Integrand size = 27, antiderivative size = 186 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {4 \sqrt {c-a^2 c x^2}}{a^3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {2 x \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^3 \sqrt {c-a^2 c x^2}}{4 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^4 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Output:
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*(-a^2*c*x^2+c)^(1/2)*ln(-a*x +1)/a^4/(1-1/a^2/x^2)^(1/2)/x
Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.40 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (\frac {4 x}{a^3}+\frac {2 x^2}{a^2}+\frac {x^3}{a}+\frac {x^4}{4}+\frac {4 \log (1-a x)}{a^4}\right )}{\sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[E^(3*ArcCoth[a*x])*x^2*Sqrt[c - a^2*c*x^2],x]
Output:
(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)
Time = 0.93 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.41, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {6746, 6747, 25, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \sqrt {c-a^2 c x^2} e^{3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^3dx}{x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int -\frac {x^2 (a x+1)^2}{1-a x}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \frac {x^2 (a x+1)^2}{1-a x}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \left (-a x^3-3 x^2-\frac {4 x}{a}-\frac {4}{a^2 (a x-1)}-\frac {4}{a^2}\right )dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \left (-\frac {4 \log (1-a x)}{a^3}-\frac {4 x}{a^2}-\frac {a x^4}{4}-\frac {2 x^2}{a}-x^3\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[E^(3*ArcCoth[a*x])*x^2*Sqrt[c - a^2*c*x^2],x]
Output:
-((Sqrt[c - a^2*c*x^2]*((-4*x)/a^2 - (2*x^2)/a - x^3 - (a*x^4)/4 - (4*Log[ 1 - a*x])/a^3))/(a*Sqrt[1 - 1/(a^2*x^2)]*x))
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(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]))
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[(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]
Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)/(x_)^2)^(p_.), x_Symb ol] :> Simp[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] && !Inte gerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) && IntegersQ[2*p, p + n/2]
Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.45
method | result | size |
default | \(\frac {\left (a^{4} x^{4}+4 a^{3} x^{3}+8 a^{2} x^{2}+16 a x +16 \ln \left (a x -1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{4 a^{3} \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(83\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*x^2*(-a^2*c*x^2+c)^(1/2),x,method=_RETURNVER BOSE)
Output:
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)
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.26 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (a^{4} x^{4} + 4 \, a^{3} x^{3} + 8 \, a^{2} x^{2} + 16 \, a x + 16 \, \log \left (a x - 1\right )\right )} \sqrt {-a^{2} c}}{4 \, a^{4}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm= "fricas")
Output:
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
\[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^{2} \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)*x**2*(-a**2*c*x**2+c)**(1/2),x)
Output:
Integral(x**2*sqrt(-c*(a*x - 1)*(a*x + 1))/((a*x - 1)/(a*x + 1))**(3/2), x )
\[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} x^{2}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm= "maxima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)*x^2/((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.66 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {1}{4} \, \sqrt {-c} {\left (\frac {a^{3} x^{4} + 4 \, a^{2} x^{3} + 8 \, a x^{2} + 16 \, x}{a^{2} \mathrm {sgn}\left (a x + 1\right )} + \frac {a^{5} x^{4} \mathrm {sgn}\left (a x + 1\right ) + 4 \, a^{4} x^{3} \mathrm {sgn}\left (a x + 1\right ) + 8 \, a^{3} x^{2} \mathrm {sgn}\left (a x + 1\right ) + 16 \, a^{2} x \mathrm {sgn}\left (a x + 1\right )}{a^{4}} + \frac {32 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{3} \mathrm {sgn}\left (a x + 1\right )}\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x^2*(-a^2*c*x^2+c)^(1/2),x, algorithm= "giac")
Output:
1/4*sqrt(-c)*((a^3*x^4 + 4*a^2*x^3 + 8*a*x^2 + 16*x)/(a^2*sgn(a*x + 1)) + (a^5*x^4*sgn(a*x + 1) + 4*a^4*x^3*sgn(a*x + 1) + 8*a^3*x^2*sgn(a*x + 1) + 16*a^2*x*sgn(a*x + 1))/a^4 + 32*log(abs(a*x - 1))/(a^3*sgn(a*x + 1)))
Timed out. \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\int \frac {x^2\,\sqrt {c-a^2\,c\,x^2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] Input:
int((x^2*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((x^2*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.26 \[ \int e^{3 \coth ^{-1}(a x)} x^2 \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, i \left (-32 \,\mathrm {log}\left (\sqrt {-a x +1}\right )-a^{4} x^{4}-4 a^{3} x^{3}-8 a^{2} x^{2}-16 a x +29\right )}{4 a^{3}} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*x^2*(-a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*i*( - 32*log(sqrt( - a*x + 1)) - a**4*x**4 - 4*a**3*x**3 - 8*a**2 *x**2 - 16*a*x + 29))/(4*a**3)