Integrand size = 25, antiderivative size = 152 \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {4 \sqrt {c-a^2 c x^2}}{a^2 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {3 x \sqrt {c-a^2 c x^2}}{2 a \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {x^2 \sqrt {c-a^2 c x^2}}{3 \sqrt {1-\frac {1}{a^2 x^2}}}+\frac {4 \sqrt {c-a^2 c x^2} \log (1-a x)}{a^3 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Output:
4*(-a^2*c*x^2+c)^(1/2)/a^2/(1-1/a^2/x^2)^(1/2)+3/2*x*(-a^2*c*x^2+c)^(1/2)/ a/(1-1/a^2/x^2)^(1/2)+1/3*x^2*(-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^3/(1-1/a^2/x^2)^(1/2)/x
Time = 0.08 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.43 \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c-a^2 c x^2} \left (a x \left (24+9 a x+2 a^2 x^2\right )+24 \log (1-a x)\right )}{6 a^3 \sqrt {1-\frac {1}{a^2 x^2}} x} \] Input:
Integrate[E^(3*ArcCoth[a*x])*x*Sqrt[c - a^2*c*x^2],x]
Output:
(Sqrt[c - a^2*c*x^2]*(a*x*(24 + 9*a*x + 2*a^2*x^2) + 24*Log[1 - a*x]))/(6* a^3*Sqrt[1 - 1/(a^2*x^2)]*x)
Time = 0.83 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.47, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6746, 6747, 25, 86, 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.
\(\displaystyle \int x \sqrt {c-a^2 c x^2} e^{3 \coth ^{-1}(a x)} \, dx\) |
\(\Big \downarrow \) 6746 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int e^{3 \coth ^{-1}(a x)} \sqrt {1-\frac {1}{a^2 x^2}} x^2dx}{x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 6747 |
\(\displaystyle \frac {\sqrt {c-a^2 c x^2} \int -\frac {x (a x+1)^2}{1-a x}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \frac {x (a x+1)^2}{1-a x}dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \int \left (-a x^2-3 x-\frac {4}{a}-\frac {4}{a (a x-1)}\right )dx}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {c-a^2 c x^2} \left (-\frac {4 \log (1-a x)}{a^2}-\frac {a x^3}{3}-\frac {4 x}{a}-\frac {3 x^2}{2}\right )}{a x \sqrt {1-\frac {1}{a^2 x^2}}}\) |
Input:
Int[E^(3*ArcCoth[a*x])*x*Sqrt[c - a^2*c*x^2],x]
Output:
-((Sqrt[c - a^2*c*x^2]*((-4*x)/a - (3*x^2)/2 - (a*x^3)/3 - (4*Log[1 - a*x] )/a^2))/(a*Sqrt[1 - 1/(a^2*x^2)]*x))
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
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 = 76, normalized size of antiderivative = 0.50
method | result | size |
default | \(\frac {\left (2 a^{3} x^{3}+9 a^{2} x^{2}+24 a x +24 \ln \left (a x -1\right )\right ) \sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (a x -1\right )}{6 a^{2} \left (a x +1\right )^{2} \left (\frac {a x -1}{a x +1}\right )^{\frac {3}{2}}}\) | \(76\) |
Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*x*(-a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBO SE)
Output:
1/6*(2*a^3*x^3+9*a^2*x^2+24*a*x+24*ln(a*x-1))*(-c*(a^2*x^2-1))^(1/2)*(a*x- 1)/a^2/(a*x+1)^2/((a*x-1)/(a*x+1))^(3/2)
Time = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.28 \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {{\left (2 \, a^{3} x^{3} + 9 \, a^{2} x^{2} + 24 \, a x + 24 \, \log \left (a x - 1\right )\right )} \sqrt {-a^{2} c}}{6 \, a^{3}} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="f ricas")
Output:
1/6*(2*a^3*x^3 + 9*a^2*x^2 + 24*a*x + 24*log(a*x - 1))*sqrt(-a^2*c)/a^3
\[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x \sqrt {- c \left (a x - 1\right ) \left (a x + 1\right )}}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/((a*x-1)/(a*x+1))**(3/2)*x*(-a**2*c*x**2+c)**(1/2),x)
Output:
Integral(x*sqrt(-c*(a*x - 1)*(a*x + 1))/((a*x - 1)/(a*x + 1))**(3/2), x)
\[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int { \frac {\sqrt {-a^{2} c x^{2} + c} x}{\left (\frac {a x - 1}{a x + 1}\right )^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="m axima")
Output:
integrate(sqrt(-a^2*c*x^2 + c)*x/((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.61 \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {1}{6} \, \sqrt {-c} {\left (\frac {2 \, a^{2} x^{3} + 9 \, a x^{2} + 24 \, x}{a \mathrm {sgn}\left (a x + 1\right )} + \frac {48 \, \log \left ({\left | a x - 1 \right |}\right )}{a^{2} \mathrm {sgn}\left (a x + 1\right )} + \frac {2 \, a^{4} x^{3} + 9 \, a^{3} x^{2} + 24 \, a^{2} x}{a^{3} \mathrm {sgn}\left (a x + 1\right )}\right )} \] Input:
integrate(1/((a*x-1)/(a*x+1))^(3/2)*x*(-a^2*c*x^2+c)^(1/2),x, algorithm="g iac")
Output:
1/6*sqrt(-c)*((2*a^2*x^3 + 9*a*x^2 + 24*x)/(a*sgn(a*x + 1)) + 48*log(abs(a *x - 1))/(a^2*sgn(a*x + 1)) + (2*a^4*x^3 + 9*a^3*x^2 + 24*a^2*x)/(a^3*sgn( a*x + 1)))
Timed out. \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\int \frac {x\,\sqrt {c-a^2\,c\,x^2}}{{\left (\frac {a\,x-1}{a\,x+1}\right )}^{3/2}} \,d x \] Input:
int((x*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2),x)
Output:
int((x*(c - a^2*c*x^2)^(1/2))/((a*x - 1)/(a*x + 1))^(3/2), x)
Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.26 \[ \int e^{3 \coth ^{-1}(a x)} x \sqrt {c-a^2 c x^2} \, dx=\frac {\sqrt {c}\, i \left (-48 \,\mathrm {log}\left (\sqrt {-a x +1}\right )-2 a^{3} x^{3}-9 a^{2} x^{2}-24 a x +35\right )}{6 a^{2}} \] Input:
int(1/((a*x-1)/(a*x+1))^(3/2)*x*(-a^2*c*x^2+c)^(1/2),x)
Output:
(sqrt(c)*i*( - 48*log(sqrt( - a*x + 1)) - 2*a**3*x**3 - 9*a**2*x**2 - 24*a *x + 35))/(6*a**2)