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

Optimal. Leaf size=228 \[ \frac {4 x^2 \sqrt {c-a^2 c x^2}}{3 a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^4 \sqrt {c-a^2 c x^2}}{5 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x^3 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^5 x \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6192, 6193, 88} \[ \frac {x^4 \sqrt {c-a^2 c x^2}}{5 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x^3 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 x^2 \sqrt {c-a^2 c x^2}}{3 a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2}}{a^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^5 x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[c - a^2*c*x^2])/(a^4*Sqrt[1 - 1/(a^2*x^2)]) + (2*x*Sqrt[c - a^2*c*x^2])/(a^3*Sqrt[1 - 1/(a^2*x^2)]) +
(4*x^2*Sqrt[c - a^2*c*x^2])/(3*a^2*Sqrt[1 - 1/(a^2*x^2)]) + (3*x^3*Sqrt[c - a^2*c*x^2])/(4*a*Sqrt[1 - 1/(a^2*x
^2)]) + (x^4*Sqrt[c - a^2*c*x^2])/(5*Sqrt[1 - 1/(a^2*x^2)]) + (4*Sqrt[c - a^2*c*x^2]*Log[1 - a*x])/(a^5*Sqrt[1
- 1/(a^2*x^2)]*x)

Rule 88

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 6192

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 6193

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symbol] :> Dist[c^p/a^(2*p), Int[(u*(-1
+ a*x)^(p - n/2)*(1 + a*x)^(p + n/2))/x^(2*p), 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*} \int e^{3 \coth ^{-1}(a x)} x^3 \sqrt {c-a^2 c x^2} \, dx &=\frac {\sqrt {c-a^2 c x^2} \int e^{3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^4 \, dx}{\sqrt {1-\frac {1}{a^2 x^2}} x}\\ &=\frac {\sqrt {c-a^2 c x^2} \int \frac {x^3 (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^3}+\frac {4 x}{a^2}+\frac {4 x^2}{a}+3 x^3+a x^4+\frac {4}{a^3 (-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^4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 x^2 \sqrt {c-a^2 c x^2}}{3 a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x^3 \sqrt {c-a^2 c x^2}}{4 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^4 \sqrt {c-a^2 c x^2}}{5 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^5 \sqrt {1-\frac {1}{a^2 x^2}} x}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 88, normalized size = 0.39 \[ \frac {\sqrt {c-a^2 c x^2} \left (\frac {4 \log (1-a x)}{a^4}+\frac {4 x}{a^3}+\frac {2 x^2}{a^2}+\frac {a x^5}{5}+\frac {4 x^3}{3 a}+\frac {3 x^4}{4}\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(3*ArcCoth[a*x])*x^3*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 + (4*x^3)/(3*a) + (3*x^4)/4 + (a*x^5)/5 + (4*Log[1 - a*x])/a^4))
/(a*Sqrt[1 - 1/(a^2*x^2)]*x)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 58, normalized size = 0.25 \[ \frac {{\left (12 \, a^{5} x^{5} + 45 \, a^{4} x^{4} + 80 \, a^{3} x^{3} + 120 \, a^{2} x^{2} + 240 \, a x + 240 \, \log \left (a x - 1\right )\right )} \sqrt {-a^{2} c}}{60 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^5*x^5 + 45*a^4*x^4 + 80*a^3*x^3 + 120*a^2*x^2 + 240*a*x + 240*log(a*x - 1))*sqrt(-a^2*c)/a^5

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^3*(-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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.06, size = 92, normalized size = 0.40 \[ \frac {\left (12 x^{5} a^{5}+45 x^{4} a^{4}+80 x^{3} a^{3}+120 a^{2} x^{2}+240 a x +240 \ln \left (a x -1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{60 a^{4} \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*x^5*a^5+45*x^4*a^4+80*x^3*a^3+120*a^2*x^2+240*a*x+240*ln(a*x-1))*(-c*(a^2*x^2-1))^(1/2)*(a*x-1)/a^4/(
a*x+1)^2/((a*x-1)/(a*x+1))^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-a^{2} c x^{2} + c} x^{3}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3\,\sqrt {c-a^2\,c\,x^2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________