x\right )} \sqrt {-a^{2} c}}{12 \, a} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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