\(\int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx\) [683]

Optimal result

Integrand size = 27, antiderivative size = 186 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {4 \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^3 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6327, 6328, 90} \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {x^2 \sqrt {c-a^2 c x^2}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^3 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^4 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[In]

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

[Out]

(4*Sqrt[c - a^2*c*x^2])/(a^3*Sqrt[1 - 1/(a^2*x^2)]) + (2*x*Sqrt[c - a^2*c*x^2])/(a^2*Sqrt[1 - 1/(a^2*x^2)]) +
(x^2*Sqrt[c - a^2*c*x^2])/(a*Sqrt[1 - 1/(a^2*x^2)]) + (x^3*Sqrt[c - a^2*c*x^2])/(4*Sqrt[1 - 1/(a^2*x^2)]) + (4
*Sqrt[c - a^2*c*x^2]*Log[1 - a*x])/(a^4*Sqrt[1 - 1/(a^2*x^2)]*x)

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.40 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (\frac {4 x}{a^3}+\frac {2 x^2}{a^2}+\frac {x^3}{a}+\frac {x^4}{4}+\frac {4 \log (1-a x)}{a^4}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} x} \]

[In]

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

[Out]

(Sqrt[c - a^2*c*x^2]*((4*x)/a^3 + (2*x^2)/a^2 + x^3/a + x^4/4 + (4*Log[1 - a*x])/a^4))/(Sqrt[1 - 1/(a^2*x^2)]*
x)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45

method	result	size
default	\(\frac {\left (a^{4} x^{4}+4 a^{3} x^{3}+8 a^{2} x^{2}+16 a x +16 \ln \left (a x -1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{4 a^{3} \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\)	\(83\)

[In]

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

[Out]

1/4*(a^4*x^4+4*a^3*x^3+8*a^2*x^2+16*a*x+16*ln(a*x-1))*(-c*(a^2*x^2-1))^(1/2)*(a*x-1)/a^3/(a*x+1)^2/((a*x-1)/(a
*x+1))^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.26 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 16 \, a x + 16 \, \log \left (a x - 1\right )\right )} \sqrt {-a^{2} c}}{4 \, a^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F(-2)]

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

[In]

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

[In]

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

[Out]

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